Class 12 Chemistry Chapter 2: Electrochemistry worksheet with answer key
A 20-mark worksheet on Electrochemistry for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (b) Na
The lower the standard electrode potential, the stronger the metal as a reducing agent; sodium has the lowest E° of the four.
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Answer: (d) Ni
Only a metal with a negative E° reduces H⁺ (E° = 0.00 V) to H₂; of the four, only nickel has a negative E°.
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Answer: (b) 0.177 V
Each electrode has E = −0.059 pH. The pH 5 electrode (lower E) is the anode: Ecell = (−0.059 × 2) − (−0.059 × 5) = 0.059 × 3 = 0.177 V. 0.413 V adds the two pH values; 0.295 V uses pH 5 alone.
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Answer: (d) 0.281 V
E = E° − (0.059/2) log(1/[Cu²⁺]) = 0.34 − 0.0295 × log 100 = 0.34 − 0.059 = 0.281 V. Adding the term gives 0.399 V; using log 10 gives 0.311 V; dropping the 2 gives 0.222 V.
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Answer: (b) 1.41 V
E°cell = E°right − E°left = −0.25 − (−1.66) = 1.41 V. E° is intensive: the factor 2 or 3 in the balanced equation does not multiply it.
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In aqueous solution water (H⁺, E° 0.00 V) is reduced in preference to Na⁺ (E° −2.71 V), so H₂ is evolved at the cathode, not sodium. Sodium is got by electrolysing molten (fused) NaCl: Na⁺ + e⁻ → Na at the cathode, with Cl₂ at the anode.
- water is reduced before Na⁺ (higher E°), giving H₂ (1)
- electrolysis of fused NaCl, Na⁺ + e⁻ → Na (1)
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Rusting needs both water (moisture) and air (oxygen). Rust is hydrated ferric oxide, Fe₂O₃·xH₂O.
- water and oxygen (air) both needed (1)
- Fe₂O₃·xH₂O (1)
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A primary battery reacts only once and cannot be recharged, e.g. the dry cell or the mercury cell. A secondary battery can be recharged by passing current through it in the opposite direction and used again, e.g. the lead storage battery or the nickel–cadmium cell.
- primary: used once, not rechargeable, with an example (1)
- secondary: recharged by reverse current, with an example (1)
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(i) Anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻. (ii) Cathode: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l). (iii) Overall Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O: sulphuric acid is used up and water formed, so the acid becomes more dilute and its density falls as the battery discharges.
- anode reaction (1)
- cathode reaction (1)
- H₂SO₄ consumed and water formed, so density falls (1)
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(i) MnO₄⁻ reduced, Cl⁻ oxidised: E°cell = 1.51 − 1.36 = +0.15 V, positive, so MnO₄⁻ oxidises Cl⁻ to Cl₂ and part of the KMnO₄ is used up on the acid, spoiling the titre. (ii) Pb²⁺ reduced, Ni oxidised: E°cell = −0.13 − (−0.25) = +0.12 V, positive, so the nickel dissolves as Ni²⁺ and a layer of lead deposits on the stirrer. (iii) Zn²⁺ reduced, Fe oxidised: E°cell = −0.76 − (−0.44) = −0.32 V, negative, so iron cannot displace zinc and no zinc is recovered.
- (i) E°cell = +0.15 V, Cl⁻ is oxidised by MnO₄⁻ (1)
- (ii) E°cell = +0.12 V, lead deposits and nickel dissolves (1)
- (iii) E°cell = −0.32 V, no reaction, zinc not recovered (1)
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Ka = cα²/(1 − α): 0.010α² + 1.29 × 10⁻³α − 1.29 × 10⁻³ = 0, so α = [−1.29 × 10⁻³ + √(1.66 × 10⁻⁶ + 5.16 × 10⁻⁵)]/0.020 = 0.30. Λ°m = λ°(H⁺) + λ°(ClCH₂COO⁻) = 349.6 + 40.4 = 390.0 S cm² mol⁻¹ (Kohlrausch); Λm = αΛ°m = 0.30 × 390.0 = 117 S cm² mol⁻¹. κ = Λm × c/1000 = 117 × 0.010/1000 = 1.17 × 10⁻³ S cm⁻¹.
- α = 0.30 from the quadratic in Ka = cα²/(1 − α) (1)
- Λ°m = 390.0 and Λm = 117 S cm² mol⁻¹ (1)
- κ = 1.17 × 10⁻³ S cm⁻¹ (1)