SC T10 ELECTROCHEMISTRY

ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

10.1 & 10.2 : Galvanic cell & Nernst equation

1. (a) Definition Measure of the half-cell to attract electrons. Its unit is volt.
i. Voltage associated with a reduction process occur at an electrode, and
ii. measured under 1 atm pressure at 25°C and concentration 1 M.

(b) How to obtained SRP from SHE

• Standard reduction potential is the voltage associated with a reduction process occur at an
electrode, and measured under 1 atm pressure at 25°C and concentration 1 M.

• Standard hydrogen electrode is consist of hydrogen gas that is bubbled into a 1.0 M
hydrochloric acid solution at 25oC. Under standard condition, the hydrogen potential is
zero.

• If one half-cell is connected to SHE, the measured emf of the cell is the sum of electrical
potentials between two electrodes.

• If one of these electrode potential is zero, the other electrode potential could be obtained.

(c) i. cell diagram Voltmeter
V
e

Al(anode) Pb(cathode)

Salt
bridge

Al(NO3)3 (aq,1.0 M) Pb(NO3)2 (aq,1.0 M)

ii. overall cell reaction
Anode : Al(s)→ Al3+ (aq) + 3e
Cathode: Pb2+(aq) + 2e→Pb(s)
overall cell reaction: 2Al(s) + 3Pb2+(aq) → 2Al3+ (aq) + 3Pb(s)

iii. Eocell

Eo = (−0.13) − (−1.66) = +1.53 V
cell

iv. Electrode that increase in weight
Cathode (Pb) because Pb2+ ions in the solution undergo reduction to form Pb
atoms that deposit on the cathode.

2. (a) oxidizing agent
I2(s) + 2e → 2I-(aq)
Fe3+ (aq) + e → Fe2+ (aq)

Eo = (+0.77) − (+0.53) = +0.24 V
cell

spontaneous reaction Fe3+ can oxidise I-

(b) Oxidizing agent
i. Au3+
ii. O2 in acidic medium

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

(c) reducing agent
i. Li
ii. Fe2+

3. (a) function of salt bridge
To complete the circuit so that electric current can flow and to maintain electrical
neutrality.

(b) half-cell equations
Anode : Ag(s) → Ag+ (aq) + e
Cathode : M3+ (aq) + 3e → M (s)

(c) balance equation

3Ag(s) + M 3+ (aq) → 3Ag+ (aq) + M (s)

(d) oxidizing agent
M3+

(e) Eocell

( ) ( )Eo
cell
= Eo − +0.8
M 3+ M

Eo

M 3+ M = +1.8 V

4. (a) equation
Sn2+(aq) + Ni(s) → Sn(s) + Ni2+(aq)

(b) cell notation
Ni(s) | Ni2+ (aq, 1M) || Sn2+ (aq, 1M) | Sn (s)

(c) Eocell

Eo = (−0.136) − (−0.250) = +0.114 V
cell

(d) Ecell

Ecell = (+0.114) − 0.0592 log  0.01  + 0.173 V
2  1.0  =

(e) Increase Ecell
• increasing the concentration of reactant (Sn2+)

• decreasing the concentration of product (Ni2+)

5. (a) Overall cell reaction
anode : Zn(s) → Zn2+(aq) + 2ē
cathode: 2H+(aq) + 2ē → H2(g)
overall : Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

cell notation ; Zn(s) Zn2+ (aq) H + (aq) H2 (g) Pt(s)

(b) Spontaneity

Eo = (−0.00) − (−0.76) = +0.76 V
cell

E0cell = + ve( the reaction is spontaneous)

(c) Reducing and oxidizing agent

Reductant = Zn(s) Oxidant = H+(aq)

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

(d) Ecell = (+0.76) − 0.0592  ( 0.01) (1) 
(e) Ecell 2  
 310−4 2 
( )Ecell log  = + 0.61 V



pH = 1.2
[ H+] = 0.0631 M

Ecell = (+0.76) − 0.0592 log  ( 0.01) (1)  = + 0.75 V
2  (0.0631)2 

The value of Ecellis higher.

6. (a) Overall cell reaction

anode : H2 (g) → 2H + (aq) + 2e
cathode : Cu2+ (aq) + 2e → Cu(s)

Overall cell reaction: Cu2+ (aq) + H2 (g) → Cu(s) + 2H + (aq)

(b) pH

Eo = (+0.34) − (0.00) = +0.34 V
cell

+0.48 = ( +0.34 ) − 0.0592 log  H + 2 
2  
 (1.0) (1) 


[H+] = 4.317 x 10-3 M

pH = 2.36

7. From equation,

Anode: Zn(s) → Zn2+ (aq) + 2e

Cathode: Cr3+ (aq) + 3e → Cr(s)

(a) Eocell From SRP table,

Eo = −0.76 V
Zn2+ /Zn

Eo = −0.74 V
Cr3+ /Cr

E = E − Eo o o

cell cathode anode

= E − Eo o
Cr3+ /Cr Zn2+ /Zn

= (−0.74 V) − (−0.76 V)

(b) Ecell = +0.02 V
Given:
[Zn2+] = 0.009 M, [Cr3+] = 0.10 M

Ecell = Eo − 0.0592 lg Zn2+ 3
cell n Cr3+ 2

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

= ( 0.02 V ) − 0.0592 log  (0.009 M)3 
6  (0.10 M)2 

= +0.0608 V

Cell potential

Zn(s) Zn2+(aq,0.009M ) Cr3+(aq,0.1M ) Cr(s)

(c) Equilibrium constant

At equilibrium, Ecell = 0 V

Ecell = Eo − 0.0592 lg Zn2+ 3
cell n Cr3+ 2

0 = (0.02) − 0.0592 log k

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k = 106.42

10.3: Electrolytic cell

1. (a) Comparison

Diferrence Galvanic cell Electrolytic cell

Energy Energy release by spontaneous redox reaction Electrical energy is used to drive

conversion is converted to electrical energy nonspontaneous redox reaction

Electrodes Anode is negative terminal Anode is positive terminal
Cathode is positive terminal Cathode is negative terminal
Transfer of Negative to positive Positive to negative
electrons

(b) Predict product at anode and cathode Anode
Set Cathode Answer : Cu2+ (aq)
I Answer : Cu (s) Reason : Since Cu(s) electrode is an
Reason : Cu2+(aq) which have more positive active electrode, Cu(s) will undergo
value of E°red will undergo reduction to form oxidation to produce Cu2+(aq)
Cu(s).
Cu(s) → Cu2+ (aq) + 2e
Cu2+ (aq) + 2e → Cu(s)
Answer : O2 (g)
II Answer : Cu (s) Reason : SO42- cannot undergo oxidation
Reason : Cu2+(aq) which have more positive because the oxidation number of S is
value of E°red will undergo reduction to form maximum. Therefore, H2O will undergo
Cu(s). oxidation and produce O2.

Cu2+ (aq) + 2e → Cu(s) 2H2O(l) → O2 (g) + 4H + (aq) + 4e

III Answer : Cu (s) Answer : Cu2+ (aq)
Reason : Cu2+(aq) which have more positive Reason : Since Cu(s) electrode is an
value of E°red will undergo reduction to form active electrode, Cu(s) will undergo

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Cu(s). ANSWER TOPIC 10
SK026
Cu2+ (aq) + 2e → Cu(s)
ELECTROCHEMISTRY
IV Answer : Pb (s)
Reason : For PbBr2 (l), Pb2+ are attracted to oxidation to produce Cu2+(aq)
cathode and undergo reduction.
Cu(s) → Cu2+ (aq) + 2e
Pb+ (l) + 2e → Pb(s)
Answer : Br2 (I)
Reason : Br- are attracted to anode and
undergo oxidation.

2Br− (l) → Br2 (l) + 2e

2. (a) Faraday’s first law
The amount of a substance produced or consumed during electrolysis is directly
proportional to the amount of electricity that passes through the electrical circuit of the cell.

(b) i. mass of silver

Ag+ (aq) + e → Ag(s)

1 mol Ag ≡ 1 mol e ≡ 1 F ≡ 96500 C

96500 C ≡ 1 mol Ag
1000 C ≡ 0.0104 mol Ag
∴ Mass of Ag = 1.12 g

ii. time needed
n Ag = 0.0185 mol
1 mol Ag ≡ 96500 C
0.0185 mol Ag ≡ 1789 C
1789 C = (2)t
t = 895 s

3. Anode: 2H2O(l) → O2 (g) + 4H + (aq) + 4e
Cathode: Au3+ (aq) + 3e → Au(s)
Overall: 6H2O(l) + 4Au3+ (aq) → 3O2 (g) +12H + (aq) + 4Au(s)

(a) Volume
n Au = 0.04701 mol
4 mol Au ≡ 3 mol O2
0.04701 mol Au ≡ 0.0353 mol O2

( 0.0353) ( 0.08206 ) ( 296.15)
V = (0.983) = 0.872 L

(b) current
1 mol O2 ≡ 4 mol e ≡ 4F ≡ 4(96500 C)
0.0353 mol O2 ≡ 13625.8 C

13625.8 C = Q(7200 s)
Q = 1.89 A

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

4. time

Anode: 2H2O(l) → O2 (g) + 4H + (aq) + 4e
Cathode: 2H2O(l) + 2e → H2 (g) + 2OH − (aq)

( 0.924) (10.00 )
n H2 = (0.08206)(295.15) = 0.382 mol

1 mol H2 ≡ 2 mol e ≡ 2F ≡2(96500 C)
0.382 mol H2 ≡ 73726 C

73726 C = (12)t
t = 6143.83 s

5. Volume

Anode: 2H2O(l) → O2 (g) + 4H + (aq) + 4e
Cathode: Ag+ (aq) + e → Ag(s)
Overall: 2H2O(l) + 4Ag+ (aq) → O2 (g) + 4H + (aq) + 4Ag(s)

n Ag = 9.453 x 10-3 mol
4 mol Ag ≡ 1 mol O2
9.453 x 10-3 mol Ag ≡2.363 x 10-3 mol O2

(2.36310−3 )(0.08206)(298.15)

V = (0.971) = 59.5 mL

MEKA 10

1. (a) cell notation
i. From overall equation,

Anode: Sn(s) → Sn2+ (aq) + 2e
Cathode: Cu2+ (aq) + e → Cu+ (aq)
Sn(s) Sn2+ (aq) Cu2+ (aq) Cu+ (aq) Pt(s)

ii. Based on SRP table,

Eo = +1.36 V (Cathode)
Cl2 /Cl−

Eo = +0.77 V (Anode)
Fe3+ /Fe2+

Anode: Fe2+ (aq) → Fe3+ (aq) + e

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

Cathode: Cl2(g) + 2e → 2Cl− (aq)
Pt(s) Fe2+ (aq), Fe3+ (aq) Cl2 (aq) Cl− (g) Pt(s)

(b) Eocell Based on cell notation,
i.
Anode: Cd(s) → Cd 2+ (aq) + 2e

Cathode: Sn2+ (aq) + 2e → Sn(s)

From SRP table,

Eo = −0.40 V
Cd2+ /Cd

Eo = −0.14 V
Sn2+ /Sn

E = E − Eo o o

cell cathode anode

= E − Eo o
Sn2+ /Sn Cd 2+ /Cd

= (-0.14 V) – (-0.40 V)

= 0.26 V

ii. Based on equation,

Anode: Ni(s) → Ni2+ (aq) + 2e

Cathode: Ag+ (aq) + e → Ag(s)

From SRP table,

Eo = +0.80 V
Ag+ /Ag

Eo = −0.25 V
Ni2+ / Ni

E = E − Eo o o

cell cathode anode

= E − Eo o
Ag+ / Ag Ni2+ / Ni

= (+0.80) − (−0.25)

= 1.05 V

(c) Spontaneity
Based on equation,

Anode: Au(s) → Au3+ (aq) + 3e

Cathode: Ca2+ (aq) + 2e → Ca(s)

From SRP table,

Eo = +1.50 V
Au3+ /Au

Eo = −2.87 V
Ca2+ /Ca

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

E = E − Eo o o

cell cathode anode

= E − Eo o
Ca2+ /Ca Au3+ / Au

= (−2.87 V) − (+1.50 V)

= -4.37 V

∴ E0cell = - ve (the reaction is non-spontaneous)

2. (a) Oxidizing agent: substance that undergo reduction reaction
Reducing agent: substance that undergo oxidation reaction.

i. Cl2(g) will accept electron to form Cl−(aq): oxidizing agent

ii. MnO−4 (aq) will accept electron to from Mn2+(aq): oxidizing agent

iii. Ba(s) will released electrons to form Ba2+(aq): reducing agent
iv. Mg(s) will released electrons to from Mg2+(aq): reducing agent

(b) Ascending order of oxidizing agent: Based on SRP value, (-ve to +ve)

Eo = +0.80 V Eo = +1.33 V
Ag+ /Ag CrO72− /Cr3+

Eo = +0.96 V
NO−3 /NO

Ag+(aq)< NO3−(aq) < Cr2O72−(aq)

3. cell potential
From cell notation,

( )Anode: Al(s) → Al3+ (aq) + 3e  2
( )Cathode: Ni2+ (aq) + 2e → Ni(s)  3

Overall equation: 2Al(s) + 3Ni2+ (aq) → 2Al3+ (aq) + 3Ni(s)

E = E − Eo o o

cell cathode anode

= E − Eo o
Ni2+ / Ni Al3+ / Al

= (−0.25 V) − (−1.66 V)

= +1.41 V

Ecell = Eo − 0.0592 lg Al3+ 2
cell n Ni2+ 3

= ( +1.41 V ) − 0.0592 log  (0.01 M)2 
6  (1.0 M)3 

= 1.45 V

4. equilibrium constant
From equation,

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

Anode: Fe(s) → Fe2+ (aq) + 2e
Cathode: Sn2+ (aq) + 2e → Sn(s)

From SRP table,

Eo = −0.44 V
Fe2+ /Fe

Eo = −0.14 V
Sn2+ /Sn

E = E − Eo o o

cell cathode anode

= E − Eo o
Sn2+ /Sn Fe2+ /Fe

= (-0.14 V) – (-0.44 V)

= 0.30 V

At equilibrium, Ecell = 0 V

Ecell = Eo − 0.0592 lg Fe2+ 
cell n Sn2+ 

0 = (0.30 V) − 0.0592 log k

2

k = 1.37 x 1010

5. half-cell equations and product
(a) Molten NaCl,
The electrolyte contain Na+(l) and Cl−(l)

Cl−(l) will be attracted to anode. While Na+(l) will be attracted to cathode.

Anode : 2Cl− (l) → Cl2 (g) + 2e

chlorine gas evolved

Cathode : Na+ (l) + e → Na(l)

sodium liquid

(b) The electrolyte only contains H2O, Na+(aq) and Cl-(aq).

Both H2O, and Cl-(aq) will be attracted to anode, but H2O will be selectively oxidized at anode

because Eo of water is less positive than Cl-.
reduction

Anode : 2H2O(l) → O2 (g) + 4H + (aq) + 4e

oxygen gas evolved

Both H2O, and Na+(aq) will be attracted to anode, but H2O will be selectively reduced at

cathode because Eo of water is higher than Na+.
reduction

Cathode : 2H2O(l) + 2e → H2 (g) + 2OH − (aq)

hydrogen gas evolved

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

KUMBE 10

1. (a) Draw diagram
From SRP table,

Eo = +0.80 V (cathode)
Ag+ /Ag

Eo = +0.34 V (anode)
Cu2+ /Cu

Voltmeter

e V e
e Salt bridge e
e e

Cu(s) Ag(s)

Cu2+(aq) 10

Ag+(aq)

ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

(b) salt bridge
KCl or NH4NO3

(c) cell notation

Anode: Cu(s) → Cu2+ (aq) + 2e

( )Cathode: Ag + (aq) + e → Ag(s) 2

Cu(s) Cu2+(aq) Ag+(aq) Ag(s)

(d) Eocell

E = E − Eo o o

cell cathode anode

= E − Eo o
Ag+ / Ag Cu2+ /Cu

= (+0.80) − (+0.34)

= +0.46 v

2. (a) Concentration
From cell notation,

Anode: Fe2+ (aq) → Fe3+ (aq) + e

Cathode: Ag+ (aq) + e → Ag(s)

Overall Equation: Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag(s)

From SRP table,

Eo = +0.80 V
Ag+ /Ag

Eo = +0.77 V
Fe3+ /Fe2+

Given: Ecell = 0.015 V, [Fe2+] = 0.125 M, [Fe3+] = 0.068 M

E = E − Eo o o

cell cathode anode

= E − Eo o
Fe3+ /Fe2+ Ag+ / Ag

= (+0.8 V) − (+0.77 V)

= 0.03 V

Ecell = Eo − 0.0592 lg Fe3+ 
cell n Fe2+  Ag+ 

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

0.015 V = ( 0.03 V ) − 0.0592 log  ( (0.068 M) +  
1  0.125 M) 
 Ag

[Ag+] = 0.3035 M

(b) pH

From cell notation,

Anode: Zn(s) → Zn2+ (aq) + 2e

Cathode: 2H + (aq) + 2e → H2(g)
Overall Equation: Zn(s) + 2H + (aq) → Zn2+ (aq) + H2 (g)

( )Given:
Eo = +0.76 V, Ecell = 0.54 V, [Zn2+] = 1 M, [H+] = x M, PH2 = 1 atm
cell

( )Ecell
= Eo − 0.0592 lg Zn2+  PH2
cell n H+ 2

0.54 V = ( 0.76 V ) − 0.0592 log  (1.0 M)(1.0 atm ) 
2  
 H + 2 
 

[H+] = 1.922×10-4 M

pH = - log [H+]
= - log (1.922×10-4 M)
= 3.72

3. Effect on Ecell
(a) decrease
(b) decrease
(c) increase
(d) remain unchanged

4. (a) Cell diagram,

+-

anode cathode
carbon(s)

dilute Na2SO4(aq)

(b) Equation and product

The electrolyte only contains H2O, Na+(aq) and SO42− (aq) .

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

Both H2O, and SO42− (aq) will be attracted to anode, but H2O will be selectively oxidized at

anode because Eo of water is less positive than SO42− (aq) and sulphur atom in
reduction

SO42− (aq) cannot be oxidised because the sulphur atom in anion is already at the maximum

oxidation state +6.

Anode: 2H2O(l) → O2 (g) + 4H + (aq) + 4e

Oxygen gas evolved

Both H2O, and Na+(aq) will be attracted to cathode, but H2O will be selectively reduced at

cathode because Eo of water is higher than Na+.
reduction

Cathode: 2H2O(l) + 2e → H2 (g) + 2OH − (aq)

Hydrogen gas evolved

5. (a) mass
Given:
I = 1.73 A
t = 2.05 hours = 7380 s

Q = It
= (1.73 A)(7380 s)
= 12767.4 C

Ag+ (aq) + e → Ag(s)

96500 C ≡ 1 mol e
12767.4 C ≡ 0.132 mol e

1 mol e ≡ 1 mol Ag
0.132 mol e ≡ 0.132 mol Ag

∴ mass of Ag = (0.132 mol)(107.9 g/mol)
= 14.28 g

(b) current

Given:
Mass of Ag = 0.212 g, nAg = 1.96×10−3 mol

t = 1435 s

Ag+ (aq) + e → Ag(s)

1 mol Ag ≡ 1 mol e
1.96×10−3 mol Ag ≡ 1.96×10−3 mol e

1 mol e ≡ 96500 C
1.96×10−3 mol e ≡ 189.603 C

Q = It
(189.603 C) = (I)(1435 s)
I = 0.132 A

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ANSWER TOPIC 10
SK026

ELECTROCHEMISTRY

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