Redox Reactions and Electrochemistry — Practice

Question 1

The oxidation number of manganese in KMnO₄ is:
(a) +7 (b) +2 (c) +4 (d) +6

Accepted Answer

(a) +7.

K = +1, O = −2 (×4 = −8). For a neutral molecule: +1 + Mn − 8 = 0 ⇒ Mn = +7.

Question 2

The oxidation number of oxygen in hydrogen peroxide (H₂O₂) is:
(a) −1 (b) −2 (c) 0 (d) +1

Accepted Answer

(a) −1.

In peroxides oxygen is −1 (not the usual −2). Check: 2(+1) + 2(−1) = 0. ✓

Question 3

In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, the oxidising agent is:
(a) Cu²⁺ (b) Zn (c) SO₄²⁻ (d) Zn²⁺

Accepted Answer

(a) Cu²⁺.

Cu²⁺ is reduced to Cu (gains electrons), so it is the oxidising agent. Zn is oxidised, acting as the reducing agent.

Question 4

Which reaction is a disproportionation?
(a) 2H₂O₂ → 2H₂O + O₂ (b) Zn + Cu²⁺ → Zn²⁺ + Cu (c) C + O₂ → CO₂ (d) 2KClO₃ → 2KCl + 3O₂

Accepted Answer

(a) 2H₂O₂ → 2H₂O + O₂.

Oxygen in H₂O₂ (−1) is simultaneously reduced to −2 (in H₂O) and oxidised to 0 (in O₂) — the same element in one species going both up and down defines disproportionation.

Question 5

In a galvanic cell, oxidation occurs at the:
(a) anode (b) cathode (c) salt bridge (d) both electrodes

Accepted Answer

(a) anode.

By definition oxidation occurs at the anode (negative terminal in a galvanic cell) and reduction at the cathode.

Question 6

Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, the standard EMF of the Daniell cell is:
(a) +1.10 V (b) −1.10 V (c) +0.42 V (d) −0.42 V

Accepted Answer

(a) +1.10 V.

E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V (Cu is cathode, Zn is anode). Positive → spontaneous.

Question 7

Which metal can displace copper from a copper sulphate solution?
(a) Zn (b) Ag (c) Au (d) Pt

Accepted Answer

(a) Zn.

Zinc has a more negative reduction potential (−0.76 V) than copper (+0.34 V), so it is more reactive and displaces Cu²⁺ from solution. Ag, Au and Pt are all less reactive than Cu.

Question 8

For the Daniell cell (n = 2, E° = 1.10 V, F = 96500 C/mol), ΔG° is approximately:
(a) −212 kJ/mol (b) −106 kJ/mol (c) +212 kJ/mol (d) −21.2 kJ/mol

Accepted Answer

(a) −212 kJ/mol.

ΔG° = −nFE° = −2 × 96500 × 1.10 = −212300 J ≈ −212 kJ/mol. The large negative value confirms a strongly spontaneous reaction.

Question 9

In the Nernst equation E = E° − (0.0591/n)log Q, increasing the product-ion concentration will:
(a) decrease the cell potential (b) increase the cell potential (c) not affect it (d) reverse the anode and cathode

Accepted Answer

(a) decrease the cell potential.

Raising product concentration increases Q, so the −(0.0591/n)log Q term becomes more negative and E falls.

Question 10

The mass of copper (atomic mass 63.5, n = 2) deposited by passing 2 A for 965 s through CuSO₄ solution is:
(a) 0.635 g (b) 1.27 g (c) 0.318 g (d) 6.35 g

Accepted Answer

(a) 0.635 g.

q = It = 2 × 965 = 1930 C; m = (M/nF)·q = (63.5/(2×96500))×1930 = 0.635 g.

Question 11

The number of faradays required to deposit 1 mole of aluminium from Al³⁺ is:
(a) 3 (b) 1 (c) 2 (d) 96500

Accepted Answer

(a) 3.

Al³⁺ + 3e⁻ → Al needs 3 electrons per atom, so 3 faradays (3 × 96500 C) per mole of aluminium.

Question 12

On dilution of an electrolyte solution, the molar conductivity:
(a) increases (b) decreases (c) stays constant (d) becomes zero

Accepted Answer

(a) increases.

Molar conductivity increases on dilution because the ions move more freely (and weak electrolytes dissociate more), even though the conductivity κ falls as ion concentration drops.

Question 13

Kohlrausch's law is used to determine the limiting molar conductivity of:
(a) weak electrolytes (b) strong electrolytes only (c) metals (d) gases

Accepted Answer

(a) weak electrolytes.

A weak electrolyte's Λ°m cannot be found by extrapolation, so Kohlrausch's law of independent ion migration builds it from strong-electrolyte values; the degree of dissociation is then α = Λm/Λ°m.

Question 14

Galvanising protects iron from rusting because zinc:
(a) is more reactive and acts as a sacrificial anode (b) is less reactive than iron (c) forms an unreactive gas (d) increases the iron's reduction potential

Accepted Answer

(a) is more reactive and acts as a sacrificial anode.

Zinc has a more negative reduction potential, so it is oxidised in preference to iron (cathodic protection), shielding the iron even where the coating is scratched.

Question 15

The product formed in a hydrogen–oxygen fuel cell is:
(a) water (b) carbon dioxide (c) hydrogen peroxide (d) ozone

Accepted Answer

(a) water.

The H₂–O₂ fuel cell combines hydrogen and oxygen to produce electricity, with water as the only product — clean and efficient, which is why it powered spacecraft.

Practice — Redox Reactions and Electrochemistry

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