Question

Write the reactions taking place at anode and cathode in a dry cell.

Hint:

Zinc is oxidised at the anode. MnO2 with NH4+ is reduced at the graphite cathode to give MnO(OH) and NH3.

Answer 1

Step 1: Understanding the Concept:
A dry cell (Leclanche cell) is a primary cell. Its container is a zinc cup, which acts as the anode. The cathode is a graphite rod surrounded by a paste of \( \mathrm{MnO_2} \) and carbon. The electrolyte is a moist paste of \( \mathrm{NH_4Cl} \) and \( \mathrm{ZnCl_2} \).

Step 2: Anode reaction (oxidation):
Zinc loses electrons and goes into solution as zinc ions.
\[ \mathrm{Zn(s)} \rightarrow \mathrm{Zn^{2+}(aq)} + 2e^- \]

Step 3: Cathode reaction (reduction):
Manganese in \( \mathrm{MnO_2} \) is reduced from +4 to +3. The ammonium ions supply the hydrogen.
\[ \mathrm{MnO_2(s)} + \mathrm{NH_4^{+}(aq)} + e^- \rightarrow \mathrm{MnO(OH)(s)} + \mathrm{NH_3(g)} \]
The ammonia formed combines with \( \mathrm{Zn^{2+}} \) to give \( [\mathrm{Zn(NH_3)_4}]^{2+} \), which prevents a build-up of gas.

Final Answer:
Anode: \( \mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + 2e^- \). Cathode: \( \mathrm{MnO_2} + \mathrm{NH_4^{+}} + e^- \rightarrow \mathrm{MnO(OH)} + \mathrm{NH_3} \).