Pra-U STPM Chemistry Penggal 2 2019 CB039349b

Chemistry Term 2 STPM Chapter 8 Electrochemistry 43 8 9 Oxygen has an oxidation number of –2, except in peroxides (such as barium peroxide, BaO2), where it has an oxidation number of –1. Compound Na2O MgO SO3 P4O10 Oxidation number of oxygen –2 –2 –2 –2 Compound Hydrogen peroxide, H2O2 Barium peroxide, BaO2 Oxidation number of oxygen –1 –1 However, oxygen exhibits oxidation number of +2 in its combination with fluorine to form fluorine oxide, F2O. Example 8.2 Determine the oxidation number of the underlined element in the following species. (a) NiO2 (c) H2CO3 (e) CH3Cl (b) MnO4 – (d) S2O3 2– (f) NH3 Solution (a) Let the oxidation number of Ni = x x + 2(–2) = 0 x = +4 (b) Let the oxidation number of Mn = x x + 4(–2) = –1 x = +7 (c) Let the oxidation number of C = x 2(+1) + x + 3(–2) = 0 2 + x – 6 = 0 x = +4 (d) Let the oxidation number of S = x 2x + 3(–2) = –2 2x = +4 x = +2 (e) Let the oxidation state of carbon = x x + 3(+1) + (–1) = 0 ∴ x = –2 (f) Let the oxidation state of nitrogen be x. x + 3(+1) = 0 ∴ x = –3 O– O– xx x x x • x x •• •• • • Lewis diagram of the peroxide ion, O2 2–: Info Chem Actually this is the average oxidation state of S. The two sulphur atoms in S2O3 2– have oxidation numbers of –2 and +6 respectively. The average –2 + 6 = ——— = +2. 2 Quick Check 8.2 Determine the oxidation numbers of the following underlined elements. 1 SO3 6 S4O6 2– (Determine the average oxidation number) 2 FeO4 2– 7 Fe(CN)6 4– 3 ClO4 – 8 Cu2SO4 4 VO2+ 9 MnO4 2– 5 VO2 + 10 K4Fe(CN)6

Chemistry Term 2 STPM Chapter 8 Electrochemistry 44 8 Oxidation Number and Redox Reactions 1 Redox reactions can also be defined in terms of oxidation number. 2 Oxidation is defined as a reaction where there is an increase in the oxidation number of an atom/element. Reduction is defined as a reaction where there is a decrease in the oxidation number of an atom/element. 3 Examples of oxidation are: Al ⎯→ Al3+ + 3e– Oxidation number 0 +3 Fe2+ ⎯→ Fe3+ + e– Oxidation number +2 +3 2SO4 2– ⎯→ S2O8 2– + 2e– Oxidation number +6 +7 4 Examples of reduction are: Cl2 + 2e– ⎯→ 2Cl– Oxidation number 0 –1 H2O2 + 2H+ + 2e– ⎯→ 2H2O Oxidation number –1 –2 Definition of oxidation and reduction in terms of oxidation number Quick Check 8.3 Determine whether the underlined elements in the following equation have undergone oxidation or reduction, and name the oxidising and reducing agent. (a) Cr2O7 2– + 8H+ + 3NO2 – + 3H2O → 2Cr3+ + 7H2O + 3NO3 – (b) 2MnO4 – + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O (c) 10HNO3 + I2 → 10NO2 + 2HIO3 + 4H2O (d) H2SO4 + 6HI → 3I2 + S + 4H2O (e) 4Co3+ + 2H2O → 4Co2+ + O2 + 4H+ (f) 5H2O2 + 2IO3 – + 2H+ → I2 + 5O2 + 6H2O Disproportionation 1 Disproportionation is a redox reaction where the same substance gets oxidised and reduced simultaneously. 2 Take the following redox reaction: 2Fe3+ + Sn2+ → 2Fe2+ + Sn4+ The two half-equations are: Fe3+ + e– → Fe2+ Sn2+ → Sn4+ + 2e– In the above half-equations: (a) Fe3+ undergoes reduction, i.e. Fe3+ is the oxidising agent (electron acceptor). (b) Sn2+ undergoes oxidation, i.e. Sn2+ is the reducing agent (electron donor). Definition of disproportionation

Chemistry Term 2 STPM Chapter 8 Electrochemistry 45 8 3 Now, let’s look at another redox reaction: 2Cu2+ ⎯→ Cu2+ + Cu Oxidation number +1 +2 0 The two half-equations are: Cu+ ⎯→ Cu2+ + e– Cu+ + e– ⎯→ Cu In the above half-equations: (a) Cu+ is oxidised to Cu2+, i.e. Cu+ is the reducing agent (electron donor). (b) Cu+ is reduced to Cu, i.e. Cu+ is the oxidising agent (electron acceptor). (c) the oxidation number of copper is increased and decreased at the same time. (d) Cu+ is oxidised and reduced at the same time. (e) Cu+ is the oxidising agent as well as the reducing agent. (f) Such reaction is called disproportionation. Example 8.3 Determine if the following reaction is disproportionation. 2H2O2 → 2H2O + O2 Solution 2H2O2 → 2H2O + O2 Oxidation number –1 –2 0 The oxidation number of oxygen in H2O2 is decreased to –2 (in H2O) and increased to 0 (in O2). Hence, it is disproportionation. H2O2 acts as an oxidising agent as well as a reducing agent in the reaction. The reverse reaction: Cu2+ + Cu → 2Cu+ is not disproportion because Cu2+ is the oxidising agent, while Cu is the reducing agent. Quick Check 8.4 Determine which of the following reactions are disproportionation. 1 Cl2 + 2OH– → Cl– + ClO– + H2O 2 3BrO– → 2Br– + BrO3 – 3 2CrO4 2– + 2H+ → Cr2O7 2– + H2O 4 C12H22O11 → 12C + 11H2O 5 I2 + H2O → HI + HIO 6 Na2S2O3 + 2H+ → 2Na+ + SO2 + S + H2O 7 3BrO– → 2Br– + BrO3 – 8 3Fe2+ → Fe + 2Fe3+ 9 5Cl– + ClO3 – + 3H2O → 3Cl2 + 6OH– 10 3MnO4 2– + 4H+ → 2MnO4 – + MnO2 + 2H2O

Chemistry Term 2 STPM Chapter 8 Electrochemistry 46 8 Constructing Redox Equations 1 Redox equations, like any other chemical equations, must be balanced both in terms of mass and charge. 2 There are two main methods for balancing redox equations: The oxidation number method and the ion-electron method. The Oxidation Number Method 1 Redox reactions involve transfer of electrons from one chemical species to the other. 2 The same number of electrons that is released by a reducing agent must be accepted by an oxidising agent. 3 The total increase in the oxidation number of one species = the total decrease in the oxidation numbers of the other species. 4 The net change in the oxidation numbers of all the species in a balanced redox equation is zero. 5 In places where there is not enough oxygen atoms, it is balanced by adding ‘H2O’. 6 To balance hydrogen atoms, add ‘H+’ to the appropriate side of the equation. 7 Consider a general redox reaction: aA + bB ⎯→ Products Let the change in the oxidation number of A in the reaction be +x unit, and the change in the oxidation number of B be –y unit, then: a(+x) + b(–y) = 0 8 The following example serves to illustrate this method. H2O2 + Fe2+ ⎯→ Fe3+ + H2O (a) Let the equation be: aH2O2 + bFe2+ ⎯→ cFe3+ + dH2O Oxidation number –1 +2 +3 –2 (b) The change in the oxidation state of oxygen is –1. The change in the oxidation state of iron is +1. (c) The total change in the oxidation state of oxygen is 2a(–1), since there are 2a atoms of oxygen. The total change in the oxidation number of iron is b(+1). (d) Hence,2a(–1) + b(+1) = 0 b – 2a = 0 If, a = 1 Then, b = 2 (e) Substitute into the equation, we get: H2O2 + 2Fe2+ ⎯→ cFe3+ + dH2O (f) Balancing both sides: c = 2 and d = 2 H2O2 + 2Fe2+ ⎯→ 2Fe3+ + 2H2O (g) The left-hand side of the equation is short of 2 hydrogen atoms. Exam Tips Exam Tips Always equate the unknown with the largest coefficient equal to 1.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 47 8 This is balanced by adding 2 H+ to the left-hand side of the equation. (h) The balanced equation is: H2O2 + 2H+ + 2Fe2+ ⎯→ 2Fe3++ 2H2O Example 8.4 Balance the following equations using the oxidation number method. (a) Cr2O7 2– + Fe2+ → Cr3+ + Fe3+ (b) ClO3 – + I– → Cl– + I2 Solution (a) Let the equation be: aCr2O7 2– + bFe2+ → cCr3+ + dFe3+ Oxidation number +6 +2 +3 +3 The total change in the oxidation number of Cr = 2a(3 – 6) = –3(2a) The total change in the oxidation number of iron = b(+1) ∴ –6a + b = 0 If, a = 1 Then, b = 6 Substitute into the equation: Cr2O7 2– + 6Fe2+ → cCr3+ + dFe3+ Solving: c = 2 and d = 6 Hence, the equation becomes: Cr2O7 2– + 6Fe2+ → 2Cr3+ + 6Fe3+ Add 7H2O to the right-hand side to balance the oxygen atoms: Cr2O7 2– + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O Add 14H+ on the left-hand side to balance the hydrogen atoms: Cr2O7 2– + 6Fe2+ + 14H+ →2Cr3+ + 6Fe3+ + 7H2O (b) Let the equation be: aClO3 – + bI – → cCl– + dI2 Oxidation number +5 –1 –1 0 The change in the oxidation number of Cl = a[(–1) – (+5)] = –6a Total change in the oxidation number if iodine = b[(0) – (–1)] = +b ∴ b + (–6a) = 0 If, a = 1 Then, b = 6 Substitute into the equation: ClO3 – + 6I– → cCl– + dI2 Balancing both sides: c = 1 and d = 3

Chemistry Term 2 STPM Chapter 8 Electrochemistry 48 8 The equation becomes: ClO3 – + 6I– → Cl– + 3I2 Adding 3H2O to the right-hand side to balance the oxygen atoms: ClO3 – + 6I– → Cl– + 3I2 + 3H2O Adding 6H+ to the left-hand side to balance the hydrogen atoms: ClO3 – + 6I– + 6H+ → Cl– + 3I2 + 3H2O The Ion-electron Method 1 In this method, the redox equation is divided into two half-equations, one for oxidation and one for reduction. 2 Balance each of the half-equations by following the steps below: (a) Balance all other elements except hydrogen and oxygen. (b) Balance oxygen by adding H2O to the appropriate side of the equation. (c) Balance hydrogen by adding H+ to the appropriate side of the equation. (d) Balance the charge by adding electrons to the appropriate side. (e) Finally, combine the two half-equations such that the electrons cancel one another. 3 Let us look at the oxidation of iron(II) by dichromate(VI) ions: Cr2O7 2– + Fe2+ → Cr3+ + Fe3+ (a) Divide the equation into two half-equations: Fe2+ → Fe3+ (Oxidation) Cr2O7 2– → Cr3+ (Reduction) (b) The first half-equation is balanced by adding 1 electron to the right-hand side. Fe2+ → Fe3+ + e– (c) For the second half-equation, we first balance Cr: Cr2O7 2– → 2Cr3+ Then add 7H2O to the right-hand side to balance the oxygen atoms: Cr2O7 2– → 2Cr3+ + 7H2O Then add 14H+ to the left-hand side to balance the hydrogen atoms: Cr2O7 2– + 14H+ → 2Cr3+ + 7H2O Finally add 6 electrons to the left-hand side to balance the charge: Cr2O7 2– + 14H+ + 6e– → 2Cr3+ + 7H2O (d) Multiply the first half-equation by 6 add to the second halfequation: 6Fe2+ → 6Fe3+ + 6e– Cr2O7 2– + 14H+ + 6e– → 2Cr3+ + 7H2O ————————————————————– Cr2O7 2– + 14H+ + 6Fe2+ → 6Fe3+ + 2Cr3+ + 7H2O 4 The advantage of the ion-electron method is that we do not need to calculate the oxidation numbers of the species involved. Since, a wrong oxidation number would cause the equation to be incorrect. Info Chem Advantage: Do not have to determine the oxidation states of the species involved.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 49 8 Example 8.5 Balance the equations below using the ion-electron method. (a) HNO3 + I2 → HIO3 + NO2 (b) Cu + HNO3 → Cu2+ + NO Solution (a) Divide the equation into two half-equations and balance them separately. (i) HNO3 → NO2 HNO3 → NO2 + H2O HNO3 + H+ → NO2 + H2O HNO3 + H+ + e– → NO2 + H2O (ii) I2 → HIO3 I2 → 2HIO3 I2 + 6H2O → 2HIO3 I2 + 6H2O → 2HIO3 + 10H+ I2 + 6H2O → 2HIO3 + 10H+ + 10e– (iii) Equation (i)  10 and add to the equation (ii) 10HNO2 + 10H+ + 10e– → 10NO2 + 10H2O I2 + 6H2O → 2HIO3 + 10H+ + 10e– ——————————————————– 10HNO3 + I2 → 10NO2 + 2HIO3 + 4H2O (b) (i) Cu → Cu2+ Cu → Cu2+ + 2e– (ii) HNO3 → NO HNO3 → NO + 2H2O HNO2 + 3H+ → NO + 2H2O HNO3 + 3H+ + 3e– → NO + 2H2O (iii) [(i)  3] + [(ii)  2] 3Cu → 3Cu2+ + 6e– 2HNO3 + 6H+ + 6e– → 2NO + 4H2O ———————————————————— 3Cu + 2HNO3 + 6H+ → 3Cu2+ + 2NO + 4H2O Balancing Redox Equations in Basic Solutions 1 Certain redox reactions are carried in basic solution such as in aqueous sodium hydroxide. 2 To balance such equation, we first balance the equation by either the oxidation number method or the ion-electron method as discussed before. 3 Add OH– to the appropriate side to combine with any H+ present to form H2O. 4 Add an equal amount of OH– to the opposite side to balance the equation. 5 For example: The reaction between chlorine and Cr3+ to produce CrO4 2– in basic medium. Cl2 + Cr3+ → Cl– + CrO4 2– 6 We will use the ion-electron method (which saves us the trouble of working out the oxidation number of the species involved) to balance the equation: Info Chem Add OH– to cancel out H+.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 50 8 (i) Cl2 → Cl– Cl2 → 2Cl– Cl2 + 2e– → 2Cl– (ii) Cr3+ → CrO4 2– Cr3+ + 4H2O → CrO4 2– Cr3+ + 4H2O → CrO4 2– + 8H+ Cr3+ + 4H2O → CrO4 2– + 8H+ + 3e– (iii) [(i)  3] + [(ii)  2]: 3Cl2 + 6e– → 6Cl– 2Cr3+ + 8H2O → 2CrO4 2– + 16H+ + 6e– ———————————————————— 3Cl2 + 2Cr3+ + 8H2O → 2CrO4 2– + 16H+ + 6Cl– (iv) Since there are 16H+ on the right-hand side of the equation, we add 16OH– to both sides to eliminate the H+: 3Cl2 + 2Cr3+ + 8H2O + 16OH– → 2CrO4 2– + 16H + 16OH– + 6Cl– ⏐ ↓ 16H2O (v) The equation now becomes: 3Cl2 + 2Cr3+ + 8H2O + 16OH– → 2CrO4 2– + 16H2O + 6Cl– (vi) Cancelling H2O from both sides of the equation, we arrived at the final balanced equation: 3Cl2 + 2Cr3+ + 16OH– → 2CrO4 2– + 8H2O + 6Cl– Quick Check 8.5 1 Determine the oxidation number of the underlined elements in the following species. (a) ClO4 – (c) VO2 + (e) Fe(CN)6 4– (g) MnO4 2– (b) VO2+ (d) IF3 (f) Cu2SO4 2 Balanced the following equations. (a) MnO4 – + C2O4 2– → Mn2+ + CO2 (f) Co + IO3 – → Co2+ + I2 (b) VO3 – + Fe2+ → V3+ + Fe3+ (g) MnO2 + H2O2 → MnO4 – + H2O (c) S2O3 2– + I2 → S4O6 2– + I– (h) MnO4 – + Pb → MnO2 + PbO2 (In alkaline medium) (d) Cr2O7 2– + H2O2 → Cr3+ + O2 (i) IO3 – + Cl2 → IO4 – + Cl– (In alkaline medium) (e) N2H5 ++ + Cl2 → Cl– + N2 8.2 Standard Electrode Potential 1 All metals (due to their relatively low ionisation energies) are reducing agents. This means they have the tendency to lose electrons during chemical reactions. For example, Cu → Cu2+ + 2e– Zn → Zn2+ + 2e– Cr → Cr3+ + 3e– The metals act as reducing agents.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 51 8 2 On the other hand, the metal cations have the tendency to accept electrons. They are oxidising agents. For example, Fe3+ + 3e– → Fe Ni2+ + 2e– → Ni 3 Now, let us see what happens if a piece of metal, M (the reducing agent) is immersed in an aqueous solution containing the Mn+ ions (the oxidising agent). 4 M, being a reducing agent, has the tendency to lose electrons and goes into the solution as Mn+ aqueous ion, leaving behind the electrons. This makes the metal rod negatively charged with respect to its aqueous solution. M(s) → Mn+(aq) + ne– M(s) M(s) - - - Mn+(aq) Mn+ Mn+ Mn+(aq) 5 Conversely, the Mn+ ions in the solution can ‘snatch’ electrons from the metal atoms on the metal rod and get reduced and deposit as M atoms. This makes the metal rod positively charged with respect to its aqueous solution. Mn+(aq) + ne– → M(s) M(s) M(s) + + + Mn+(aq) Mn+(aq) Mn+ M Mn+ 6 At equilibrium, the rate of dissolution of M is equal to the rate of deposition of Mn+. M(s) Mn+(aq) + ne– Thissets up a potential difference between the metal and its aqueous ions. This potential difference is called the electrode potential. 7 The position of equilibrium depends on the nature of the metal. For a very reactive metal such as magnesium, the equilibrium lies more to the right-hand side. As a result, the magnesium rod acquires a net negative charge with respect to the solution. Mg(s) Mg2+(aq) + 2e– The Mg2+/Mg half-cell is said to have a negative electrode potential. The metal cations act as oxidising agents. Mn+(aq) M(s) Dissolution of metal M Deposition of M Mg2+(aq) Mg(s) – – The Mg rod is negatively charged with respect to its solution. This gives rise to a negative electrode potential.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 52 8 8 For a less reactive metal such as silver, the equilibrium is more to the left-hand side. The silver rod acquires a net positive charge with respect to the solution. Ag(s) Ag+(aq) + e– The Ag+/Ag half-cell is said to have a positive electrode potential. 9 The electrode potential is a measure of the ease of loss of electrons by a metal. The more negative the electrode potential of a half-cell, the easier it is to lose electrons (oxidation). The more positive the electrode potential of a half-cell, the easier it is for its ion to accept electrons (reduction). 10 Going back to the Daniell cell, we would conclude that the electrode potential of the zinc half-cell is more negative than that of the copper half-cell. Conversely, the electrode potential of the copper half-cell is more positive than that of the zinc half-cell. 11 It is this difference that provides the driving force that drives electrons through the connecting wires from the zinc half-cell to the copper half-cell. The potential difference between the two half-cells is called the cell potential, Ecell. 12 The electrode potential of half-cells depends on several factors. Among them are: (a) The nature of the electrode (b) Temperature (c) Pressure (d) Concentration of ions in the electrolyte 13 By international agreement, the following sets of conditions are accepted as the standard conditions for measuring electrode potentials of half-cells: (a) Temperature is fixed at 25 °C or 298 K (b) Pressure is fixed at 101 kPa or 1 atm (c) Concentration of ions in the electrolytes is fixed at 1.00 mol dm–3 14 Under such conditions, the electrode potential of a half-cell is called the standard electrode potential, SEP, and given a symbol of E° with the unit of volt. 15 In the standard electrode potential series, the reactions for the halfcells are written as reduction with a reversible sign. The standard electrode potential is also known as standard reduction potential. 16 The e.m.f. of an electrochemical cell is given by: E.m.f. = E°(positive terminal) – E°(negative terminal) Or E.m.f. = E°(cathode) – E°(anode) 17 But how do we go about measuring the standard electrode potential of a half-cell? Like all measurement, a standard must be chosen as a reference. The reference half-cell chosen for this purpose is the standard hydrogen electrode. Ag+(aq) Ag(s) + + The Ag rod is positively charged with respect to its solution. This gives rise to a positive electrode potential. The more negative the electrode potential, the easier it is to lose electrons. Standard electrode potential: Mn+(aq, 1.0 M) M(s) Temperature = 25 °C Pressure = 101 kPa The positive terminal of a cell is called the cathode. Exam Tips Exam Tips Anode is the electrode where oxidation takes place. Cathode is the electrode where reduction takes place.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 53 8 Standard Hydrogen Electrode 1 The e.m.f. of the Daniell cell under standard conditions is 1.10 V. This represents the difference between the standard electrode potentials of the copper half-cell and the zinc half-cell. 1.10 = E°(+ve terminal) – E°(–ve terminal) 1.10 = E°(Copper half-cell) – E°(Zinc half-cell) 2 If we know the absolute E° value of the copper half-cell, we can then calculate the absolute standard electrode potential of the zinc halfcell. E°(Zinc half-cell) = E°(Copper half-cell) – 1.10 3 The absolute value of the standard electrode potential of a half-cell cannot be measured because we need another half-cell to complete the circuit. So, what is measured is always the difference between the two half cells. 4 However, if a particular half-cell is chosen as reference and arbitrarily assigned its standard electrode potential as zero, then the e.m.f. of a cell consisting of the half-cell and the reference half-cell would give the standard electrode potential of the half-cell under investigation. 5 Assuming that the reference electrode forms the anode/negative terminal of the cell, then E.m.f. = E°(Half-cell) – E°(Reference half-cell) E.m.f. = E°(Half-cell) – 0 and E°(Half-cell) = E.m.f. 6 On the other hand, if the reference electrode forms the cathode/ positive terminal of the cell, then E.m.f. = E°(Reference half-cell) – E°(Half-cell) E.m.f. = 0 – E°(Half-cell) and E°(Half-cell) = – E.m.f. 7 The half-cell chosen as reference is the standard hydrogen electrode, which consists of a platinum electrode (coated with finely divided platinum oxide), immersed in an aqueous solution containing H+(aq) ions of concentration 1.0 mol dm–3 (For example, 1.0 M HCl or 0.50 M H2SO4). A stream of hydrogen gas (at 1 atm) is bubbled over the surface of the platinum electrode. The temperature is kept constant at 298 K. 8 The function of platinum oxide (also called platinum black) is to facilitate the adsorption of hydrogen gas on the platinum electrode, and at the same time acts as a catalyst for the half-cell reaction. 9 By convention, the half-cell equation for the standard hydrogen electrode is: 2H+(aq, 1 M) + 2e– H2(g, 1 atm) E° = 0.00 V [The half-cell reaction is written as reduction from left to right] The reference electrode VIDEO Standard Hydrogen Electrode The standard hydrogen electrode: [H+ ] (aq,1.0 M) Pt wire [H2 ] (g,1 atm) Pt foil Platinum oxide facilitates the adsorption of hydrogen gas and at the same time catalyses the equilibrium.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 54 8 Measuring Standard Electrode Potential 1 The half-cell whose standard electrode potential is to be measured is connected to the standard hydrogen electrode to form an electrochemical cell. 2 The cell e.m.f. (measured using a high resistant voltmeter or a potentiometer) and the direction of flow of current are then noted. 3 The standard electrode potential of the half-cell under investigation can then be calculated. Half-cells Consisting of a Solid Electrode and Its Aqueous Ions 1 The set-up for the determination of the standard electrode potential of the zinc half-cell is shown below: Salt bridge V Zn [H2 ] (g,1 atm) [H+] (aq,1.0 M) Pt Zn2+ (aq, 1.0 M) 2 When the circuit is closed, the high resistant voltmeter reads 0.76 V, and the current is flowing from the hydrogen electrode to the zinc electrode. 3 Thus, 0.76= E°(Hydrogen electrode) – E°(zinc half-cell) = 0 – E°(zinc half-cell) ∴ E°(zinc half-cell) = –0.76 V 4 Following the IUPAC convention, the half-cell equation corresponding to the E° value of –0.76 V is: Zn2+(aq, 1.0 M) + 2e– Zn(s) E° = –0.76 V or simply as, Zn2+(aq) + 2e– Zn(s) E° = –0.76 V or: E°(Zn2+/Zn) = –0.76 V 5 Note that the value of the E° of a half-cell does not depend on the stoichiometry of the equation. For example: Zn2+(aq) + 2e– Zn(s) E° = –0.76 V 2Zn2+(aq) + 4e– 2Zn(s) E° = –0.76 V 6 However, the E° value for the reverse reaction: Zn(s) Zn2+(aq) + 2e– is –(–0.76) = +0.76 V 7 When the zinc half-cell is replaced by the silver half-cell, the e.m.f. of the cell is 0.80 V, and silver forms the positive terminal of the cell. Hence, 0.80 = E°(silver half-cell) – E°(hydrogen electrode) ∴ E°(silver half-cell) = +0.80 V That is, Ag+(aq, 1.0 M) + e– Ag(s) E° = +0.80 V Exam Tips A high resistant voltmeter is used to ensure that only a small current is drawn out from the cell so that the change in the concentrations of the ions is negligible. 2007/P2/Q7(a) Hydrogen half-cell is the positive terminal (cathode). VIDEO Measuring Standard Reduction Potential Info Chem Ag(s)  Ag+(aq) + e– E° = –0.80 V

Chemistry Term 2 STPM Chapter 8 Electrochemistry 55 8 Half-cells Consisting of a Gas and Its Aqueous Ions 1 Take the chlorine half-cell as example. The chlorine half-cell consists of the following equilibrium: Cl2(g) + 2e– 2Cl– (aq) 2 Since, there is no solid involved in the half-cell, an inert electrode such as platinum is used to connect the half-cell to the standard hydrogen electrode as shown below: Salt bridge V Cl– H2 H+ Platinum Cl2 (g, 1 atm) (aq, 1 M) (g, 1 atm) (aq, 1 M) 3 The e.m.f. of the cell is 1.36 V and the chlorine half-cell is the positive terminal. Hence, 1.36 = E°(chlorine half-cell) – E°(hydrogen electrode) ∴ E°(chlorine half-cell) = +1.36 V That is, Cl2(g, 1 atm) + 2e– 2Cl– (aq, 1.0 M) E° = +1.36 V Half-cells Consisting of Aqueous Ions 1 For example, we measure the standard electrode potential of the following half-cell. Fe3+(aq) + e– Fe2+(aq) 2 First, an aqueous mixture containing 1.0 mol dm–3 of Fe2+ and Fe3+ respectively is prepared by dissolving 1 mole of iron(II) chloride and 1 mole of iron(III) chloride in water to make up 1 dm3 of solution. 3 The half-cell is then connected to the standard hydrogen electrode as shown below. V Fe3+ (aq, 1 M) H2 (g, 1 atm) H+ (aq, 1 M) Platinum Fe2+ (aq, 1 M) 4 The e.m.f. of the cell is 0.77 V and the Fe3+/Fe2+ half-cell forms the positive terminal of the cell. 0.77 = E°(Fe3+/Fe2+) – E°(hydrogen electrode) ∴ E°(Fe3+/Fe2+) = +0.77 V Or, Fe3+(aq, 1.0 M) + e– Fe2+(aq, 1.0 M) E° = +0.77 V 2012/P1/Q6(a)

Chemistry Term 2 STPM Chapter 8 Electrochemistry 56 8 Quick Check 8.6 1 By means of fully labelled diagrams, show how to measure the standard electrode potential of the following half-cells. (All ions are aqueous ions.) (a) MnO4 –+ 8H+ + 5e– Mn2+ + 4H2O (b) AgCl + e– Ag + Cl– (c) H2O2 + 2H+ + 2e– 2H2O (d) Hg2Cl2 + 2e– Hg + 2Cl– (This is known as the calomel electrode.) 2 Explain why the standard electrode potential of the following half-cell cannot be measured directly. K+(aq) + e– K(s) 3 Why is a high resistant voltmeter used in the measurement of standard electrode potentials? Standard Electrode Potential Series 1 When all the half-cells are arranged in increasing order of their standard electrode potentials (from the most negative to the most positive), the standard electrode potential series is formed. 2 Part of the standard electrode potential series is given below. Electrode reaction E°/V Ag+ + e– Ag +0.80 AgCl + e– Ag + Cl– +0.22 [Ag(NH3)2] + + e– Ag + 2NH3 +0.37 Al3+ + 3e– Al –1.66 Ba2+ + 2e– Ba –2.90 Be2+ + 2e– Be –1.85 —1 2 Br2 + e– Br – +1.07 Ca2+ + 2e– Ca –2.87 —1 2 Cl2 + 2e– Cl– +1.36 3 In the series: (a) all ions are aqueous ions, (b) all half-cell reactions are written as reduction from left to right, (c) oxidising agents (electron acceptors) are found on the left-hand side of the half-equation, (d) reducing agents (electron donors) are found on the right-hand side of the half-equation. Oxidising agent + electron reducing agent 4 For example, Mg2+ + 2e– Mg Al3+ + 3e– Al Cu2+ + 2e– Cu Cl2 + 2e– 2Cl– Cu+ + e– Cu Cu2+ + e– Cu+ Oxidising agent Reducing agent The standard electrode potential series is also known as standard reduction potential series. Refer to the APPENDIX for a complete list of the standard electrode potential series. Exam Tips Exam Tips The E° value does not depend on the number of moles of substance present. For example, Ag+ + e– Ag +0.80 V 2Ag+ + 2e– 2Ag +0.80 V But, Ag Ag+ + e– –0.80V

Chemistry Term 2 STPM Chapter 8 Electrochemistry 57 8 (a) Mg2+, Al3+, Cu2+, Cl2 are all oxidising agents. (b) Mg, Al, Cu, Cl– are all reducing agents. (c) Cu+ is both an oxidising agent as well as a reducing agent. 5 The standard electrode potential series can be used to: (a) compare the strength of oxidising agents and reducing agents, (b) predict feasibility of redox reactions, (c) predict the stability of aqueous ions, (d) calculate the e.m.f. of an electrochemical cell, (e) to predict the products of electrolysis. Construction of Electrochemical Cells 1 Electrochemical cells (such as the Daniell cell) are devices that convert chemical energy (from spontaneous redox reactions) into electrical energy. 2 An electrochemical cell can be constructed by combining two halfcells with different standard electrode potentials. 3 The half-cell with a more positive E° value will form the positive terminal (or cathode) of the cell. 4 The half-cell with a more negative E° value will form the negative terminal (or anode) of the cell. 5 The reaction that takes place at the positive terminal (cathode) is reduction. 6 The reaction that takes place at the negative terminal (anode) is oxidation. 7 Electrons flow from the negative terminal (anode) to the positive terminal (cathode), while current flows from the positive terminal (cathode) to the negative terminal (anode). Electrons Positive terminal (cathode) Current Negative terminal (anode) 8 The e.m.f. of the cell = E°(positive terminal) – E°(negative terminal) = E°(cathode) – E°(anode) = E°(more positive value) – E°(more negative value ) 9 An electrochemical cell can be represented by a cell diagram (or cell notation) which takes the following form: Electrons Anode/negative terminal Cathod/positive terminal Electrode(s) Electrolyte(aq) Electrolyte(aq) Electrode(s) In the cell diagram, ions or species with lower oxidation states are written closest to their respective electrodes. Cu+ has ‘amphoteric’ nature. 2007/P1/Q17 2010/P1/Q45 2016/P2/Q4; Q18(a) 2013/P2/Q5, Q16, 2018/P2/Q4 2009/P1/Q16 2010/P2/Q6(b) 2017/P2/Q4; Q16 2014/P2/Q4; Q6; Q16(b) Cathode is defined as the electrode where reduction occurs. Anode is defined as the electrode where oxidation occurs. Exam Tips Exam Tips A half-cell with more negative E° has a higher tendency to release electrons. Hence, it forms the negative terminal of the cell. A half-cell with more positive E° has higher tendency to accept electrons. Hence, it forms the positive terminal of the cell. Exam Tips Exam Tips One way to remember how to draw a cell diagram is ‘ABC’. Anode, Bridge, Cathode. Info Chem Positive terminal: Cathode Negative terminal: Anode

Chemistry Term 2 STPM Chapter 8 Electrochemistry 58 8 10 Take the example of a cell constructed from the following two halfcells: Mg2+(aq) + 2e– 2Mg(s) –2.38 V Cu2+(aq) + 2e– Cu(s) +0.34 V (a) The positive terminal (cathode) is Cu (E° more positive). (b) The negative terminal (anode) is Mg (E° more negative). (c) The cell diagram is: Mg(s) | Mg2+(aq) || Cu2+(aq) | Cu(s) 11 Consider the following two half-cells: Fe3+(aq) + e– Fe2+(aq) +0.77 V Sn4+(aq) + 2e– Sn2+(aq) +0.15 V (a) The Fe3+/Fe2+ is the positive terminal (cathode). (b) The Sn4+/Sn2+ half-cell is the negative terminal (anode). (c) The cell diagram is: Pt(s) | Sn2+(aq), Sn4+(aq) || Fe3+(aq); Fe2+(aq) | Pt(s) And not: Pt(s) | Sn4+(aq), Sn2+(aq) || Fe3+(aq); Fe2+(aq) | Pt(s) or: Pt(s) | Sn2+(aq), Sn4+(aq) || Fe2+(aq); Fe3+(aq) | Pt(s) 12 To predict the half-cell reactions, we proceed from the left of cell diagram towards the right, adding electrons where appropriate. 13 For example, Mg(s) | Mg2+(aq) || Cu2+(aq) | Cu(s) Direction of reactions The magnesium half-cell: Mg(s) ⎯→ Mg2+(aq) + 2e– The copper half-cell: Cu2+(aq) + 2e– ⎯→ Cu(s) The overall cell reaction: Mg(s) + Cu2+(aq) ⎯→ Mg2+(aq) + Cu(s) 14 Similarly, for the following cell: Pt(s) | Sn2+(aq), Sn4+(aq) || Fe3+(aq); Fe2+(aq) | Pt(s) Direction of reactions The left-hand half-cell: Sn2+(aq) ⎯→ Sn4+(aq) + 2e– The right-hand half-cell: Fe3+(aq) + e– ⎯→ Fe2+(aq) The overall cell reaction: Sn2+(aq) + 2Fe3+(aq) ⎯→ Sn4+(aq) + 2Fe2+(aq) Exam Tips Exam Tips Right-hand half-cell: Reduction Exam Tips Exam Tips The species with the lowest oxidation state is written closest to the electrodes. INFO Electrochemical Cell and Cell Notation

Chemistry Term 2 STPM Chapter 8 Electrochemistry 59 8 Example 8.6 An electrochemical cell is constructed from the following two halfcells: 2H+(aq) + 2e– H2(g) 0.00 V Zn2+(aq) + 2e– Zn(s) –0.76 V (a) Draw the cell diagram. (b) Write equations for the reactions taking place at the (i) anode, and (ii) cathode (c) What is the overall cell reaction? (d) Calculate the e.m.f. of the cell under standard conditions. Solution (a) The hydrogen electrode is the positive terminal, while the zinc electrode is the negative terminal. Zn(s) | Zn2+(aq) || H+(aq) | H2(g), Pt(s) (b) (i) Anode (Negative terminal): Zn(s) → Zn2+(aq) + 2e– (ii) Cathode (positive terminal): 2H+(aq) + 2e– → H2(g) (c) Overall cell reaction: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) (d) E.m.f. = E°(H+/H2) – E°(Zn2+/Zn) = 0 – (–0.76) = 0.76 V Exam Tips Exam Tips Right-hand side of the cell diagram undergoes reduction. Quick Check 8.7 For each of the following cells, 1 Mg2+/Mg and Zn2+/Zn 4 I2/I– and S4O6 2–/S2O3 2– 2 Sn4+/Sn2+ and Cl2/Cl– 5 Cr3+/Cr and Fe2+/Fe 3 MnO4 – /Mn2+ and Br2/Br– (a) draw the cell diagram. (b) write equations for the reactions taking place at the (i) anode, and (ii) cathode. (c) write the overall cell reaction. (d) calculate the e.m.f. of the cell under standard conditions. The Power of Oxidising Agents and Reducing Agents 1 For oxidising agents, the more positive the E° values, the stronger is its oxidising power. 2 For reducing agents, the more negative the E° values, the stronger is the reducing power. 3 When comparing the strength of oxidising agents and reducing agents using the E° values form the SEP series, the sign of the E° value should not be changed. 4 For example, Zn2+(aq) + 2e– Zn(s) E° = –0.76 V 2012/P1/Q17 2013/P2/Q4 2015/P2/Q4 2014/P2/Q16(a) 2018/P2/Q11 The more positive the E°, the stronger the oxidising agent. The more negative the E°, the stronger the reducing agent. Info Chem The e.m.f. of an electrochemical cell must always be positive. Otherwise, no electricity will be produced.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 60 8 This means that, (a) the oxidising strength of Zn2+ (compared to H+) is –0.76 V (b) the reducing strength of Zn (compared to H2) is –0.76 V 5 Considering the following two half-cells: Mg2+ + 2e– Mg E° = –2.38 V Cl2 + 2e– 2Cl– E° = +1.36 V (a) Cl2 is a stronger oxidising agent than Mg2+. (b) Mg is a stronger reducing agent than Cl– . (c) A strong oxidising agent will have a weak conjugate reducing agent and vice versa. 6 Alternatively, we can use the method describe below: Mg2+ + 2e– Mg E° = –2.38 V Cl2 + 2e– 2Cl– E° = +1.36 V (a) To compare the reducing strength of Mg and Cl– : (Note: Reducing agents are electron donors.) Mg ⎯→ Mg2+ + 2e– E° = +2.38 V* (i) 2Cl– ⎯→ Cl2 + 2e– E° = –1.36 V* (ii) (*The E° values are reversed because the half-equations are written as oxidation.) The E° value for reaction (i) is more positive. This means that Mg has a higher tendency to donate electrons compared to Cl– ions. Therefore, Mg is a stronger reducing agent than Cl– . (b) To compare the oxidising strength of Mg2+ and Cl2: (Note: Oxidising agents are electron acceptors.) Mg2+ + 2e– ⎯→ Mg E° = –2.38 V (iii) Cl2 + 2e– ⎯→ 2Cl– E° = +1.36 V (iv) The E° value for reaction (iv) is more positive. This means that Cl2 has a higher tendency to accept electrons compared to Mg2+ ions. Therefore, Cl2 is a stronger oxidising agent than Mg2+. Example 8.7 Arrange the following species in order of (a) increasing oxidising power (b) increasing reducing power Zn2+ + 2e– Zn –0.76 V Cu2+ + 2e– Cu +0.34 V Ag+ + e– Ag +0.80 V Cl2 + 2e– 2Cl– +1.36 V Solution (a) Oxidising strength: Zn2+  Cu2+  Ag+  Cl2 (b) Reducing strength: Cl–  Ag  Cu  Zn Oxidising agent: Cl2  Mg2+ Reducing agent: Mg  Cl– Info Chem The sign of the E° value is not changed because the half-equation is still written as reduction. Info Chem The signs of the E° values are reversed because the half-equations are written as oxidation.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 61 8 Quick Check 8.8 1 Arrange the following species in the order of increasing oxidising strength and reducing strength. MnO4 – + 8H+ + 5e– Mn2+ + 4H2O +1.52 V Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O +1.33 V Fe3+ + e– Fe2+ +0.77V H2O2 + 2H+ + 2e– 2H2O +1.77 V F2 + 2e– 2F– +2.87 V 2 The E° values of four half-cells are given below: Fe3+ + e– Fe2+ +0.77 V Ag+ + e– Ag +0.80 V I2 + 2e– 2I– +0.54 V Cl2 + 2e– 2Cl– +1.36 V Cr3+ + 3e– Cr –0.74 V Arrange the appropriate species in the order of (a) increasing oxidising strength, and (b) increasing reducing strength. The Feasibility of Redox Reactions 1 Consider the following two half-cells: MnO4 – + 8H+ + 5e– Mn2+ + 4H2O Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O Both the MnO4 – ion and the Cr2O7 2– ion are oxidising agents. Can MnO4 – oxidise Cl– to Cl2? Can Cr2O7 2– oxidise Cl– to Cl2? 2 There are two ways to predict the feasibility of a redox reaction. (a) By calculating the e.m.f. of the reaction, E° (b) By applying the anti-clockwise rule The E.M.F. of Redox Reactions 1 A redox reaction is energetically feasible under standard conditions if the e.m.f. of the reaction is positive. 2 However, if the e.m.f. of the reaction is negative, it is not feasible under standard conditions. 3 In calculating the e.m.f. of the reaction, the sign of the E° values have to be changed according to the direction of the half-reactions. 4 Consider the following reaction: Zn + Cu2+ ⎯→ Zn2+ + Cu The two half-reactions that are involved are: Zn ⎯→ Zn2+ + 2e– E° = –(–0.76)* V Cu2+ + 2e– ⎯→ Cu E° = +0.34 V ———————————————————————– Zn + Cu2+ ⎯→ Zn2+ + Cu E° = –(–0.76) + (+0.34) = +1.10 V A positive E° value means that the reaction is energetically feasible under standard conditions. * The sign of the E° value is reverse because the equation is now written as oxidation.] 2009/P1/Q45 2012/P2/Q6(b) 2012/P1/Q17

Chemistry Term 2 STPM Chapter 8 Electrochemistry 62 8 5 A simpler way to solve this problem is: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) –(–0.76) +0.34 E°cell = (+0.34) + [–(–0.76)] = +1.10 V The E° value for the zinc half-cell is reversed because the reaction is now oxidation. Quick Check 8.9 Predict the feasibility of the following reactions under standard conditions by calculating the E° values. 1 Cr2O7 2– + 14H+ + 6Cl– → 2Cr3+ + 3Cl2 + 7H2O 4 MnO2 + 2Cl– + 4H+ → Mn2+ + 2H2O + Cl2 2 MnO4 – + 5Fe2+ +8H+ → Mn2+ + 5Fe3+ + 4H2O 5 H2 + Cu2+ → 2H+ + Cu 3 H2O2 + 2H+ + 2Cl– → 2H2O + Cl2 6 2Cu+ → Cu2+ + Cu Anticlockwise Rule 1 In this method, the two half-equations are arranged in order of increasing E° values, then applying the anticlockwise rule. This method is useful that we need not have to calculate the E° values of the reactions (such as answering objective questions). 2 The following example illustrates how this method is employed: To determine the feasibility of the reaction: Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu(s) We arrange the two half-cell equations as follows and then applying the anticlockwise rule: –0.76 V +0.34 V Increasing E° values Zn2+ + 2e– Zn Cu2+ + 2e– Cu The diagram shows that Zn will reduce Cu2+ to Cu and itself gets oxidise to Zn2+. So the reaction is feasible. 3 Consider the example below that takes place under standard conditions: 3Ag(s) + Al3+(aq) ⎯→ 3Ag+(aq) + Al(s) The half-cells are: –1.66 V +0.80 V Al3+ + 3e– Al Ag+ + e– Ag The anticlockwise rule shows that Al will reduce Ag+ to silver and not the other way round. Thus, the reaction is not feasible. Arrange the half-cells in order of increasing E° values.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 63 8 Quick Check 8.10 Use the anticlockwise rule to predict the feasibility of the following reactions under standard conditions. 1 Cr2O7 2– + 14H+ + 6Cl– → 2Cr3+ + 3Cl2 + 7H2O 2 MnO4 – + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O 3 H2O2 + 2H+ + 2Cl– → 2H2O + Cl2 4 MnO2 + 2Cl– + 4H+ → Mn2+ + 2H2O + Cl2 5 H2 + Cu2+ → 2H+ + Cu 6 2Cu+ → Cu2+ + Cu Important Points in Using The E° Values in Determining Feasibility of Redox Reactions 1 A positive E° value indicates that the redox reaction is energetically feasible under standard conditions. Standard conditions refer to: (a) the temperature at 25°C, (b) pressure at 1 atm (or 101 kPa), (c) concentration of aqueous ions at 1.0 mol dm–3. 2 However, it does not give any information regarding the rate of the reaction. 3 For example, the reduction of copper(II) aqueous ions by hydrogen gas: H2(g) + Cu2+(aq) ⎯→ 2H+(aq) + Cu(s) The E° value for the reaction is +0.34 V. 4 However, when hydrogen gas (at 1 atm) is bubbled through aqueous copper(II) sulphate of 1.0 mol dm–3 at 25 °C, no copper is formed. 5 This is because the activation energy of the reaction is high. Under standard conditions, the particles do not have enough energy to overcome the high activation energy. Thus, no reaction occurs. 6 In such cases, the reaction is said to be energetically feasible but not kinetically feasible. 7 The diagram below shows the energy profile of the reaction. Very high Ea Cu2+ (aq) + H2 (g) Cu(s) + 2H ΔH = negative + (aq) E° values refer to standard conditions. Though the E° is positive, the reaction is not feasible. Quick Check 8.11 1 Consider the following reaction: MnO2(s) + 2Cl– (aq) + 4H+(aq) → Mn2+(aq) + Cl2(g) + 2H2O(l) (a) By using calculation, show that the reaction is not feasible under standard conditions. (b) Suggest how you would change the conditions of the reaction to make the reaction feasible.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 64 8 2 The E° of the following reaction: 2MnO4 – + MnO2 + 4OH– → 3MnO4 2– + 2H2O is –0.04 V. (a) Determine the oxidation state of manganese in the three manganese containing species. (b) Is this reaction disproportionation? Explain your answer. (c) Suggest the changes you would make to make the reaction feasible. The Stability of Aqueous Ions 1 Certain elements (especially the transition elements) can exhibit variable oxidation states in their compounds. 2 For example, iron can exhibit oxidation states of +2 and +3 in Fe2+ and Fe3+ respectively. Other examples are: Titanium : Ti2+, Ti3+, Ti4+ Cobalt : Co2+; Co3+ Vanadium : V2+; V3+; VO2+; VO2 + 3 However, the relative stabilities of the ions in aqueous solution are not the same. 4 Take the case of cobalt: (a) When cobalt(III) chloride is added to water, effervescence occurs and the pink cobalt(II) aqueous ions are produced. (b) This can be explained by considering the following E° values: O2(g) + 4H+(aq) + 4e– 2H2O(l) +1.23 V Co3+(aq) + e– Co2+(aq) +1.82 V The anticlockwise rule shows that water will reduce Co3+(aq) to Co2+(aq) and gets itself oxidised to oxygen gas (effervescence). We can also say that Co3+(aq) oxidises H2O to O2 and itself reduced to Co2+(aq). (c) The E°cell of the reaction is +0.59 V. 4Co3+(aq) + 2H2O(l) ⎯→ 4Co2+(aq) + 4H+(aq) + O2(g) (d) Hence, in aqueous solution, Co2+ is more stable than Co3+. Example 8.8 Predict what happens when iron is added to dilute sulphuric acid (a) in the absence of air (b) in the presence of air Solution The half-cell equations needed are: Fe2+ + 2e– Fe –0.44 V 2H+ + 2e– H2 0.00 V Fe3+ + e– Fe2+ +0.77 V O2(g) + 4H+(aq) + 4e– 2H2O(l) +1.23 V Info Chem When copper(I) sulphate, a white solid, is added to water, a blue solution is formed and a reddish-brown precipitate sinks to the bottom of the container. This is because Cu+(aq) ion is unstable in aqueous solution and will disproportionate to Cu2+(aq) and metallic copper. 2Cu+ ⎯→ Cu2+ + Cu –0.15 V +0.52 V E°(Reaction) = 0.52 – 0.15 = + 0.37 V

Chemistry Term 2 STPM Chapter 8 Electrochemistry 65 8 (a) In the absence of air, iron will dissolve in dilute sulphuric acid to form iron(II) ions and hydrogen gas is liberated. Fe + 2H+ → Fe2+ + H2 E°cell = +0.44 V Or, Fe2+ + 2e– Fe –0.44 V 2H+ + 2e– H2 0.00 V (b) However, in the presence of air (oxygen), the iron(II) will be oxidised to iron(III). 4Fe2+ + O2 + 4H+ → 4Fe3+ + 2H2O E°cell = +0.46 V Or, Fe3+ + e– Fe2+ +0.77 V O2(g) + 4H+(aq) + 4e– 2H2O(l) +1.23 V Quick Check 8.12 1 When manganese(III) chloride is added to water, effervescence occurs and a faint pink solution is produced. (Mn3+ + e– Mn2+ E° = +1.51 V) (a) What ion is responsible for the pink colour of the solution? (b) Explain the above observations by quoting suitable data from the Data Booklet. 2 The graph shows the E° values for the M3+/M2+ and M2+/M half-cells for some transition metals. (a) List the M2+ aqueous ions that are stable in the presence of oxygen. (b) List the M3+ aqueous ions that are stable in the presence of oxygen. O2 + 4H+ + 4e– 2H2O 1.23 0 Ti V Cr Mn E°/V Fe Co Factors Affecting Standard Electrode Potential 1 The standard electrode potentials of half-cells are affected by: (a) concentration of the aqueous ions, (b) pressure, (c) pH, (d) formation of complexes, (e) temperature. 2 The E° values in the standard electrode potential series refer to standard conditions where the concentration of all the aqueous ions involved are at 1.0 mol dm–3. 3 When the concentration of the ions is changed, it will affect the electrode potential values.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 66 8 Effect of Concentration: Qualitative Aspect 1 The Le Chatelier’s Principle is used to predict qualitatively how the electrode potential of a half-cell changes with a change in concentration. 2 Consider the copper half-cell: Cu2+(aq, 1.0 M) + 2e– Cu(s) E° = +0.34 V (a) When the concentration of Cu2+(aq) is increased, according to Le Chatelier’s Principle, the equilibrium will shift to the righthand side. (b) This means that more Cu2+ from the solution will ‘take’ electrons from the copper atom in the copper electrode and get deposited as copper atoms. (c) As a result, the copper electrode becomes more positively charged, and the electrode potential becomes more positive. (d) When the concentration of Cu2+(aq) is decreased, the equilibrium will shift to the left-hand side. (e) More copper atom from the copper electrode will dissolve to form Cu2+(aq) ions leaving behind their electrons. (f) As a result, the copper electrode becomes more negatively charged, and the electrode potential decreases (becomes less positive). (g) The table below lists the electrode potentials for the copper halfcell at three different Cu2+(aq) concentration at 298 K. [Cu2+]/mol dm–3 0.10 1.00 10.0 E/V +0.31 +0.34 +0.37 3 Consider the following half-cell: Cl2(g, 1 atm) + 2e– 2Cl– (aq, 1.0 M) E° = +1.36 V (a) Increasing the concentration of the Cl– (aq) ions, will cause the equilibrium to shift to the left-hand side. (b) More Cl– ions will get oxidised to Cl2 leaving their electrons on the platinum electrode. (c) As a result, the platinum electrode becomes more negatively charged and the electrode potential decreases. (d) Conversely, when the concentration of the Cl– (aq) ion is decreased, more Cl2 will accept electrons and get reduced and dissolves to form Cl– (aq) ion. (e) As a result, the platinum electrode becomes more positively charged (as electrons are removed) and the electrode potential increases. (f) The table below lists the electrode potential of the chlorine halfcell at three different concentrations of the Cl– ions. [Cl– ]/mol dm–3 0.10 1.00 10.0 E/V +1.42 +1.36 +1.30 4 It can be generalised that: (a) When the concentration of the reactant ion is increased, or when the concentration of the product ion is decreased, the electrode potential becomes more positive. Increasing the concentration of Cu2+ increases the E° value and vice versa. E/V [Cu2+]/mol dm–3 1.0 0.34 Increasing the concentration of Cl– decreases the E° value. E/V [Cl– ]/mol dm–3 1.0 1.36

Chemistry Term 2 STPM Chapter 8 Electrochemistry 67 8 (b) When the concentration of the product ion is increased or when the concentration of the reactant ion is decreased, the electrode potential becomes more negative. (c) Any change that shifts the equilibrium of the half-cell to the right will cause an increase (more positive) in the electrode potential. (d) Any change that shifts the equilibrium of the half-cell to the left will cause a decrease (more negative) in the electrode potential. Oxidising agent + electrons reducing agent E° becomes more positive E° becomes more negative Example 8.9 Given the following half-cell: Fe3+(aq) + e– Fe2+(aq) E° = +0.77 V Predict qualitatively how the electrode potential will change on (a) increasing the Fe3+ concentration (b) decreasing the Fe2+ concentration (c) increasing the Fe2+ concentration Solution (a) Equilibrium shifts to the right. Hence, electrode potential increases (more positive than +0.77 V). (b) Equilibrium shifts to the right. Electrode potential increases. (c) Equilibrium shifts to the left. Electrode potential decreases (less positive than +0.77 V). Increasing the reactant ion increases the E° value. Increasing the product ion decreases the E° value. Quick Check 8.13 1 Mg2+ + 2e– Mg E° = –2.38 V Predict the change in the electrode potential of the above half-cell if (a) the concentration of Mg2+ is increased (b) the concentration of Mg2+ is decreased 2 VO2+ + 2H+ + e– V3+ + H2O E° = +0.34 V Predict the change in the electrode potential of the above half-cell if (a) the concentration of H+ is decreased (b) the concentration of V3+ is decreased (c) the concentration of VO2+ is increased 3 Sn4+ + 2e– Sn2+ E° = +0.15 V Predict the change (if any) in the electrode potential of the above half-cell if (a) the concentration of Sn4+ is increased (b) the concentration of Sn4+ is decreased (c) the concentrations of Sn4+ and Sn2+ are increased by the same factor

Chemistry Term 2 STPM Chapter 8 Electrochemistry 68 8 Effect of Concentration on the E.M.F. of a Cell 1 Consider the Daniell cell: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) 2 The two half-cells are: Cu2+(aq) + 2e– Cu(s) E° = +0.34 V Zn2+(aq) + 2e– Zn(s) E° = –0.76 V The overall cell reaction is: Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu(s) The e.m.f. of the cell under standard conditions is: E°cell = E°(Cu) – E°(Zn) = (+0.34) – (–0.76) = 1.10 V 3 If the concentration of the Cu2+(aq) ion is increased, the electrode potential of the copper half-cell will increase (more positive). Then, E°(Cu) – E°(Zn)  1.10 V 4 If the concentration of Zn2+(aq) ion is increased, the electrode potential of the zinc half-cell will increase (less negative). Then, E°(Cu) – E°(Zn)  1.10 V 5 Hence, for an electrochemical cell, (a) when the concentration of the reactant ions increases, the e.m.f. of the cell will increase (b) when the concentration of the product ions increase, the e.m.f. of the cell will decrease Example 8.10 Consider the following electrochemical cell: Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s) (a) Calculate the e.m.f. of the cell under standard conditions. (b) Write an equation for the overall cell reaction. (c) Predict how the e.m.f. of the cell will change, if (i) the concentration of Al3+ is decreased (ii) the concentration of Cu2+ is increased Solution (a) E°cell = (+0.34) – (–1.66) = 2.00 V (b) Al(s) → Al3+(aq) + 3e– Cu2+(aq) + 2e– → Cu(s) Overall cell reaction is: 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s) (c) (i) Equilibrium shifts to the right. Thus, e.m.f. increases. (ii) Equilibrium shifts to the right. Thus, e.m.f. increases. 2016/P2/Q18(a) Any change that shifts the equilibrium to the right will increase the e.m.f. of the cell. Any change that shifts the equilibrium to the left will decrease the e.m.f. of the cell. Quick Check 8.14 1 An electrochemical cell is constructed from the following two half-cells: Zn2+ + 2e– Zn Fe3+ + e– Fe2+ (a) Write the cell diagram for the electrochemical cell. (b) Calculate the e.m.f. of the cell under standard conditions.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 69 8 (c) Write an equation for the overall cell reaction. (d) Predict the change (if any) in the e.m.f. if, (i) the concentration of Fe3+ is increased (ii) the concentration of Fe2+ is decreased (iii) the concentration of Zn2+ is decreased (iv) the concentration of Zn2+ is increased 2 Consider the following electrochemical cell: Pt(s) | Fe2+(aq), Fe3+(aq) || Ag+(aq) | Ag(s) (a) Name the two half-cells that are involved in the construction of the cell. (b) Indicate the direction of the flow of electrons in the external circuit. (c) Name the cathode and anode of the cell. (d) Calculate the e.m.f. of the cell under standard conditions. (e) Suggest two ways whereby the e.m.f. of the cell can be increased at constant temperature. 8.3 Non-standard Cell Potentials 1 The quantitative relationship between the electrode potential and concentration is given by the Nernst equation: RT [Reduced form]x E = E° – —– ln ——————— zF [Oxidised form]y where: E = Electrode potential E° = Standard electrode potential R = Gas constant (8.31 J mol–1 K–1) T = Absolute temperature z = No. of moles of electrons that are involved F = The Faraday’s constant (96 500 C mol–1) x, y = Stoichiometric coefficient of the species concerned 2 Under standard conditions, the Nernst equation becomes: 8.31  298 [Reduced form]x E = E° – ————– ln ——————— z  96 500 [Oxidised form]y 0.0257 [Reduced form]x E = E° – ——— ln ——————— z [Oxidised form]y Changing to log10: 0.0257 [Reduced form]x E = E° – (2.303)——— ln ——————— z [Oxidised form]y 0.059 [Reduced form]x E = E° – ——– log ——————— z [Oxidised form]y or: 0.059 [Oxidised form]y E = E° + ——– log ——————— z [Reduced form]x 3 For heterogeneous systems involving solids, the concentration of the solid is not included in the Nernst equation. Mathematical relationship between electrode potential and concentration Info Chem Sometimes the Nernst equation is written as: E = E° + —–– RT ZF ln [oxidised form]y —————————–– [reduced form]x 2013/P2/Q6 2018/P2/Q5

Chemistry Term 2 STPM Chapter 8 Electrochemistry 70 8 4 Take the zinc half-cell as example: Zn2+(aq) + 2e– Zn(s) (The oxidised form is Zn2+ and Zn is the reduced form.) The Nernst equation takes the following form: 0.059 1 E = E° – ——–log ——– 2 [Zn2+] 5 For the following half-cell: Sn4+(aq) + 2e– Sn2+(aq) The Nernst equation is: 0.059 [Sn2+] E = E° – ——–log——– 2 [Sn4+] 6 The Nernst equation for the following half-cell: VO3 – (aq) + 4H+(aq) + e– VO2+(aq) + 2H2O is 0.059 [VO2+] E = E° – ——–log——–——– 1 [VO3 – ][H+]4 Example 8.11 Calculate the electrode potential of the following half-cell: Cl2(g, 1 atm) + 2e– 2Cl– (aq, 0.60 M) Solution 0.059 [Cl– ]2 E = E° – ——–log ——– 2 1 0.059 = +1.36 – ——– log (0.60)2 = +1.37V 2 Reduced form Sn2+ Oxidised form Sn4+ Info Chem [H2O] is not included in the expression because it is the solvent and is present in a large quantity. Decreasing the [Cl– ] will shift the equilibrium to the right. As a result, E becomes more positive (1.36 V). Quick Check 8.15 1 Calculate the electrode potential of the following half-cells. (a) Cu2+(aq, 3.4 M) + 2e– Cu(s) (b) 2H+(aq, 1  10–3 M) + 2e– H2(g) (c) Fe3+(aq, 0.8 M) + e– Fe2+(aq, 1.6 M) (d) Cr2O7 2–(aq, 0.6 M) + 14H+(aq, 1.0 M) + 6e– 2Cr2+(1.2 M) + 7H2O(l) 2 The electrode potential of the following half-cell was found to be +0.19 V. Sn4+(aq) + 2e– Sn2+(aq) [Sn4+] Calculate the ratio of ——–. [Sn2+] 3 Calculate the concentration of Fe2+ if the electrode potential of the following half-cell is –0.46 V. Fe2+(aq) + 2e– Fe(s) 4 Consider the zinc half-cell: Zn2+(aq) + 2e– Zn(s) E° = –0.76 V Sketch a graph to show the variation of the electrode potential with (a) [Zn2+(aq)] (b) log[Zn2+(aq)] 5 For the half-cell: Cl2(g) + 2e– 2Cl– (aq) Sketch a graph to show the variation of the electrode potential with (a) [Cl– (aq)] (b) log[Cl– (aq)] Info Chem The concentration of solids is equal to 1.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 71 8 Effect of Pressure on the Electrode Potential 1 Consider the chlorine half-cell: Cl2(g, 1 atm) + 2e– 2Cl– (aq, 1.0 M) E° = +1.36 V 2 If the pressure of the chlorine gas is increased, the equilibrium will shift to the right-hand side. As a result, the electrode potential increases. Cl2(g, P  1 atm) + 2e– 2Cl– (aq, 1.0 M) E  +1.36 V 3 If the pressure of the chlorine gas is reduced, the equilibrium will shift to the left-hand side. As a result, the electrode potential decreases. Cl2(g, P  1 atm) + 2e– 2Cl– (aq, 1.0 M) E  +1.36 V 4 The table below lists the electrode potential of the chlorine half-cell under different pressure of chlorine gas. P(Cl2)/atm 0.10 1.00 10.0 E/V +1.331 +1.360 +1.390 5 The Nernst equation for the chlorine half-cell is: 0.059 [Cl– ] E = E° – ——– log——– 2 P[Cl2] Under standard conditions, [Cl– ] = 1.0 mol dm–3 and P(Cl2) = 1 atm Hence, E = E° 6 Let us calculate the electrode potential of the chlorine half-cell under non-standard conditions: Cl2(g, 5 atm) + 2e– 2Cl– (aq, 1.0 M) Using, 0.059 [Cl– ]2 E = E° – ——– log——– 2 P[Cl2] 0.059 1 E = + 1.36 – ——– log— = +1.38 V 2 5 Info Chem The unit for pressure in the Nernst equation is atm. Increasing the pressure will shift the equilibrium to the right. As a result, E becomes more positive. Quick Check 8.16 1 Calculate the electrode potential of the following half-cells at 298 K. (a) 2H+(aq, 1.0 M) + 2e– H2(g, 10 atm) (b) 2H+(aq, 1.0  10–4 M) + 2e– H2(g, 2 atm) (c) Cl2(g, 0.05 atm) + 2e– 2Cl– (aq, 0.01 M) Nernst Equation and Electrochemical Cell 1 Assume that the reaction in an electrochemical cell can be represented by the following equation: aA + bB cC + dD 2 The Nernst equation will take the following form: 0.059 [C]c [D]d E = E° – ——– log———– z [A]a [B]b where, E = e.m.f. of the cell E° = e.m.f. of the cell under standard conditions z = No. of moles of electrons involved in the reaction [ ]= Molar concentration of the species Info Chem The unit for the partial pressure of a gas in the Nernst equation is atm. 2012/P2/Q6(a)(iii) 2014/P2/Q5 2016/P2/Q3

Chemistry Term 2 STPM Chapter 8 Electrochemistry 72 8 [C]c [D]d 3 The ratio ———– is called the reaction quotient, Q. [A]a [B]b 4 In the simplified form: 0.059 E = E° – ——– log Q z 5 Under standard conditions, Q = 1 and E = E°. 6 Consider the Daniell cell: Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu The Nernst equation takes the form: 0.059 [Zn2+] E = 1.10 – ——– log——– 2 [Cu2+] 7 For the following cell: Sn2+(aq) + 2Fe3+(aq) ⎯→ 2Fe2+(aq) + Sn4+(aq) E° = (+0.77) – (+0.15) = 0.62 V The Nernst equation is, 0.059 [Fe2+]2 [Sn4+] E = 0.62 – ——– log ——–——– 2 [Sn2+][Fe3+]2 Example 8.12 Calculate the e.m.f. of the following cell. Zn(s) | Zn2+(aq, 1.10 M) || Cu2+(aq, 0.55 M) | Cu(s) Solution The overall cell reaction is: Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) E°cell = (+0.34) – (–0.76) = 1.10 V 0.059 [Zn2+] ∴ E = 1.10 – ——–log——– 2 [Cu2+] 0.059 1.10 = 1.10 – ——– log—— 2 0.55 = 1.091 V Under stardard conditions, [Zn2+] ——— = 1. [Cu2+] [Zn2+] 1.10 But in this case, ——— = ——– = 2 [Cu2+] 0.55 ∴The equilibrium will shift to the left to reduce [Zn2+] and at the same time increase [Cu2+] causing the e.m.f. to decrease. Quick Check 8.17 1 Zn(s) | Zn2+(aq, x M) || Cu2+(aq, 1.3 M) | Cu(s) The e.m.f. of the above cell is 1.18 V. Calculate the molar concentration of the Zn2+(aq) ions. 2 Calculate the e.m.f. of the following electrochemical cells. (a) Zn(s) | Zn2+(aq, 1.0 M) || (H+(aq, 0.025 M), H2(g) | Pt(s) (b) Cu(s) | Cu2+(aq, 0.10 M) || Cu2+(aq, 1.5 M) | Cu(s) (c) Pt(s) | Fe2+(aq, 0.86 M); Fe3+(aq, 1.10 M) || Ag+(aq, 1.35 M) | Ag(s) 3 The e.m.f. of the following cell is 3.17 V. Mg(s) | Mg2+(aq, 10.0 M) || Fe3+(aq), Fe2+(aq) | Pt(s) [Fe3+] Calculate the ratio of ——– in the right-hand half-cell. [Fe2+] 4 For the Daniell cell: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Sketch a graph to show the variation of the e.m.f. of the cell with (a) log[Zn2+] (b) log[Cu2+]

Chemistry Term 2 STPM Chapter 8 Electrochemistry 73 8 Nernst Equation and Equilibrium Constant 1 The standard e.m.f. of the Daniell cell is 1.10 V. Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu(s) E°cell = E°(Cu) – E°(Zn) = (+0.34) – (–0.76) = 1.10 V 2 A graph of the e.m.f. of the cell against time is shown below: Time E.m.f./V 1.10 3 The variation of the e.m.f. with time can be explained as follows: (a) When electricity is drawn from the cell, the concentration of Cu2+ decreases while the concentration of Zn2+ increases. (b) This will cause the electrode potential of copper to decrease. Cu2+(aq) + 2e– Cu(s) +0.34 V Electrode potential [Cu2+ ] [Zn2+] Electrode potential (c) However, the electrode potential of zinc will increase. Zn2+(aq) +2e– Zn(s) –0.76 V (d) The difference between the two electrode potential decreases. As a result, the e.m.f. of the cell also decreases (see the graph on the right). (e) A time is reached when the electrode potential of the copper half-cell is equal to the electrode potential of the zinc half-cell. (f) At this point, the difference between the two electrode potential is zero and the e.m.f. of the cell becomes zero. (g) The system has reached a state of dynamic equilibrium and the ratio of [Zn2+] ——— [Cu2+] is the same as the equilibrium constant, Kc of the reaction: Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) 4 For the Daniell cell: 0.059 [Zn2+] E = 1.10 – ——– log——– 2 [Cu2 ] At equilibrium, E = 0 and [Zn2+] ——— [Cu2+] = Kc By substitution: 0.059 0 = 1.10 – ——– log Kc 2 ∴ Kc = 1.94  1037 Info Chem The e.m.f. of the cell decreases gradually and eventually becomes zero. E(Cu) = E(Zn) Time E(Cu) E(Zn) +0.34 –0.76 1.10 Electrode potential/V 2015/P2/Q16(a)

Chemistry Term 2 STPM Chapter 8 Electrochemistry 74 8 5 A large value of Kc indicates that the reaction is almost 100% complete from the left to the right. 6 For a general redox reaction: 0.059 E° = ——– log Kc z Example 8.13 Calculate the equilibrium constant for the following reaction. Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq) Solution E° cell = (+0.77) – (+0.15) = 0.62 V 0.059 Using: E° = ——– log Kc z 0.059 0.62 = ——– log Kc 2 ∴ Kc = 1.04  1021 Quick Check 8.18 1 Calculate the equilibrium constant, Kc for the following reaction. (a) Sn2+(aq) + 2Ag+(aq) Sn4+(aq) + 2Ag (b) Zn(s) + 2Fe3+(aq) Zn2+(aq) + 2Fe2+(aq) (c) 2Ti2+(aq) + Ni2+(aq) Ni(s) + 2Ti3+ (d) 2Cu+(aq) Cu2+(aq) + Cu(s) (e) Zn(s) + Br2(aq) 2Br– (aq) + Zn2+(aq) 2 Consider the following cell: Pb(s) + Sn2+(aq) → Pb2+(aq) + Sn(s) Calculate the ratio of [Sn2+] ——— [Pb2+] when the e.m.f. of the cell equals to zero. Nernst Equation and Solubility Product, Ksp 1 Silver chloride is sparingly soluble salt in water. AgCl(s) + aq Ag+(aq) + Cl– (aq) Ag+ (aq) + Cl – (aq) AgCl(s) 2 In a saturated solution of silver chloride, the ionic product: [Ag+][Cl– ]= constant. (Where [ ] = equilibrium concentration) The constant is called the solubility product, Ksp, of silver chloride. 3 The solubility product of silver chloride can be determined by setting up an electrochemical cell made up of the standard hydrogen electrode and a saturated silver chloride half-cell. Pt(s) |H2(1 atm), H+(1 M) || Ag+(aq, saturated) | Ag(s) The e.m.f. of the cell as given by the voltmeter is 0.51 V. A saturated solution is a solution where the solid is in equilibrium with its aqueous ions. AgCl(aq) AgCl(s) H+ (aq) H2 (g) V

Chemistry Term 2 STPM Chapter 8 Electrochemistry 75 8 4 The standard e.m.f. of the cell: H2(g, 1 atm) + 2Ag+(aq, 1.0 M) ⎯→ 2H+(aq, 1.0 M) + 2Ag(s) is 0.80 V. 5 Using the Nernst equation, we can calculate the [Ag+] in the saturated solution of silver chloride that gives an e.m.f. of 0.51 V. 0. 059 [H+]2 Ecell = E°cell – ——– log ——–– 2 [Ag+]2 0.059 [Ag+]2 0.51 = 0.80 + ——– log ——– 2 1 [Ag+]saturated = 1.22  10–5 mol dm–3 In a saturated solution, [Ag+] = [Cl– ] ∴ Ksp = [Ag+][Cl– ] = (1.22  10–5)2 = 1.49  10–10 mol2 dm–6 6 The solubility product of silver chloride can also be calculated from the E° values of the following two half-cells: AgCl(s) + e– Ag(s) + Cl– (aq) E° = +0.22 V Ag+(aq) + e– Ag(s) E° = +0.88 V (a) Combining the two equations: Ag(s) + Cl– (aq) AgCl(s) + e– E° = –0.22 V Ag+(aq) + e– Ag(s) E° = +0.80 V —————————————————————— Ag+(aq) + Cl– (aq) AgCl(s) + e– E°cell = 0.58 V (b) Applying the Nernst equation: 1 Ecell = E°cell – (0.059) log ———— [Ag+][Cl– ] At equilibrium, Ecell = 0.00 V, and [Ag+][Cl– ] = Ksp 1 ∴ 0 = 0.58 – (0.059) log —– Ksp Ksp = 1.48  10–10 mol2 dm–6 Example 8.14 The e.m.f. of the following cell at 298 K is 0.36 V. Calculate the solubility product of lead sulphate. V Pb Pb2+(aq, x M) H2 (g, 1 atm) [H+ ](aq, 1 M ) SO 2– (aq, 1 M) 4 PbSO4 (s) Solution The e.m.f. of the cell = 0.0 – E(lead half-cell) 0.36 = 0.0 – E(lead half-cell) ∴ E(lead half-cell) = –0.36 V Pb2+(aq, 1 M) + 2e– Pb(s) E° = –0.13 V

Chemistry Term 2 STPM Chapter 8 Electrochemistry 76 8 Using the Nernst equation: 0.059 1 – 0.36 = – 0.13 – ——– log ——– 2 [Pb2+] ∴ [Pb2+] = 1.60  10–8 mol dm–3 Thus, Ksp of PbSO4 = [Pb2+][SO4 2–] = (1.60  10–8)2 = 2.56  10–16 mol2 dm–6 In a saturated solution: [Pb2+] = [SO4 2–] Quick Check 8.19 1 The e.m.f. of the following cell is 0.296 V at 298 K. V Ag Ag+ (aq, 1 M) Ag AgCl(s) AgCl(aq) (a) What is the function of solid silver chloride in the left-hand side half-cell? (b) Calculate the electrode potential of the left-hand side half-cell. (c) Calculate the concentration of Ag+ in the left-hand side half-cell. (d) Calculate the solubility product of silver chloride at 298 K. Effect of pH on the Electrode Potential 1 pH is a measure of the concentration of H+ ions in an aqueous solution. pH = –log[H+] 2 The standard electrode potentials of half-cells refers to the [H+] = 1.0 mol dm–3. That is at pH = 0. 3 Several half-cells that involve H+ ions are: MnO4 – + 8H+ + 5e– ~Mn2+ + 4H2O E° = +1.52 V Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O E° = +1.33 V VO3 – + 4H+ + 4e– VO2+ + 2H2O E° = +1.00 V When altering the pH of the solutions, their electrode potentials will be affected. 4 Let us look at the MnO4 – /Mn2+ half-cell: MnO4 – + 8H+ + 5e– Mn2+ + 4H2O E° = +1.52 V (a) Increasing the pH (the same as decreasing [H+]) will cause the equilibrium to shift to the left. This will cause the electrode potential to decrease. (b) Decreasing the pH (the same as increasing [H+]) will cause the equilibrium to shift to the right. This will cause the electrode potential to increase. (c) This is summarised in the graph shown. Exam Tips Exam Tips Exam Tips Exam Tips Changing the pH is the same as changing the concentration of the H+ ions. pH ~ 1 [H+]

Chemistry Term 2 STPM Chapter 8 Electrochemistry 77 8 E/V 1.52 V 0 pH 5 The table below lists the electrode potentials of the MnO4 – /Mn2+ half-cell at various pH values. pH 0 1 2 3 [H+]/mol dm 1.0 1.0  10–1 1.0  10–2 1.0  10–3 E/V 1.52 1.43 1.33 1.24 6 Another example involves the Fe3+/Fe2+ half-cell: In acidic solution (low pH): Fe3+(aq) + e– Fe2+(aq) E° = +0.77 V In alkaline medium (high pH): Fe(OH)3 + e– Fe(OH)2 + OH– E° = –0.56 V Example 8.15 Calculate the electrode potential of the following half-cell at pH 5. MnO4 – + 8H+ + 5e– Mn2+ + 4H2O E° = +1.52 V Solution Using the Nernst equation: 0.059 [Mn2+] E = E° – ——– log —————— 5 [MnO4 – ][H+]5 0.059 1 = 1.52 – ——– log —————– 5 (1.0  10–5)5 = 1.52 – 0.295 = +1.225 V Example 8.16 Can acidified potassium manganate(VII) oxidise Cl– to Cl2 at the following [H+] concentration? (a) 1.0 mol dm–3 (b) 1.0  10–3 mol dm–3 Solution The E° for the chlorine half-cell is +1.36 V. Cl2 + 2e– 2Cl– E° = +1.36 V Any E° value of the MnO4 2–/Mn2+ half-cell more positive than +1.36 V will be able to oxidise Cl– to Cl2. (a) At [H+] =1.0 mol dm–3, E° = +1.52 V. It can oxidise Cl– to Cl2. (b) At [H+] = 1.0  10–3 mol dm–3, E = +1.24 V (Refer to the table above) It cannot oxidise Cl– to Cl2. pH increases, ∴[H+] decreases. Equilibrium shifts to the left. ∴ E decreases.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 78 8 Quick Check 8.20 1 Calculate the electrode potentials of the following half-cells at the pH indicated. (a) Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O (at pH = –1) (b) VO3 – + 4H+ + 4e– VO2+ + 2H2O (at pH = 6.5) 2 The e.m.f. of the following cell is 0.88 V at 298 K. Pt(s) | H2(g, 1 atm); H+(aq) || Ag+(aq, 1.0 M) | Ag(s) Calculate the pH of the acid in the hydrogen electrode. Effect of Complex Formation on the Electrode Potential 1 A complex ion is an ion consisting of a central metal cation bonded to a group of atoms, molecules or ions (called ligands) via coordinate bonds. 2 Most metal cations exist in aqueous solution as the hexaaquacomplex, [M(H2O)6]n+. For example, Fe3+(aq) exists as [Fe(H2O)6]3+. Fe2+(aq) exists as [Fe(H2O)6]2+. Co3+(aq) exists as [Co(H2O)6]3+. 3 The ligand in the hexaaqua-complex is the H2O molecule. When the ligands of the aqueous cations are changed, the E° values will be affected. 4 Take the case of the Fe3+(aq)/Fe2+(aq) system: [Fe(H2O)6]3+(aq) + e– [Fe(H2O)6]2+ E° = +0.77 V However, if the H2O ligands are replaced by the CN– ligands: [Fe(CN)6]3–(aq) + e– [Fe(CN)6]4–(aq) E° = +0.36 V 5 The oxidising strength of the Fe3+ ions is reduced in the presence of CN– ligands. This can be easily demonstrated. (a) When aqueous iron(III) is added to aqueous potassium iodide, the iodide ions are oxidised to iodine, and the solution turns brown. 2Fe3+(aq) + 2I– (aq) → 2Fe2+(aq) + I2(aq) E° = (+0.77) – (+0.54) = +0.23 V (b) However, when aqueous hexacyanoferate(III) is added to aqueous potassium iodide, nothing happens. This is because the hexacyanoferate(III) ion is too weak as an oxidising agent to oxidise I– to I2. 2[Fe(CN)6]3–(aq) + 2I– (aq) → 2[Fe(CN)6]4–(aq) + I2(aq) E° = (+0.36) – 0.54 = –0.18 V M 3+ OH2 OH2 OH2 OH2 H2 O H2 O [Fe(CN)6] 3– is a weaker oxidising agent because it is more difficult to add an electron to an anion. I2(aq) + 2e– 2I– (aq) E° = +0.54 V Exam Tips Exam Tips Complexing is a convenient way of stabilising an aqueous ios

Chemistry Term 2 STPM Chapter 8 Electrochemistry 79 8 Example 8.17 Given the following standard electrode potential values: [Co(NH3)6]3+ + e– [Co(NH3)6]2+ +0.10 V O2(g) + 4H+(aq) + 4e– 2H2O(l) +1.23 V [Co(H2O)6]3+(aq) + e– [Co(H2O)6]2+(aq) +1.84 V (a) When solid cobalt(III) chloride is added to water, effervescence occurs and a solution containing cobalt(II) chloride remains. However, when hexaamminecobalt(III) chloride is added to water, nothing happens. Explain the observations. (b) Predict what happens when excess ammonia is added to aqueous cobalt(II) chloride. The mixture is then left standing in air. Solution (a) The cobalt(III) cation is a powerful oxidising agent. In aqueous solution, it will oxidise water to oxygen (the effervescence): 4[Co(H2O)6]3+ + 2H2O → 4[Co(H2O)6]2+ + O2 + 4H+ E° = (+1.84) – (+1.23) = +0.61 V However, the hexaamminecobalt(III) ion is a weak oxidising agent and cannot oxidise water to oxygen. 4[Co(NH3)6]3+ + 2H2O → 4[Co(NH3)6]2+ + O2 + 4H+ E° = (+0.10) – (+1.23) = –1.13 V (b) Cobalt(II) ion reacts with aqueous ammonia to give a precipitate of cobalt(II) hydroxide, which dissolves in excess of ammonia to form the hexaamminecobalt(II) complex: Co2+(aq) + 2NH3(aq) + 2H2O(l) → Co(OH)2(s) + 2NH4 +(aq) Co(OH)2(s) + 6NH3(aq) → [Co(NH3)6]2+(aq) + 2OH– (aq) When the solution is exposed to air, the hexaamminecobalt(II) ion is oxidised to hexaamminecobalt(III). 4[Co(NH3)6]2+ + O2 + 4H+ → 4[Co(NH3)6]3+ + 2H2O E° = (+1.23) – (+0.10) = +1.13 V 8.4 Fuel Cells 1 A fuel cell is a special kind of electrochemical cell which converts the chemical energy from a continuous supply of reactants into electrical energy. 2 The reactants are not contained in the cell, but are supplied continuously from an external source. 3 A fuel (such as hydrogen gas or methane gas) is supplied to one electrode while an oxidising agent (usually oxygen) to the other. 4 The direct combustion of hydrogen in oxygen produces heat energy and water. 2H2(g) + O2(g) ⎯→ 2H2O(l) ∆H = –572 kJ 5 However, in the fuel cell, the energy released from the electrolytic oxidation of hydrogen, is converted directly into electricity. 2H2(g) + O2(g) ⎯→ 2H2O(l) E° = 1.23 V Info Chem In the presence of ammonia as ligand, Co3+ is more stable than Co2+. A fuel cell, in theory, can supply electricity indefinitely as long as there is a continuous supply of the oxidant and reductant.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 80 8 The Hydrogen Fuel Cell 1 The simplified diagram of the hydrogen fuel cell is shown below: H2 H2 O2 H2 O H2 O O2 V (Anode) –ve (Cathode) NaOH(aq) +ve Electrons Porous graphite coated with nickel 2 The electrodes in the cell are porous graphite rods coated with finely divided nickel. The nickel acts as catalyst for the electrode reactions. 3 The electrolyte is hot, concentrated aqueous sodium hydroxide. 4 Hydrogen gas and oxygen gas (at a pressure of about 30 atm) are pumped into the anode and cathode compartments respectively. 5 The gases diffuse through the porous electrodes where the following reactions occur: At the anode (negative terminal): H2(g) + 2OH– (aq) ⎯→ 2H2O(l) + 2e– E° = +0.83 V At the cathode (positive terminal): O2(g) + 2H2O(l) + 4e– ⎯→ 4OH– (aq) E° = +0.40 V Overall cell reaction: 2H2(g) + O2(g) ⎯→ 2H2O(l) E° = 0.83 + 0.40 = 1.23 V 6 The advantages of fuel cells are: (a) They are very efficient. About 75% of the chemical energy can be harvested as electrical energy. (b) They are pollution free. For example, the by-product of the hydrogen fuel cell is nothing but water. (c) Theoretically, they can supply electrical energy indefinitely as long as the supply of fuels is not stopped. 7 The disadvantages of fuel cells are: (a) The fuels must be pure. Any impurities would poison the catalyst used in the cell. (b) They are bulky as the gases have to be stored in strong containers. 2010/P2/Q6(c) Advantages of fuel cells Disadvantages of fuel cells

Chemistry Term 2 STPM Chapter 8 Electrochemistry 81 8 The Aluminium/Air Fuel Cell 1 The simplified diagram of the aluminium/air fuel cell is shown below: Anode (–) Cathode (+) Air Porous platinum NaOH(aq) Aluminium 2 At the cathode (platinum electrode): Oxygen undergoes reduction. O2(g) + 2H2O(l) + 4e– ⎯→ 4OH– (aq) E° = +0.40 V 3 At the anode (aluminium electrode): Aluminium undergoes oxidation: Al(s) ⎯→ Al3+(aq) + 3e– E° = +1.66 V The Al3+(aq) ions then combine with aqueous sodium hydroxide to form aluminium hydroxide. Al3+(aq) + 3OH– (aq) ⎯→ Al(OH)3(s) 4 The overall cell reaction is: 4Al(s) + 3O2(g) + 6H2O(l) ⎯→ 4Al(OH)3(s) E° = 0.40 + 1.66 = 2.06 V 5 During the cell reaction, aluminium dissolves, and water is being used up. The cell can be recharged by (a) replacing the aluminium electrode (b) adding water (c) filter off the aluminium hydroxide Batteries for Electric Cars 1 Internal combustion engine uses petrol to provide the energy required to move the vehicle. 2 However, burning of petrol or diesel gives rise to environmental pollution from the exhaust fumes. 3 The exhaust fumes contain pollutants such as carbon monoxide, carbon dioxide, oxides of nitrogen and unburned petrol. 4 Carbon monoxide is poisonous when inhaled because it will react with haemoglobin and destroys the ability of haemoglobin as oxygen carrier. 5 Too much carbon dioxide (a greenhouse gas) in the atmosphere will lead to global warming. 6 Oxides of nitrogen dissolve in water and contribute to acid rain. 7 Unburned petrol contributes to chemical smog which causes respiratory illness. 8 In order to reduce these pollutions, steps are taken to replace petrolburning vehicles with one that runs on electricity. Recharging the cell Carbon monoxide is produced from the imcomplete combustion of petrol. Oxides of nitrogen are produced via the combination of oxygen and nitrogen in the air intake.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 82 8 9 Research into making batteries for electric cars is still in its infant state. The batteries for electric cars should have the following features: (a) It must lightweight. (b) It should be small in size. (c) It should be able to provide a high voltage. (d) It should be economical. (e) It should be rechargeable. (f) It should be able to store electric charge for a long period of time. 8.5 Electrolysis 1 When an electric current flows through an electrolyte, chemical reactions occur at the electrodes. This is called electrolysis which is defined as the decomposition of an electrolyte by an electric current. 2 Electrolysis takes place in an electrolytic cell. Electrolytic Cells and Electrochemical Cells 1 An electrochemical cell converts chemical energy (derived from spontaneous redox reactions) to electrical energy. 2 An electrolytic cell uses electrical energy to produce redox reactions which are non-spontaneous. 3 A simplified diagram of an electrolytic cell is shown below: – + Anion Cation Cathode (–) Electrolyte Anode (+) Current + – Electrons 4 The electrode that is connected to the positive terminal of the battery is called the anode. 5 The electrode that is connected to the negative terminal of the battery is called the cathode. 6 Oxidation takes place at the anode. Reduction takes place at the cathode. 7 The battery acts as an electron pump. 8 The table below shows the comparison between an electrolytic cell and an electrochemical cell. Electrolytic cell Electrochemical cell Uses electricity to cause a nonspontaneous redox reaction to occur. Uses spontaneous redox reactions to produce electricity. The cathode is negatively charged. The cathode is positively charged. The anode is positively charged. The anode is negatively charged. Electrons flow from the cathode to the anode. Electrons flow from the anode to the cathode. Important features of batteries for electric cars Using electricity to force a nonspontaneous reaction to occur. For example: 2H2O(l) ⎯→ 2H2(g) + O2(g) 2010/P1/Q17 The battery functions as an electron pump. Anode Cathode Electrolytic cell Electrochemical cell Cathode + – Anode + – e– e– 2015/P2/Q5

Chemistry Term 2 STPM Chapter 8 Electrochemistry 83 8 9 In the electrolysis of molten lead(II) bromide (m.p. = 373 °C), platinum is used as the electrodes. Cathode Crucible Anode + – Heat Pb2+ Br– 10 In the molten state, lead(II) bromide dissociates into lead(II) and bromide ions. PbBr2(s) Pb2+(l) + 2Br– (l) 11 The lead(II) ion (cation) migrates to the cathode where it accepts electrons (from the battery) and gets reduced to molten lead (melting point = 327 °C). Pb2+(l) + 2e– ⎯→ Pb(l) 12 The bromide ion (anion) migrates to the anode where it gives up electrons and gets oxidised to bromine, a reddish brown vapour (boiling point = 59 °C). 2Br– (l) ⎯→ Br2(g) + 2e– Factors Affecting Products of Electrolysis 1 The products of electrolysis of molten electrolytes are straight forward because the electrolytes consist of one type of cation and one type of anion only. 2 However, it is more complicated when the electrolyte is an aqueous solution of a salt. This is because water too can be oxidised to oxygen at the anode or gets reduced to hydrogen gas at the cathode. The standard electrode potentials for the two half-cells are: O2(g) + 4H+(aq) + 4e– 2H2O(l) E° = +1.23 V 2H2O(l) + 2e– H2(g) + 2OH– (aq) E° = –0.83 V 3 For example, the electrolysis of copper(II) chloride using platinum electrodes. CuCl2(aq) ⎯→ Cu2+(aq) + 2Cl– (aq) 4 Possible reactions occurring at the cathode are reduction of copper(II) and of water. Cu2+(aq) + 2e– ⎯→ Cu(s) 2H2O(l) + 2e– ⎯→ H2(g) + 2OH– (aq) 5 The two possible reactions occurring at the anode are the oxidation of Cl– and water. 2Cl– (aq) ⎯→ Cl2(g) + 2e– 2H2O(l) ⎯→ O2(g) + 4H+(aq) + 4e– 6 What will the actual products of the electrolysis be? The products of electrolysis depend on several factors. Among them are: (a) the standard electrode potential of the competing species (b) the concentration of the aqueous ions (c) the type of electrode used At the anode: 2H2O → O2 + 4H+ + 4e– At the cathode: 2H2O + 2e– → H2 + 2OH– Competing species

Chemistry Term 2 STPM Chapter 8 Electrochemistry 84 8 The Standard Electrode Potential Values 1 During electrolysis, the reaction taking place at the cathode is reduction. Species that get reduced are oxidising agents: Oxidising agent + electrons ⎯→ product at cathode 2 The stronger an oxidising agent, the greater is its tendency to accept electrons. As a result, the ease to be discharged at the cathode increases. 3 In other words, species with more positive E° values will be selectively discharged first at the cathode. 4 The reaction taking place at the anode during electrolysis is oxidation. Species that get oxidised are reducing agents. Reducing agent ⎯→ product at anode + electrons 5 The stronger the reducing agent, the higher the tendency to lose electrons. As a result, the ease to be discharged at the anode increases. 6 In other words, species with more negative E° values will be selectively discharged at the anode. 7 Looking at our example of electrolysis of aqueous copper chloride: (a) The two half-cells that are involved in the cathodic reaction are: Cu2+(aq) + 2e– Cu(s) +0.34 V ................(i) 2H2O(l) + 2e– H2(g) + 2OH– (aq) –0.83 V ................(ii) Reaction (i) has a more positive E° value. This means, the ability of Cu2+ to accept electrons is higher than H2O. Thus, Cu2+ will get reduced in preference to H2O. (b) The two half-cells that are involved in the anodic reaction are: Cl2(g) + 2e– 2Cl– +1.36 V .............. (iii) O2(g) + 4H+(aq) + 4e– 2H2O(l) +1.23 V............... (iv) Reaction (iv) has a more negative E° value. This means, the ability of H2O to release electrons is higher than Cl– . Thus, H2O will be oxidised in preference to Cl– . (c) The products of electrolysis are: Electrode Cathode Anode Product Copper Oxygen gas 8 Alternatively, we can write the equation for the 'actual' reactions that might occur at the electrodes together with the E° values (reversing the signs for the reactions at the anode, which is oxidation). The reaction with the most positive (or least negative) E° value will occur first. (Note: The E° values given in the standard electrode potential series are for reduction.) 9 For example, the electrolysis of aqueous copper(II) chloride. (a) The two possible reactions at the anode are: 2Cl– ⎯→ Cl2 + 2e– E° = –1.36 V*............... (i) 2H2O ⎯→ O2 + 4H+ + 4e– E° = –1.23 V*............... (ii) (* The signs for the two E° values are reversed because the equations are written as oxidation.) The E° value for reaction (ii) is more positive (or less negative). As a result, H2O will be oxidised in preference to Cl– . The product at the cathode is oxygen gas. 2007/P2/Q7(b) Cu2+ is a stronger oxidising agent than H2O. Hence, it is easier to be reduced. Info Chem The E° value refers to the standard reduction potential. Another way to predict the products of electrolysis

Chemistry Term 2 STPM Chapter 8 Electrochemistry 85 8 (b) The two possible reactions at the cathode are: Cu2+ + 2e– ⎯→ Cu E° = +0.34 V............... (iii) 2H2O + 2e– ⎯→ H2 + 2OH– E° = –0.83 V............... (iv) (In this case, there is no need to reverse the signs of the E° values because the reactions are written as reduction.) The E° value for reaction (iii) is more positive. As a result, Cu2+ will be reduced in preference to H2O. The product at the cathode is copper. Quick Check 8.21 Predict the products in the following electrolysis experiment using platinum electrodes. 1 Aqueous sodium chloride 3 Aqueous sodium iodide 2 Aqueous zinc sulphate 4 Dilute sulphuric acid Concentration of the Aqueous Ions 1 For example, Cu2+(aq) and Ag+(aq) ions. Cu2+(aq) + 2e– Cu(s) E° = +0.34 V Ag+(aq) + e– Ag(s) E° = +0.80 V 2 Let us ignore water which might participate in the electrolysis process. 3 Under standard conditions (where the concentrations of the ions are 1.0 mol dm–3), Ag+ will be discharged at the cathode since it has a more positive E° value. 4 However, if we increase the concentration of Cu2+(aq) ions, we can cause Cu2+ to be discharged in preference to Ag+. 5 When the concentration of Cu2+(aq) ions increases, the electrode potential will also increase. At high enough concentration, the electrode potential of the copper half-cell can become more positive than +0.80 V. Then it will be discharged instead of Ag+. E°/V [Cu2+ ]/mol dm–3 Ag+ /Ag Cu2+ /Cu > +0.80 V Cu2+ /Cu < +0.80 V +0.34 1.0 +0.80 6 However, if the difference in the standard electrode potentials of the two competing species is too large, or the product is unstable in water, then concentration is no longer an important factor. 7 For example, if an aqueous solution contains K+(aq) ion. K+(aq) + e– K(s) E° = –2.92 V 2H2O(l) + 2e– H2(g) + 2OH– (aq) E° = –0.83 V

Chemistry Term 2 STPM Chapter 8 Electrochemistry 86 8 The difference between the two E° values is too large (2.09 V). It is not possible to increase the concentration of K+(aq) until its electropotential is more positive than –0.83 V. 8 Furthermore, if K+(aq) is discharged, the potassium metal produced is unstable in aqueous solution and will react immediately to form K+(aq) ions. 2K(s) + 2H2O(l) ⎯→ 2K+(aq) + 2OH– (aq) + H2(g) Hydrogen is liberated. Example 8.18 Predict the product of electrolysis of a concentrated solution of sodium chloride using platinum electrode. Solution NaCl(aq) → Na+(aq) + Cl– (aq) Cathode: Na+(aq) + e– → Na(s) E° = –2.71 V 2H2O(l) + 2e– → H2(g) + 2OH– E° = –0.83 V Even though the concentration of Na+ is high, it is not discharged because the difference between the two E° values is too large. Furthermore, sodium is unstable in water. 2Na(s) + 2H2O(l) → 2Na+(aq) + 2OH– (aq) + H2(g) As a result, hydrogen gas will be liberated at the cathode. Anode: 2Cl– (aq) → Cl2(g) + 2e– E° = –1.36 V 2H2O(l) → O2(g) + 4H+ + 4e– E° = –1.23 V Under standard conditions, H2O would be oxidise to oxygen(because its E° value is less negative). However, due to the high concentration of the Cl– ion, chlorine is liberated instead. Thus, the products of electrolysis are hydrogen (at the cathode) and chlorine (at the anode). –0.83 H2 O/H2 –2.92 K+ (1 M)/K E/V The difference is too large. Quick Check 8.22 1 The products of electrolysis at the anode of aqueous sodium fluoride are the same, whether a dilute solution or a concentrated solution of sodium fluoride is used. Explain this observation. 2 When dilute sodium chloride is electrolysed using platinum electrodes, oxygen gas is liberated at the anode in the beginning. However, as time passes, chlorine gas is evolved instead. Explain this observation. Nature of the Electrode 1 Certain substances are more difficult to be discharged at a particular electrode compared to another electrode. 2 In these cases, the voltage required to discharge the species is higher than expected. This phenomenon is called overpotential or overvoltage. Electrolysis of concentrated NaCl: Anode Cathode Cl2 H2 Info Chem The theoretical voltage required for electrolysis to occur is called, the 'decomposition potential'. (Refer to Example 8.19)

Chemistry Term 2 STPM Chapter 8 Electrochemistry 87 8 3 Overpotential is the electric potential that must be applied in an electrolytic cell in addition to the theoretical potential required to liberate a given substance at an electrode. The value depends on the type of electrode and on the current density. 4 This is especially true when gases are involved. The gas bubbles adsorbed on the surface of the electrode are slow to be released from its surface. The built-up of layers of gas bubbles acts as an insulating layer. As a result, higher voltage has to be applied to continue the electrolysis process. 5 The overpotentials of some gases at different electrodes are given below: Gas Electrode Overpotential/V H2 Copper 0.81 H2 Platinum 0.30 H2 Silver 0.90 H2 Zinc 1.10 H2 Mercury 1.50 O2 Platinum 0.44 O2 Lead 0.32 Cl2 Platinum 0.70 6 In the electrolysis of concentrated sodium chloride using platinum cathode, hydrogen gas is liberated. 7 However, if the cathode is changed to mercury, sodium will be formed instead. This is due to the high overpotential of hydrogen gas at the mercury electrode (1.50 V). Furthermore, the sodium produced can react with mercury to form sodium/mercury amalgam which is stable in water. Hg(l) Na+(aq) + e– ⎯⎯→ Na/Hg(l) Electrolysis of concentrated NaCl: Nature of cathode Platinum Mercury Product Hydrogen Sodium 8 In the electrolysis of aqueous copper(II) sulphate using platinum anode, oxygen is liberated. However, if platinum is substituted by copper as the anode, the copper anode will dissolve and no oxygen gas is liberated. 9 This is because copper is an active electrode (as oppose to platinum which is an inert electrode) and will take part in the electrolysis process. There are now three possible reactions occurring at the anode: 2SO4 2–(aq) ⎯→ S2O8 2–(aq) + 2e– 2H2O(l) ⎯→ O2(g) + 4H+(aq) + 4e– Cu(s) ⎯→ Cu2+(aq) + 2e– 10 Considering the E° values for the three reactions listed below: S2O8 2–(aq) + 2e– 2SO4 2–(aq) E° = +2.01 V O2(g) + 4H+(aq) + 4e– 2H2O(l) E° = +1.23 V Cu2+(aq) + 2e– Cu(s) E° = +0.34 V Of the three species, Cu is the strongest reducing agent (with the least positive E° value), and hence will be discharged first in preference to SO4 2– and H2O. Definition of overpotential The adsorbed gas bubbles from an insulating layer. Gas bubbles Electrode Copper is a stronger reducing agent than H2O and SO4 2–, thus will be discharged first.

Chemistry Term 2 STPM Chapter 8 Electrochemistry 88 8 11 Alternatively, the possible reactions occurring at the anode are: 2SO4 2– ⎯→ S2O8 2– + 2e– E° = –2.01 V............... (i) 2H2O ⎯→ O2 + 4H+ + 4e– E° = –1.23 V............... (ii) Cu ⎯→ Cu2+ + 2e– E° = –0.34 V............... (iii) The E° for reaction (iii) is the most positive (or the least negative). As a result, copper will go into solution as Cu2+. Electrolysis of CuSO4: Nature of anode Platinum Copper Product Oxygen gas Anode dissolves Example 8.19 An aqueous solution of sodium bromide is electrolysed using platinum electrodes. (a) Determine the products of the electrolysis. (b) Calculate the decomposition potential for the electrolysis. (c) Determine whether the following electrochemical cell is suitable for the electrolysis experiment under standard conditions. Zn | Zn2+ || Cu2+ | Cu Solution (a) The possible reactions at the anode are: 2H2O ⎯→ O2 + 4H+ + 4e– E° = –1.23 V............... (i) 2Br– ⎯→ Br2 + 2e– E° = –1.07 V............... (ii) Since the E° for reaction (ii) is or less negative (or more positive), Br– will be oxidised to Br2. The possible reactions at the cathode are: 2H2O + 2e– ⎯→ H2 + 2OH– E° = –0.83 V............... (iii) Na+ + e – ⎯→ Na E° = –2.71 V............... (iv) Since the E° for reaction (iii) is less negative (or more positive), water will be reduced to hydrogen. Products of electrolysis are: Anode : Bromine Cathode : Hydrogen (b) The overall reaction of the electrolysis is: 2H2O + 2Br– ⎯→ Br2 + 2OH– + H2 The E° for the reaction = –1.07 + (–0.83) = –1.90 V (Note: The E° is negative because it is not a spontaneous reaction.) ∴ The decomposition potential = +1.90 V (Note: Decomposition potential is the minimum voltage that must be supplied before electrolysis can take place.)

Chemistry Term 2 STPM Chapter 8 Electrochemistry 89 8 (c) For the electrochemical cell: Zn | Zn2+ || Cu2+ | Cu The two half-cells are: Zn2+ + 2e– Zn E° = –0.76 V............... (Anode) Cu2+ + 2e– Cu E° = +0.34 V............... (Cathode) E.m.f. = E°(Cathode) – E°(Anode) = +0.34 – (–0.76) = +1.10 V Since the e.m.f. of the cell is less than the decomposition potential, the cell is not suitable for the electrolysis for aqueous sodium bromide under standard conditions. Quick Check 8.23 1 When a warm solution of sodium sulphate is electrolysed using platinum electrodes, oxygen gas is liberated at the anode. However, when a warm solution of sodium sulphate is electrolysed using graphite electrodes, a mixture of oxygen, carbon dioxide and carbon monoxide are liberated at the anode. Explain the observations. 2 Predict the products of electrolysis of molten lead oxide using graphite electrodes. 3 Aqueous copper(II) sulphate is electrolysed using copper electrodes. What are the products of the electrolysis? Quantitative Electrolysis 1 Quantitative electrolysis deals with the amount of substances produced during electrolysis. 2 The relationship between the amount of substance produced during electrolysis and the amount of electrical charge is given by Faraday’s Law of Electrolysis. 3 The electrical charge is the charge carried by the electrons when they flow through the electrical circuit. 4 The unit for electrical charge is coulomb (symbol, C: not to be mistaken for the symbol of carbon). 1 Coulomb of electrical charge = 1 Ampere  1 second 5 The amount of electrical charge carried by 1 mole of electrons is 96 500 coulomb. This value is called one Faraday (symbol, F). 1 F = 96 500 C mol–1 6 If the charge on one electron is e C, then 1 F = Le. [Where L = Avogadro’s constant = 6.02  1023 mol–1] Faraday’s First Law of Electrolysis 1 Faraday’s first law of electrolysis states that the amount of substance produced during electrolysis is directly proportional to the amount of electrical charge that flows through the electrolyte. n ∝ Q and Q = It Where n = amount of substance, Q = amount of electrical charge, I = current in A, t = time in second 2 For example, when a current of 0.025 A flows for 25 minutes through an electrolyte, the quantity of electrical charge = 0.025  (25  60) = 37.5 C 2007/P1/Q16 2009/P1/Q17 2008/P1/Q45 2010/P1/Q18 2012/P1/Q18 2015/P2/Q5; Q18(b)

Chemistry Term 2 STPM Chapter 8 Electrochemistry 90 8 Quick Check 8.24 1 When a current of x A flows through aqueous copper(II) sulphate 1.0 mol dm–3 for 30 minutes, 0.68 g of copper is deposited at the cathode at 298 K. What is the mass of copper that would be deposited if the experiment is repeated (a) for 2 hours (b) using 2.0 mol dm–3 of copper(II) sulphate (c) using a current of x A for 1 hour (d) at 350 K (Assuming that in each of the experiment, the other variables are kept constant.) 2 Calculate the volume of gas (measured at s.t.p.) given off at the anode when 0.56 A flows through aqueous sodium sulphate for 34 minutes. 3 When a current of 1.5 A flows for 43 minutes through an aqueous solution containing gold ion, Aun+, 2.63 g of gold was deposited on the cathode. Calculate the charge on the gold ion. 4 Calculate the time required to deposit 78.0 g of gold from a solution containing AuCl4 – ions by a current of 0.50 A. What volume of chlorine gas (measured at s.t.p.) will be formed simultaneously at the anode? Faraday’s Second Law 1 Faraday’s Second Law states that the amount of substances produced during electrolysis by the same amount of electricity is inversely proportional to the charge on the ions. 2 When a certain amount of electrical charge flows through an electrolytic cell containing Mg2+(aq) ion, 2.4 g of magnesium is produced at the cathode. If the same amount of electricity now flows through an electrolytic cell containing Al3+(aq) ions, what is the mass of aluminium deposited at the cathode? This can be solved as follows: 2.4 Number of moles of magnesium deposited = —— = 0.10 mol 24 By Faraday’s second law: No. of moles of Al deposited Charge on the Magnesium ion 2 ————————————= ————————————– = –– No. of moles of Mg deposited Charge on the Aluminium ion 3 Thus, the no. of moles of Al deposited = —2 3  0.10 mol = 0.0667 mol ∴ Mass of aluminium deposited = 0.0667  27 g = 1.8 g Mg2+ (aq) Al3+ (aq) Two cells connected in series. Quick Check 8.25 1 A certain amount of electricity passes through two electrolytic cells connected in series. One contains CuSO4 and the other, AlCl3. If 1.58 g of copper is deposited at the cathode, what mass of aluminium will be deposited in the other cell. 2 The same current that liberates 2.16 g of silver was passed through a solution containing the gold ion, and 1.31 g of gold was deposited. What is the oxidation state of the gold ion?

Chemistry Term 2 STPM Chapter 8 Electrochemistry 91 8 8.6 Applications of Electrochemistry Corrosion (Rusting of Iron) 1 Corrosion is a redox reaction where a metal undergoes electrolytic oxidation. 2 An important example of corrosion is the rusting of iron. 3 For iron to rust, water and oxygen must be present. During rusting, iron is oxidised to hydrated iron(III) oxide, Fe2O3•xH2O. (The number of water of crystallisation associated with the oxide is not fixed.) 4 Consider the diagram below which shows a drop of water on the surface of iron. Fe(s) Anodic region Cathodic region Electrons Iron surface Fe2+(aq) Water droplet 2H2O + O2 + 4e– → 4OH– (aq) 5 The actual rusting process is complicated. However, the main steps are as follows: (a) The iron at the middle of the drop (where the concentration of dissolved oxygen is low) acts as anode. (b) The iron at the outer edge of the drop (where the concentration of dissolved oxygen is high) acts as cathode. (c) At the anodic region, iron undergoes oxidation to form iron(II) aqueous ions. Fe(s) ⎯→ Fe2+(aq) + 2e– (d) The electrons travel through the iron to the cathodic region where dissolved oxygen is reduced to OH– . 2H2O(l) + O2(g) + 4e– ⎯→ 4OH– (aq) (e) The Fe2+(aq) and OH– (aq) ions then diffuse from their respective poles and react to form insoluble iron(II) hydroxide. Fe2+(aq) + 2OH– (aq) ⎯→ Fe(OH)2(s) (f) The iron(II) hydroxide is further oxidised to iron(III) hydroxide. 4Fe(OH)2(s) + O2(g) + 2H2O(l) ⎯→ 4Fe(OH)3(s) (g) Decomposition of iron(III) hydroxide to hydrated iron(III) oxide which forms the ‘rust’. 2Fe(OH)3(s) ⎯→ Fe2O3•xH2O(s) + (3 – x)H2O (h) Dissolved salts accelerate the rate of rusting by supplying ions which increase the conductivity of the system. Anode: Oxidation Cathode: Reduction Info Chem Rusting always starts from the middle of the water droplet because this is the region where iron gets oxidised to iron(II) aqueous ions. The overall reaction is: 2Fe + xH2O + —3 2 O2 → Fe2O3 .xH2O 2015/P2/Q6

Chemistry Term 2 STPM Chapter 8 Electrochemistry 92 8 Prevention of Rusting Rusting of iron can be prevented or slowed down by: (a) Formation of alloy (b) The protective layer method (c) Use of sacrificial anode Formation of Alloy 1 In this method, iron is mixed with other metals/elements to form alloys that are more resistant to rusting. 2 When iron is alloyed with chromium and nickel, it forms stainless steel, where the chromium oxide layer formed on the iron surface prevents iron from rusting. 3 Other elements used are manganese and carbon. The Protective Layer Method 1 This method is also called the barrier method. 2 Iron is prevented from coming into contact with water and oxygen by covering or coating the iron surface with paint, grease, plastic, zinc, tin or other materials. 3 As long as the protective layer is intact, iron will be prevented from rusting. 4 In galvanisation, the surface of iron is coated with a layer of zinc. This prevents iron from rusting even if the layer is broken. (a) When the zinc layer is scratched, the iron underneath is exposed to moist air. However, zinc (E° = –0.76 V) is more reactive than iron (E° = –0.44 V). It is zinc that will get oxidised and not iron. Zn(s) ⎯→ Zn2+(aq) + 2e– Zn(s) O2 + 2H2O + 4e– → 4OH– (aq) Zn(s) Electron Zn2+(aq) Fe(s) (b) The reactions taking place are: Zn(s) ⎯→ Zn2+(aq) + 2e– O2(g) + 2H2O(l) + 4e– ⎯→ 4OH– (aq) Overall reaction: 2Zn(s) + O2(g) + 2H2O(l) ⎯→ 2Zn(OH)2(s) (c) As a result, iron is prevented from rusting. 5 In the making of tin cans, the iron is coated with a layer of tin. As long as this layer is not broken, iron will be prevented from rusting. (a) However, when the tin layer is scratched and the iron beneath it is exposed to moist air, iron will now rust at a faster rate. Electron Fe2+(aq) Fe(s) Iron Sn(s) O2 + 2H2O + 4e– → 4OH– (aq) Zn(s) Protective layer Iron Zinc Iron Tin Iron This is like a short-circuited cell where the two terminals (+ and –) are in direct contact. There is a direct transfer of electrons between the two terminals. Galvanisation: 2014/P2/Q16(c)