JEE Main · 2025 · Shift-IImediumEC-136

Match List-I with List-II: List-I (Applications) | List-II (Batteries/Cell) (A) Transistors | (I) Anode – Zn/Hg;…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

Match List-I with List-II:

List-I (Applications) | List-II (Batteries/Cell) (A) Transistors | (I) Anode – Zn/Hg; Cathode – HgO + C (B) Hearing aids | (II) Hydrogen fuel cell (C) Invertors | (III) Anode – Zn; Cathode – Carbon (D) Apollo spa\ce ship | (IV) Anode – Pb; Cathode – Pb | PbO₂

Choose the correct answer from the options given below:

Options
  1. a

    (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

  2. b

    (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

  3. c

    (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

  4. d

    (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Correct Answera

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Detailed Solution

Strategy: A spontaneous reaction requires a positive standard cell potential (Ecell>0E^\circ_{ \text{cell}} > 0). Calculate EcellE^\circ_{ \text{cell}} for each pairing.

Check Pairings (EcatEanE^\circ_{ \text{cat}} - E^\circ_{ \text{an}}):

  • (a) \ceMn/Fe2+\ce{Mn/Fe^{2+}}: 0.44(1.18)=+0.74 V-0.44 - (-1.18) = +0.74 \text{ V} (Spontaneous) ✓
  • (b) \ceCr/Fe2+\ce{Cr/Fe^{2+}}: 0.44(0.74)=+0.30 V-0.44 - (-0.74) = +0.30 \text{ V} (Spontaneous) ✓
  • (c) \ceZn/Ni2+\ce{Zn/Ni^{2+}}: 0.25(0.76)=+0.51 V-0.25 - (-0.76) = +0.51 \text{ V} (Spontaneous) ✓
  • (d) \ceCr/Mn2+\ce{Cr/Mn^{2+}}: 1.18(0.74)=0.44 V-1.18 - (-0.74) = -0.44 \text{ V} (Non-spontaneous) ✗

Answer: (d)\boxed{\text{Answer: (d)}}

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Match List-I with List-II: List-I (Applications) | List-II (Batteries/Cell) (A) Transistors | (I)… (JEE Main 2025) | Canvas Classes