➢ oxidation number for halogen = +ve 101 Exception: 1. H bonded to metal NaH, MgH2 oxidation number for H = -1 2. Halogen bonded to oxygen Cl2O7 Example: Example: oxidation number for Cl = +7
Example : Determine the oxidation number of iodine, I in NaIO3 Total oxidation no. = ➢ The total oxidation number of each atoms that made up the molecule is zero. +1 + I + 3 (-2) = 0 I = +5 3. In a neutral compound 102
➢ The total oxidation number of each atoms that made up the ion is equal to the net charge of the ion. Total oxidation no. = Determine the oxidation number of chromium, Cr in Cr2O7 2- Example : 4. The oxidation number of polyatomic ions 2Cr + 7(-2) = -2 Cr = +6 103
Balance the chemical equation in acidic solution using the Ion–electron Method / Half – Reaction Method 104 EXAMPLE 3: Fe2+ + Cr27 2− ⟶ Fe3+ + Cr3+
1. Separate the equation into two half reactions. Reduction : Cr27 2− ⟶ Cr3+ Oxidation : Fe2+ ⟶ Fe3+ STEPS: Oxidation Reduction 105 +2 +3 +6 +3 Fe2+ + Cr27 2− ⟶ Fe3+ + Cr3+
Cr2O7 2– → 2Cr3+ Cr2O7 2– → 2Cr3+ + 7H2O 14H+ + Cr2O7 2– → 2Cr3+ + 7H2O Fe2+ → Fe3+ (no change) 106 2. Balance the atoms other than O and H in each halfreaction. 3. Add H2O to balance O atom & add H+ to balance H atoms.
Fe2+ → Fe3+ 14H+ + Cr2O7 2– → 2Cr3+ + 7H2O + 2 + 3 – 1 + e- – 6 + 14 – 2 + 6 6e- + 107 + 2 + 6 Tips: 1e - -1 0 4. Add eto one side of each half-reaction to balance the charges on the half-reaction.
6e- + 14H+ + Cr2O7 2– → 2Cr3+ + 7H2O 6Fe2+ → 6Fe3+ + 6eFe2+ → Fe3+ + e- X 6 108 5. Multiply each half-reaction by an integer, so that number of electron lost in one half-reaction equals the number gained in the other.
Check that the number of atoms and the charges are balanced. 6e- + 14H+ + Cr2O7 2– → 2Cr3+ + 7H2O 6Fe2+ → 6Fe3+ + 6e Oxidation: - Reduction: 14H+ + Cr2O7 2– + 6Fe2+ → 6Fe3+ + 2Cr3+ + 7H2O (14 x 1) + ( – 2) + (6 x 2) = 24 = (6 x 3) + (2 x 3) Left side Right side 109 acidic solution / acidic medium 6. Combine the two half-reactions and cancel out the species that appear on both sides of the equation.
1. Add to both sides of the equation the same number of OHas there are H+ to eliminate ALL H+ 2. Combine H+ and OHto form H2O. 3. Cancel any H2O that you can. ➢ Follows the steps in acidic medium(step 1 –step 6) followed by these additional steps; 110 ❑ Balance the chemical equation in basic solution using the Ion–electron Method / Half–Reaction Method Example: For Basic Solutions Fe2+ + Cr27 2− ⟶ Fe3+ + Cr3+
8. Combine H+ and OHto form H2O. 7. Add to both sides of the equation the same number of OH- as there are H+ to eliminate ALL H+ 14 OH-+ 14H++ Cr2O7 2– + 6Fe2+→ 6Fe3+ + 2Cr3+ +7H2O+14 OH111 In this case: add 14 OH14 OH-+ 14H++ Cr2O7 2– + 6Fe2+→ 6Fe3+ + 2Cr3+ +7H2O+14 OH14H2O+ Cr2O7 2– + 6Fe2+→ 6Fe3+ + 2Cr3+ +7H2O+14 OH14H+ + Cr2O7 2– + 6Fe2+ → 6Fe3+ + 2Cr3+ + 7H2O Equation from acidic solution :
Check that the number of atoms and the charges are balanced. 9. Cancel any H2O that you can. (0) + ( – 2) + (6 x 2) = 10 = (6 x 3) + (2 x 3) + (-1x14) Left side Right side 112 14H2O+ Cr2O7 2– + 6Fe2+→ 6Fe3+ + 2Cr3+ +7H2O+14 OH7H2O + Cr2O7 2– + 6Fe2+ → 6Fe3+ + 2Cr3+ + 14 OHbasic solution / basic medium 7
MnO4 - ⎯→ Mn2+ +4H2O MnO4 - + 8H+ ⎯→ Mn2+ +4H2O -1 + 8 = +7 2 + 0 = +2 MnO4 - + 8H+ + 5e- ⎯→ Mn2+ + 4H2O MnO4 - ⎯→ Mn2+ MnO4 - + Fe2+ ⎯→ Mn2+ + Fe3+ Acidic solution Oxidation Reduction ➢ Reduction : 113 Answer EXAMPLE 4:
+2 +3 Fe2+ ⎯→ Fe3+ + e - Fe2+ ⎯→ Fe ➢ Oxidation : 3+ Balance the electron 5(Fe2+ ⎯→ Fe3+ + e -) MnO4 - + 8H+ + 5e- ⎯→ Mn2+ + 4H2O Fe2+ ⎯→ Fe3+ + e - 5Fe2+ ⎯→ 5Fe3+ + 5e114
Combine two half-equations Check for net charge and atoms MnO4 - + 8H+ + 5e- ⎯→ Mn2+ +4H2O 5Fe2+ ⎯→ 5Fe3+ + 5eMnO4 - + 5Fe2+ + 8H+ ⎯→ Mn2+ + 5Fe3+ + 4H2O -1 + 10 + 8 = +17 +2 + 15 + 0 = +17 115
MnO4 - + C2O4 2- ⎯→ MnO2 + CO3 2- MnO4 - ⎯→ MnO2 MnO4 - ⎯→ MnO2 -1 + 4 = +3 0 MnO4 - + 4H+ + 3e- ⎯→ MnO2 + 2H2O Reduction Oxidation ➢ Reduction : + 4H+ + 2H2O 116 EXAMPLE 5: Basic solution Answer
C2O4 2- ⎯→ CO3 2- C2O4 2- ⎯→ 2CO3 2- C2O4 2- ⎯→ 2CO3 2- -2 -4 + 4 = 0 C2O4 2- + 2H2O ⎯→ 2CO3 2-+4H+ + 2e- ➢ Oxidation : + 2H2O + 4H+ 117
2(MnO4 - + 4H+ + 3e- ⎯→ MnO2 + 2H2O) ➢ 2MnO4 - + 8H+ + 6e- ⎯→ 2 MnO2 + 4H2O 3(C2O4 2- + 2H2O ⎯→ 2CO3 2-+4H+ + 2e-) ➢ 3C2O4 2- + 6H2O ⎯→ 6CO3 2-+12H+ + 6e2MnO4 - + 3C2O4 2- + 2H2O ⎯→ 2 MnO2 + 6CO3 2-+ 4H+ 2 4 Check for net charge and atoms 118
2MnO4 - + 3C2O4 2- + 2H2O ⎯→ 2 MnO2 + 6CO3 2-+ 4H+ 2MnO4 - + 3C2O4 2- + 2H2O → 2 MnO2 + 6CO3 + 4O H- 2-+ 4H+ 2MnO4 - + 3C2O4 2- + 2H2O + 4OH- ⎯→ 2 MnO2 + 6CO3 2-+ 4H2O 2MnO4 - + 3C2O4 2- + 4OH- ⎯→ 2MnO2 + 6CO3 2-+ 2H2O 2 4H2O ➢ Add the same number of OHas H+ on both sides of equation + 4OH119
Balancing Redox equation MnO4 - + C2O4 2- ⎯→ MnO2 + CO3 2- Keep in mind Separate the equation into two half reaction (Oxidation and Reduction) Add H2O to balance O atoms and add H+ to balance H atoms Add electron to balance charge Combine two half-reactions and cancel out the species that appear on both sides of the equation. Balance atoms other than O and H 120
Acidic medium Basic medium Add 4OH- on both side 2MnO4 - + 3C2O4 2- + 4OH- ⎯→ 2 MnO2 + 6CO3 2-+ 2H2O 2MnO4 - + 3C2O4 2- + 2H2O ⎯→ 2 MnO2 + 6CO3 2-+ 4H+ Finished 121 ❑Difference between balancing redox reactions in acidic and basic medium
Balance the following redox equation : 1. Zn(s) + H+(aq) ⎯→ Zn2+(aq) + H2 (g) 2. S2O3 2- + I2 ⎯→ S4O6 2- + I3. C2O4 2- + MnO4 - ⎯→ CO2+ Mn2+ (acidic solution) 5. CrO2 - + ClO- ⎯→ CrO4 2- + Cl- (basic solution) 4. Cr2O7 2- + Fe2+ + H+ ⎯→ Cr3++ Fe3+ + H2O 122 EXERCISE 3:
Balance the following redox equation : 1. Zn(s) + 2H+(aq) ⎯→ Zn2+(aq) + H2 (g) 2. 2S2O3 2- + I2 ⎯→ S4O6 2- + 2I3. 5C2O4 2- + 2MnO4 - + 16H+(aq) ⎯→ 10CO2 + 2Mn2+ + 8H2O (acidic solution) 5. 2CrO2 - + 3ClO- + 2OH- ⎯→ 2CrO4 2- + 3Cl- + H2O (basic solution) 4. Cr2O7 2- + 6Fe2+ + 14H+ ⎯→ 2Cr3++ 6Fe3+ + 7H2O 123 Answer
❑ Reactants used up first in a reaction ❑ Determine the amount of products formed ❑ Reactant that is completely consumed in a reaction and limit the amount of products formed. 124 LIMITING REACTANT
9 slices of Bread 3 slices of cheese Excess Reactant: bread reactants Limiting reactant: cheese The Cheese Sandwich Analogy product 125
❑ Percentage yield is the percent of the actual yield of a product to its theoretical yield = × 126 PERCENTAGE YIELD
➢The amount of product predicted by a balanced equation is the theoretical yield. 127 ➢The amount product actually obtained in a reaction is the actual yield. ➢The theoretical yield is never obtain because: 1. The reaction may undergo side reaction. 2. Many reaction are reversible. 3. There may be impurities in the reactants. 4. The product formed may react further to form other product. 5. It may be difficult to recover all of the product from the reaction medium.
❑ Quantitative study of reactants and products in a chemical reaction. xA + yB ⎯→ zC + wD reactants products 128 STOICHIOMETRIC CALCULATIONS
xA + yB ⎯→ zC + wD Determining Limiting Reactant Stoichiometric Method 1. Write complete equation 2. Calculate moles of reactants, A and B. 3. Calculate amount of reactant B required to react completely with reactant A 4. Compare the amount of B required (needed) with that available (given) in the system. If B (given) > B (needed) If B (given) < B (needed) B = limiting reactant B = excess reactant A = limiting reactant 129
➢ Determine the limiting reactant: H2 (g) + F2 (g) → 2HF (g) Initial amount: 3 mol 2 mol 4 mol 1 mol of H2 needs 1 mol of F2 3 mol of H2 need 3 mol H2 x 1 mol F2 1 mol H2 = 3 mol F2 (needed) From the equation, 130 (given) EXAMPLE 6:
Limiting reactant = F2 Excess reactant = H2 mole of F2 needed (3 mol) > mole of F2 given (2 mol) Initial amount: 3 mol 2 mol 4 mol 131 (given) H2 (g) + F2 (g) → 2HF (g)
For the following reaction : 2NH3 (g) + 3CuO(s) ⎯→ N2 (g) + 3Cu(s) + 3H2O(l) [Ar Cu = 63.6, H = 1.0, N = 14.0, O = 16.0] 18.0 g NH3 and 90.0 g CuO are allowed to react. a) Determine the limiting reactant. b) Calculate the mass of N2 gas formed. c) Determine the mass of the excess reactant remain after the completion of the reaction. 132 EXAMPLE 7:
(a) Determine the limiting reactant. 133 Answer ➢ = () ( −1) ➢ = 18.0 17.0 −1 = . () ➢ = 90.0 79.6 −1 = . ()
2NH3 (g) + 3CuO(s) ⎯→ N2 (g) + 3Cu(s) + 3H2O(l) = 1.5885 mol CuO(needed) CuO is the limiting reactant ➢ From the equation: mole of CuO needed (1.5885 mol ) > mole of CuO given(1.131 mol) 134 1. . ×
Limiting reactant limits the amount of product formed (b) Calculate the mass of N2 gas formed. 2NH3 (g) + 3CuO(s) ⎯→ N2 (g) + 3Cu(s) + 3H2O(l) 1 mol N2 3 mol of CuO produce = 0.377 mol N2 1.131 mol of CuO produce 1 mol N2 1.131 mol CuO 3 mol CuO ➢ Mass of N2 formed = n N2 x Molar mass N2 = 0.377 mol 28 g mol-1 = 10.56 g ➢ From the equation: [ Tips: compare limiting reactant with product ] 135
(c) Determine the mass of the excess reactant remain after the completion of the reaction. Only the limiting reactant is completely consumed 3 mol of CuO react with 2 mol of NH3 = 0.754 mol NH3 (reacted) 1.131 mol of CuO react with 2 mol NH3 1.131 mol CuO 3 mol CuO 2NH3 (g) + 3CuO(s) ⎯→ N2 (g) + 3Cu(s) + 3H2O(l) ➢ From the equation: 136
Mass of NH3 remained = (18 - 12.818) g = 5.182 g ➢ Mass of NH3 reacted = n NH3 x Molar mass NH3 = 0.754 mol 17 gmol-1 = 12.818 g 137
138 EXAMPLE 8: If 3.7 g sodium metal (Na) and 4.3 g chlorine gas (Cl2 ) react to form NaCl, what is the theoretical yield? If 5.5 g NaCl was formed, what is the percentage yield?
= 0.161 mol Na (given) = 0.061 mol Cl2 (given) 139 Answer ➢ = () ( −1) = 3.7 23.0 −1 ➢ 2 = 2() 2( −1) = 4.3 71.0 −1
2Na(s) + Cl2 (g) → 2NaCl(s) . . × = 0.081 mol Cl2 (needed) mole of Cl2 needed (0.081 mol) > mole of Cl2 given (0.061 mol) Cl2 is limiting reactant From the equation: 140
➢ Mass of NaCl = n NaCl x Molar mass of NaCl = 0.122 mol x 58.5 g/mol = 7.137 g NaCl (theoretical yield) = 0.122 mol NaCl 141 2Na(s) + Cl2 (g) → 2NaCl(s) . . × ➢ From the equation:
= 77.06 % 142 ➢ = × = . . ×
Hydrogen reacts with oxygen to produce water as in the reaction equation below: 2H2 (g) + O2 (g) → 2H2O(g) Assume that before the reaction takes place, there is 20g H2 gas and 224g O2 gas. i. Determine the limiting reagent in this reaction. Show your calculation. ii. Determine the quantity in moles of H2 and O2 consumed in the reaction. iii. Calculate the mass of H2O produced at the end of the reaction. iv. Calculate the mass of the excess reagent left after the reaction has completed. Answer ii) 10, 5 iii) 180 g iv) 64g 143 EXERCISE 2:
144
GLOSSARY 145 BIL TERM SYMBOL/FORMULA DEFINE 1. ISOTOPE - Two or more atoms of the same element having same proton number but different nucleon number. Or Two or more atoms of the same element having same number of protons but different number of neutrons. 2. PROTON NUMBER Z Number of protons in the nucleus of an atom of an element. 3. NUCLEON NUMBER A Total number of protons and neutrons in the nucleus of an atom of an element. 4. RELATIVE ATOMIC MASS Ar A mass of one atom of an element compared to one twelfth mass of one atom of carbon-12 atom
GLOSSARY 146 BIL TERM SYMBOL/FORMULA DEFINE 5. RELATIVE MOLECULAR MASS Mr A mass of one molecule of a compound compared to one twelfth mass of one atom of carbon-12 atom 6. EMPIRICAL FORMULA - Formula that shows the simplest ratio of all elements in a molecule. 7. MOLECULAR FORMULA - Total number of protons and neutrons in the nucleus of an atom of an element. 8. SOLUTE - Is the substance being dissolved and present in the smaller amount 9. SOLVENT - Is the substance doing the solving and present in the larger amount
GLOSSARY 147 BIL TERM SYMBOL/FORMULA DEFINE 10. SOLUTION - Is a homogenous mixture of 2 or more substances that formed when amount of solute dissolves completely in a solvent 11. CONCENTRATION - Is the amount of solute present in a given quantity of solvent in a solution 12. MOLARITY M @ mol L −1 @ mol dm-3 The number of mole of solute dissolved in 1 L of solution. 13. MOLALITY m / mol kg−1 The number of moles of solute per 1 kg of solvent in a solution. 14. MOLE FRACTION X The ratio of the number of moles of one component to the total of number of moles present.
GLOSSARY 148 BIL TERM SYMBOL/FORMULA DEFINE 15. PERCENTAGE BY MASS %Τ The ratio of the mass of a solute to the mass of the solution, multiplied by 100 16. PERCENTAGE BY VOLUME % Τ The ratio of the volume of a solute to the volume of the solution, multiplied by 100 17. LIMITING REACTANT - Reactant that is completely consumed in a reaction and limit the amount of products formed. 18. PERCENTAGE YIELD - The percent of the actual yield of a productto its theoretical yield
Edited By : NZSH. NAAN ……………………… Revised By : MPJ ……………………… Approved By : ZA ………………………… 149