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Electrochemistry

Chemistry · Class 12NCERT pp. 31–582 sets × 25 questionsFree

Key facts and formulas for Electrochemistry, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.

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Formula sheet for this chapter

IdeaRule or fact
Cell emfE°cell = E°cathode − E°anode (E° is not multiplied by n)
Nernst equation (298 K)E = E° − (0.059/n) log Q
Gibbs energyΔG° = −nFE°; F = 96500 C mol⁻¹
Equilibrium constantlog K = nE° / 0.059
Conductivityκ = G*/R; G* = l/A; Λm = 1000 κ / M
Kohlrausch lawΛ°m = ν₊λ°₊ + ν₋λ°₋; weak electrolyte α = Λm/Λ°m
Faraday's lawsQ = It; moles deposited = Q / (nF)
Electrolysis and corrosionLower E° gets oxidised; Zn protects Fe, Sn does not
BatteriesLead battery: 38 % H₂SO₄, acid consumed on discharge; fuel cell: porous carbon, NaOH

Three sample questions

Q1. A cell is made from Mg²⁺ | Mg (E° = −2.36 V) and Ag⁺ | Ag (E° = +0.80 V). Its E°cell is
  1. 1.56 V
  2. 3.16 V
  3. −3.16 V
  4. 1.80 V
Hint and solution

Hint: Subtract the anode potential from the cathode potential. The more negative electrode is the anode.

  1. E°cell = E°cathode − E°anode — Ag has the higher E°, so silver is reduced at the cathode. Mg has the lower E°, so it is the anode.
  2. E°cell = 0.80 − (−2.36) — Put the numbers in. Subtracting a negative number adds its size.
  3. E°cell = 0.80 + 2.36 = 3.16 V — E° is an intensive value, so no multiplication by electrons is needed. The answer is 3.16 volt, Option B.
Answer: B. E°cell = E°cathode − E°anode = 0.80 − (−2.36) = 3.16 V. E° is not multiplied by n.
Q2. In the cell Cu | Cu²⁺ || Ag⁺ | Ag
  1. silver is the anode and copper is the cathode
  2. both electrodes are anodes
  3. copper is reduced
  4. copper is the anode and silver is the cathode
Hint and solution

Hint: In cell notation the anode is written on the left. Ask which metal loses electrons.

  1. Anode (oxidation) on left, cathode on right — By convention the oxidation half-cell is written on the left of the double line.
  2. Left: Cu | Cu²⁺ — Copper on the left gives up electrons: Cu → Cu²⁺ + 2e⁻.
  3. Right: Ag⁺ | Ag — Silver ions gain electrons and become silver metal at the cathode.
  4. Copper anode, silver cathode — So copper is the anode and silver is the cathode. Option D.
Answer: D. The anode is on the left, where oxidation takes place: Cu → Cu²⁺ + 2e⁻. Ag⁺ is reduced at the right-hand cathode.
Q3. Given E°(Zn²⁺/Zn) = −0.76 V, E°(Fe²⁺/Fe) = −0.44 V, E°(Cu²⁺/Cu) = +0.34 V, E°(Mg²⁺/Mg) = −2.36 V. Which metal reduces Cu²⁺ but not Zn²⁺?
  1. Fe
  2. Mg
  3. Zn
  4. Ag
Hint and solution

Hint: A metal reduces an ion if its own E° is lower than that of the ion. Look for a value between Zn and Cu.

  1. Metal reduces an ion if E°(metal) < E°(ion) — A lower E° means a stronger reducing agent that can push electrons to the ion.
  2. Order: Mg −2.36 < Zn −0.76 < Fe −0.44 < Cu +0.34 — Arrange all the E° values from lowest to highest.
  3. Need a metal above Zn but below Cu — Reducing Cu²⁺ needs E° below +0.34 V. Not reducing Zn²⁺ needs E° above −0.76 V.
  4. Only Fe fits — Fe at −0.44 V sits in that window. Option A.
Answer: A. Fe (−0.44 V) is below Cu (+0.34 V) but above Zn (−0.76 V). Mg would reduce both; Ag reduces neither.

How to use this chapter

Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.

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