Match the LIST-I with LIST-II.
| LIST-I (Cell/Battery) | LIST-II (Anodic cell reaction) |
| A. Leclanche cell | I. Zn(Hg) + 2OH− → ZnO(s) + H2O + 2e− |
| B. Mercury cell | II. Zn(s) → Zn2+ + 2e− |
| C. Lead storage battery | III. 2H2(g) + 4OH− (aq) → 4H2O(l) + 4e− |
| D. Fuel cell | IV. Pb(s) + SO42−(aq) → PbSO4 (s) + 2e− |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-II, B-I, C-IV, D-III
In every galvanic cell the anode is where oxidation takes place, so the anodic reaction always releases electrons. Each cell below is identified by its anode material and electrolyte:
A. Leclanche cell — This matches II. Zn(s) → Zn2+ + 2e−. In this dry cell the zinc container is the anode, and a paste of \( \ce{NH4Cl} \) and \( \ce{ZnCl2} \) is the electrolyte; \( \ce{MnO2} \) with carbon forms the cathode.
\( \text{Cathode: } \ce{MnO2 + NH4^+ + e^- -> MnO(OH) + NH3} \)
B. Mercury cell — This matches I. Zn(Hg) + 2OH− → ZnO(s) + H2O + 2e−. The anode is zinc-mercury amalgam, and the electrolyte is a paste of KOH and ZnO, which supplies the \( \ce{OH^{-}} \) ions.
\( \text{Cathode: } \ce{HgO + H2O + 2e^- -> Hg(l) + 2OH^-} \)
\( \text{Cell reaction: } \ce{Zn(Hg) + HgO(s) -> ZnO(s) + Hg(l)} \)C. Lead storage battery — This matches IV. Pb(s) + SO42−(aq) → PbSO4 (s) + 2e−. The anode is lead and the electrolyte is 38% sulphuric acid, so lead is oxidised to lead sulphate.
\( \text{Cathode: } \ce{PbO2(s) + SO4^2-(aq) + 4H^+(aq) + 2e^- -> PbSO4(s) + 2H2O(l)} \)
\( \text{Cell reaction: } \ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)} \)
D. Fuel cell — This matches III. 2H2(g) + 4OH− (aq) → 4H2O(l) + 4e−. In the \( \ce{H2-O2} \) fuel cell hydrogen is oxidised at the anode in concentrated aqueous NaOH, and oxygen is reduced at the cathode.
\( \text{Cathode: } \ce{O2(g) + 2H2O(l) + 4e^- -> 4OH^-(aq)} \)
\( \text{Cell reaction: } \ce{2H2(g) + O2(g) -> 2H2O(l)} \)
Shortcut: The pair D–III appears only in Option (2).