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CBSE · Class 12 · Chemistry

Electrochemistry

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks

1.

In a galvanic cell, oxidation takes place at the:

salt bridge
external wire
cathode
anode

2.

The standard emf of the cell Fe | Fe²⁺ || Pb²⁺ | Pb is: (E°: Fe²⁺/Fe = −0.44 V, Pb²⁺/Pb = −0.13 V)

0.31 V
−0.57 V
−0.31 V
0.62 V

3.

A cell reaction involves 2 electrons and has E°cell = 0.11 V. Its standard Gibbs energy change is: (F = 96,500 C/mol)

−21,230 kJ/mol
21.23 kJ/mol
−10.62 kJ/mol
−21.23 kJ/mol

4.

A current of 5 A is passed through a solution of a gold salt (gold ion charge 3+, molar mass 197 g/mol) for 3,860 s. The mass of gold deposited is: (F = 96,500 C/mol)

6.57 g
39.4 g
131.33 g
13.13 g

5.

A 0.11 M solution has conductivity 0.087 S cm⁻¹. Its molar conductivity is:

790.91 S cm² mol⁻¹
9.57 S cm² mol⁻¹
79.09 S cm² mol⁻¹
0.79 S cm² mol⁻¹

6.

The electrode potential of Sn²⁺/Sn when [Sn²⁺] = 10⁻³ M at 298 K is: (E° = −0.14 V; 2.303RT/F = 0.059 V)

−0.0515 V
−0.2285 V
−0.317 V
−0.14 V

7.

Using Kohlrausch's law, the limiting molar conductivity of HCl is: (λ° in S cm² mol⁻¹: H⁺ 349.6, Cl⁻ 76.3)

426.9 S cm² mol⁻¹
851.8 S cm² mol⁻¹
212.95 S cm² mol⁻¹
425.9 S cm² mol⁻¹

8.

Assertion (A): In the cell made of Ni²⁺/Ni (E° = −0.25 V) and Cu²⁺/Cu (E° = 0.34 V), Ni is the anode.
Reason (R): A salt bridge keeps the two half-cells electrically neutral.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true, but R is false.
A is false, but R is true.

Section B — Short answer8 × 2 = 16 marks

9.

What is the function of a salt bridge?

10.

For the cell Mg | Mg²⁺ || Ni²⁺ | Ni, find E°cell and ΔG°, and say whether the reaction is spontaneous. (E°: Mg²⁺/Mg = −2.37 V, Ni²⁺/Ni = −0.25 V) (F = 96,500 C/mol)

11.

How long must a current of 4 A flow to deposit 23.64 g of gold (ion charge 3+, M = 197 g/mol)? (F = 96,500 C/mol)

12.

Find the emf at 298 K of the cell Pb | Pb²⁺ (10⁻¹ M) || Cu²⁺ (10⁻² M) | Cu. (E°: Pb²⁺/Pb = −0.13 V, Cu²⁺/Cu = 0.34 V; 2.303RT/F = 0.059 V)

13.

The molar conductivity of 0.01 M acetic acid is 35.15 S cm² mol⁻¹. Find its degree of dissociation. (λ°: H⁺ = 349.6, CH₃COO⁻ = 40.9 S cm² mol⁻¹)

14.

A cell reaction with n = 2 has E°cell = 0.8555 V at 298 K. Find its equilibrium constant. (2.303RT/F = 0.059 V)

15.

How many faradays of charge are needed for the change of 4 mol of MnO₄⁻ to Mn²⁺? How many coulombs is that? (F = 96,500 C/mol)

16.

Find the limiting molar conductivity of K₂SO₄. (λ° in S cm² mol⁻¹: K⁺ 73.5, SO₄²⁻ 160)

Section C — Word problems4 × 4 = 16 marks

17.

Describe the Daniell cell: its set-up, electrode reactions and emf.

18.

A galvanic cell is made from Pb²⁺/Pb (E° = −0.13 V) and Cu²⁺/Cu (E° = 0.34 V). Write the cell, name the anode and cathode, and find E°cell and ΔG°. (F = 96,500 C/mol)

19.

Two electrolytic cells containing nickel and zinc salts (ion charges 2+ and 2+) are joined in series. If 8.218 g of nickel is deposited, find the mass of zinc deposited, and the charge passed. (M: nickel 58.7, zinc 65.4 g/mol) (F = 96,500 C/mol)

20.

A 0.01 M solution of acetic acid has conductivity 3.5145 × 10⁻⁴ S cm⁻¹. Find its molar conductivity, its degree of dissociation and its dissociation constant. (λ°: H⁺ = 349.6, CH₃COO⁻ = 40.9 S cm² mol⁻¹)

AmbifulWorksheets

Practice that never repeats
CBSE · Class 12 · Chemistry

Answer key

Electrochemistry · AW-12C-ELC-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

Q	Answer	Working
1	(d) anode	—
2	(a) 0.31 V	E°cell = E°(cathode) − E°(anode)
3	(d) −21.23 kJ/mol	ΔG° = −nFE° = −2 × 96,500 × 0.11 J
4	(d) 13.13 g	Q = It = 19,300 C = 0.2 F; m = M Q/(nF)
5	(a) 790.91 S cm² mol⁻¹	Λm = κ × 1000/c
6	(b) −0.2285 V	E = E° − (0.059/2) log(1/[Sn²⁺]) = −0.14 − 0.059 × 3/2
7	(d) 425.9 S cm² mol⁻¹	Λ°m = Σ ν λ°
8	(b) Both A and R are true, but R is not the correct explanation of A.	A is true: Ni has the lower reduction potential, so it is oxidised. R is true but does not explain A.
9	It completes the inner circuit by letting ions move between the half-cells, and it keeps each half-cell electrically neutral, so the cell keeps working.	—
10	E° = 2.12 V; ΔG° = −409.16 kJ/mol; spontaneous	n = 2; ΔG° = −nFE°
11	8,685 s	charge = moles × n × F = 0.12 × 3 × 96,500
12	0.4405 V	E = E° − (0.059/2) log Q, log Q = 1; E° = 0.47 V
13	α = 0.09	Λ°m = 390.5; α = Λm/Λ°m
14	K = 10²⁹	log K = nE°/0.059 = 2 × 0.8555/0.059 = 29
15	20 F; 1,930,000 C	5 mol of electrons per mol
16	307 S cm² mol⁻¹	2 × 73.5 + 160
17	A zinc rod dips in ZnSO₄ solution and a copper rod in CuSO₄ solution; the solutions are joined by a salt bridge and the rods by a wire. Anode: Zn → Zn²⁺ + 2e⁻; cathode: Cu²⁺ + 2e⁻ → Cu. Overall Zn + Cu²⁺ → Zn²⁺ + Cu. Standard emf = 0.34 − (−0.76) = 1.10 V.	—
18	Pb \| Pb²⁺ \|\| Cu²⁺ \| Cu; anode Pb (oxidised), cathode Cu; E° = 0.47 V; ΔG° = −90.71 kJ/mol (n = 2)	the lower E° is the anode
19	9.156 g; 27,020 C	the same charge passes: 0.28 F
20	Λm = 35.15 S cm² mol⁻¹; α = 0.09; Ka = 8.90 × 10⁻⁵	Λm = 1000κ/c; α = Λm/Λ°m; Ka = cα²/(1 − α)