Question

Identify the gas produced due to reduction of $NH_{4}^{+}$ ions at cathode during working of dry cell.

• A. Ammonia
• B. Hydrogen
• C. Hydrogen chloride
• D. Chlorine

Hint:

Logic Tip: The ammonia ($NH_3$) produced in this reaction does not evolve as a gas. Instead, it complexes with the $Zn^{2+}$ ions produced at the anode to form the complex ion $[Zn(NH_3)_4]^{2+}$, which helps maintain a steady concentration of ions in the paste.

Answer 1

The Correct Option is B

Concept:
A standard dry cell (Leclanché cell) consists of a zinc container (which acts as the anode) and a carbon (graphite) rod (which acts as the cathode). The cathode is surrounded by a paste of manganese dioxide ($MnO_2$) and ammonium chloride ($NH_4Cl$). During the discharge of the cell, redox reactions occur at both electrodes.
Step 1: Identify the primary reduction reaction at the cathode.
At the cathode (the carbon rod), the ammonium ions ($NH_4^+$) from the $NH_4Cl$ paste accept electrons that have traveled through the external circuit from the zinc anode. The reduction half-reaction is: $$2NH_4^+_{(aq)} + 2e^- \longrightarrow 2NH_{3(aq)} + H_{2(g)}$$
Step 2: Identify the gas produced.
Looking at the products of the reduction half-reaction, we see that ammonia ($NH_3$) and hydrogen gas ($H_2$) are formed. The primary gas initially produced through this direct electron transfer is hydrogen gas.
Step 3: Understand the role of the depolarizer (Contextual note).
If hydrogen gas were allowed to build up, it would insulate the carbon electrode (a process called polarization) and stop the cell from working. To prevent this, the $MnO_2$ acts as a depolarizer, immediately oxidizing the hydrogen gas to water: $$H_{2(g)} + 2MnO_{2(s)} \longrightarrow Mn_2O_{3(s)} + H_2O_{(l)}$$ However, the question specifically asks for the gas produced \textit{due to the reduction} of the $NH_4^+$ ions, which is definitively $H_2$.