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Practice Set · Class 12

Class 12 Electrochemistry Numericals: Practice Set with Step-by-Step Solutions

10 original problems, every numerical sub-type covered, every answer independently verified. Attempt each one before revealing the solution.

By Genelis Team 16 min read Updated 2026-08-20T09:00:00+05:30

Our Class 12 Chemistry strategy guide told you Electrochemistry numericals cluster into a handful of recurring types. This is that practice, delivered — 10 original problems, one for every distinct sub-type CBSE actually tests, each with a complete solution you can follow line by line.

How to use this page: Read each problem, attempt it fully on paper first, then tap "Reveal Solution" to check your working — not just your final answer. Every calculation on this page was computed independently and verified before publishing, so you can trust the numbers you're checking against.

Most practice sets repeat the same easy Nernst-equation problem five times. This one covers the complete landscape:

1

Nernst equation — basic EMF calculation

2

Nernst equation — find unknown concentration

3

Concentration cell (E°cell = 0)

4

Gibbs free energy from E°cell

5

Equilibrium constant from E°cell

6

Faraday's law — mass deposited

7

Faraday's law — time required

8

Comparative electrolysis — cells in series

9

Conductivity & molar conductivity

10

Kohlrausch's law & degree of dissociation

The 10 Problems

1

Nernst Equation — Basic EMF Calculation

For the cell Zn(s) | Zn²⁺(0.01 M) || Cu²⁺(1.0 M) | Cu(s), given E°cell = 1.10 V, calculate the cell potential (Ecell) at 298 K.

Reveal Solution
1

Step 1 — Identify n and write Q

Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu, so n = 2. Reaction quotient: Q = [Zn²⁺]/[Cu²⁺] (solids don't appear).

2

Step 2 — Substitute

Q = 0.01/1.0 = 0.01, so log₁₀(Q) = −2

Ecell = E°cell − (0.0591/n) log Q = 1.10 − (0.0591/2)(−2)
3

Step 3 — Calculate

Ecell = 1.10 − (0.02955 × −2) = 1.10 − (−0.0591) = 1.10 + 0.0591

✓ Ecell = 1.1591 V
⚠️ Common mistake: sign errors when Q < 1 makes log Q negative — subtracting a negative number increases Ecell. Diluting the anode's ion concentration always raises the cell potential above E°cell.
2

Nernst Equation — Find Unknown Concentration

A cell is set up as Ni(s) | Ni²⁺(x M) || Cu²⁺(1.0 M) | Cu(s). Given E°(Cu²⁺/Cu) = +0.34 V, E°(Ni²⁺/Ni) = −0.25 V, and the measured Ecell = 0.6196 V at 298 K, find [Ni²⁺].

Reveal Solution
1

Step 1 — Find E°cell

E°cell = E°cathode − E°anode = 0.34 − (−0.25) = 0.59 V

2

Step 2 — Rearrange the Nernst equation for Q

n = 2 (both are two-electron processes)

log Q = (E°cell − Ecell) × n / 0.0591
3

Step 3 — Substitute and solve

log Q = (0.59 − 0.6196) × 2 / 0.0591 = (−0.0296 × 2)/0.0591 = −1.0017

Q = 10⁻¹·⁰⁰¹⁷ ≈ 0.0996. Since Q = [Ni²⁺]/[Cu²⁺] and [Cu²⁺] = 1.0 M:

✓ [Ni²⁺] ≈ 0.0996 M ≈ 0.1 M
⚠️ Common mistake: forgetting to reverse the sign when rearranging — since Ecell here is greater than E°cell, Q must come out less than 1, meaning the numerator concentration is diluted relative to the standard state.
3

Concentration Cell (E°cell = 0)

A concentration cell is constructed as Cu(s) | Cu²⁺(0.001 M) || Cu²⁺(0.1 M) | Cu(s). Calculate its EMF at 298 K.

Reveal Solution

Step 1 — Recognise the cell type

Both electrodes are the same metal (Cu), so E°cell = 0. The entire EMF comes purely from the concentration difference. The dilute side is the anode (oxidation, higher tendency to lose electrons at lower concentration); the concentrated side is the cathode.

Step 2 — Write Q

Q = [Cu²⁺]anode/[Cu²⁺]cathode = 0.001/0.1 = 0.01, log Q = −2

Ecell = 0 − (0.0591/n) log Q = 0 − (0.0591/2)(−2)

Step 3 — Calculate

Ecell = 0.0591 V

✓ Ecell = 0.0591 V
⚠️ Common mistake: trying to look up E° for "Cu vs Cu" — there's no such standard value to find, because E°cell for a concentration cell is always exactly zero by definition. The entire answer comes from the log term.
4

Gibbs Free Energy from E°cell

For the Daniell cell reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E°cell = 1.10 V. Calculate ΔG° for the reaction.

Reveal Solution

Step 1 — Identify n

Zn → Zn²⁺ + 2e⁻, so n = 2

ΔG° = −nFE°cell

Step 2 — Substitute

ΔG° = −(2)(96500 C/mol)(1.10 V) = −2 × 96500 × 1.10

✓ ΔG° = −212300 J/mol = −212.3 kJ/mol
⚠️ Common mistake: forgetting units — F is in C/mol and E is in V (=J/C), so the product nFE gives J/mol directly. No unit conversion needed, but always show units to catch errors.
5

Equilibrium Constant from E°cell

Using the same Daniell cell data (E°cell = 1.10 V, n = 2), calculate the equilibrium constant K for the reaction at 298 K.

Reveal Solution
log K = nE°cell / 0.0591

Step 1 — Substitute

log K = (2 × 1.10) / 0.0591 = 2.20/0.0591 = 37.225

Step 2 — Take antilog

K = 10³⁷·²²⁵

✓ K ≈ 1.68 × 10³⁷
⚠️ Don't panic at the size of this number — a large, positive E°cell genuinely does correspond to an astronomically large K. This reflects that the Daniell cell reaction is essentially irreversible; it goes to completion.
6

Faraday's Law — Mass Deposited

A current of 2.0 A is passed through a CuSO₄ solution for 1 hour using platinum electrodes. Calculate the mass of copper deposited at the cathode. (Atomic mass of Cu = 63.5)

Reveal Solution

Step 1 — Identify n and convert time

Cu²⁺ + 2e⁻ → Cu, so n = 2. t = 1 hour = 3600 s

Step 2 — Find charge passed

Q = I × t = 2.0 × 3600 = 7200 C

mass = (M × I × t) / (n × F)

Step 3 — Substitute

mass = (63.5 × 2.0 × 3600) / (2 × 96500) = 457200 / 193000

✓ mass ≈ 2.37 g
⚠️ Common mistake: writing mass = M×I×t/F and forgetting to divide by n. This only works for a one-electron process — for Cu²⁺ (n=2), forgetting n doubles your answer incorrectly.
7

Faraday's Law — Time Required

What time is required to deposit 5.0 g of silver from an AgNO₃ solution using a current of 1.5 A? (Atomic mass of Ag = 108)

Reveal Solution

Step 1 — Identify n and find moles

Ag⁺ + e⁻ → Ag, so n = 1. Moles of Ag = 5.0/108 = 0.04630 mol

Step 2 — Find moles of electrons and charge

Moles of e⁻ needed = 0.04630 mol (since n=1). Q = moles_e × F = 0.04630 × 96500 = 4467.6 C

t = Q / I

Step 3 — Substitute

t = 4467.6 / 1.5

✓ t ≈ 2978.4 s ≈ 49.6 minutes
⚠️ Common mistake: mixing up which variable to solve for — set up moles → charge → time in that fixed order every time, rather than trying to rearrange the combined formula from memory under pressure.
8

Comparative Electrolysis — Cells in Series

Three electrolytic cells containing AgNO₃, CuSO₄, and ZnSO₄ solutions are connected in series. When 2.16 g of silver is deposited in the first cell, calculate: (a) the mass of copper deposited in the second cell, (b) the mass of zinc deposited in the third cell, and (c) the time taken if a steady current of 2.0 A was used. (Atomic masses: Ag=108, Cu=63.5, Zn=65.4)

Reveal Solution

Step 1 — The key idea

Cells in series carry the same charge — so find moles of electrons from the silver data first, then apply that same charge to the other two cells.

Step 2 — Moles of electrons from Ag

Ag⁺ + e⁻ → Ag (n=1). Moles Ag = 2.16/108 = 0.02 mol = moles of electrons passed.

Cu: moles = 0.02/2 = 0.01 mol → mass = 0.01 × 63.5 = 0.635 g
Zn: moles = 0.02/2 = 0.01 mol → mass = 0.01 × 65.4 = 0.654 g

Step 3 — Time from charge and current

Q = 0.02 × 96500 = 1930 C. t = Q/I = 1930/2.0 = 965 s = 16.08 min

✓ Cu = 0.635 g, Zn = 0.654 g, t ≈ 965 s ≈ 16.08 min
⚠️ Common mistake: using n=1 for Cu and Zn too. Copper and zinc are both two-electron processes even though silver is one-electron — the masses deposited are in the ratio of their equivalent weights (M/n), not their atomic masses directly.
9

Conductivity & Molar Conductivity

The resistance of a 0.1 M KCl solution in a conductivity cell is 100 Ω, and its conductivity is known to be 1.29 S/m. When the same cell is filled with 0.02 M KCl solution, the resistance is found to be 520 Ω. Calculate the conductivity and molar conductivity of the 0.02 M KCl solution.

Reveal Solution

Step 1 — Find the cell constant using 0.1 M KCl

The cell constant (G*) is fixed for a given cell, regardless of what solution fills it.

G* = κ × R = 1.29 S/m × 100 Ω = 129 m⁻¹

Step 2 — Find κ for 0.02 M KCl using the same cell constant

κ = G*/R = 129/520 = 0.2481 S/m

Step 3 — Convert to molar conductivity

Convert κ to S/cm: 0.2481/100 = 0.002481 S/cm

Λm = κ(S/cm) × 1000 / c(mol/L) = 0.002481 × 1000 / 0.02
✓ κ = 0.2481 S/m, Λm ≈ 124.04 S cm² mol⁻¹
⚠️ Common mistake: unit mismatch. Molar conductivity's standard formula uses κ in S/cm and c in mol/L — if your κ is in S/m (SI unit), you must divide by 100 first before applying the ×1000/c formula.
10

Kohlrausch's Law & Degree of Dissociation

The molar conductivity of 0.001 M acetic acid (CH₃COOH) solution is 48.15 S cm² mol⁻¹. Given λ°(H⁺) = 349.6 S cm² mol⁻¹ and λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹, calculate the degree of dissociation (α) and the dissociation constant (Ka) of acetic acid.

Reveal Solution

Step 1 — Find Λ°m using Kohlrausch's law

Λ°m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻) = 349.6 + 40.9 = 390.5 S cm² mol⁻¹

Step 2 — Find α

α = Λm / Λ°m = 48.15 / 390.5 = 0.1233

Step 3 — Find Ka

Ka = cα² / (1−α) = (0.001 × 0.1233²) / (1 − 0.1233)
✓ α = 0.1233 (12.33% dissociated), Ka ≈ 1.73 × 10⁻⁵
⚠️ Sanity check worth knowing: the real literature value for acetic acid's Ka is close to 1.8 × 10⁻⁵ — a computed answer this close confirms your method is correct. If your Ka comes out wildly different from this order of magnitude, re-check your α calculation first.

Which of These 10 Types Would You Actually Get Right Under Exam Pressure?

Reading through 10 solved problems and being able to solve 10 new ones cold are different things. The real test is whether you can identify which sub-type a fresh question belongs to and apply the right formula without the type being labelled for you.

What a Genelis weak area map looks like after working through Electrochemistry problem sets

Faraday's law — mass & time
83%
Nernst equation — basic EMF
69%
Kohlrausch's law & degree of dissociation
48%
Comparative electrolysis (series cells)
32%

Next session: comparative electrolysis (32%) — not more basic Nernst practice. Genelis tracks accuracy by sub-type, not just by chapter, so it can tell you exactly which of these 10 patterns needs more reps.

Genelis is an AI-powered personalized learning platform built on Adaptive Personalized Intelligence. The Genelis learning system generates fresh, unlabelled Electrochemistry problems across all 10 sub-types, tracks your accuracy on each specifically, and logs every wrong answer to your wrong-question notebook for reattempt — so you find out which type trips you up before the exam does.

Step 1 Attempt fresh problems
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Step 2 Sub-type gap detected
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Step 3 AI notes for weak pattern
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Step 4 Wrong Qs auto-logged
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Step 5 Reattempt that type
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Practise unlimited fresh Electrochemistry problems on Genelis — free →
💡 For the full formula reference and exam strategy behind these problem types, see the complete Class 12 Chemistry guide.
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Frequently Asked Questions

Questions Students Commonly Ask

Quick answers to the most common questions related to this guide.

What numerical types appear in CBSE Class 12 Electrochemistry?

Ten distinct numerical sub-types appear across CBSE Class 12 Electrochemistry: Nernst equation EMF calculation, Nernst equation applied to find an unknown concentration, concentration cell EMF (where E°cell = 0), Gibbs free energy from cell potential, equilibrium constant from cell potential, Faraday's law for mass deposited, Faraday's law for time required, comparative or series electrolysis across multiple cells, conductivity and molar conductivity from resistance measurements, and Kohlrausch's law combined with degree of dissociation for weak electrolytes.

What value of the Nernst equation constant should I use — 0.0591 or 0.0592?

At 298 K, the precise value of 2.303RT/F works out to approximately 0.05913, which rounds to 0.0591. This is the standard value used throughout NCERT and CBSE board solutions. Some resources round to 0.0592 instead — both are acceptable in a board exam, but 0.0591 is the more precise and more commonly used convention, and is used consistently throughout this practice set.

Why is the equilibrium constant for a cell reaction like the Daniell cell such an enormous number?

Because the relationship log K = nE°cell/0.0591 is exponential in E°cell, even a moderate cell potential produces an astronomically large equilibrium constant. For the Daniell cell (E°cell = 1.10 V, n = 2), K works out to approximately 1.68 × 10³⁷. This reflects a genuinely useful chemical fact: a favourable, spontaneous cell reaction with positive E°cell essentially goes to completion, which is exactly what a large K value represents.

What is the most common mistake students make in Faraday's law numericals?

Forgetting to divide by n, the number of electrons transferred per ion. Writing mass = (M × I × t)/F instead of mass = (M × I × t)/(nF) is the single most common error — it works only for a one-electron process like Ag+ + e- → Ag, and gives a wrong answer for anything else, such as Cu2+ + 2e- → Cu. Always identify n from the half-reaction before substituting into the formula.

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