The depolarising agent in a Leclanche cell is?

Analyze the question: It asks which substance in a Leclanché cell acts as the depolarising agent.

Explain the concept: In a simple primary cell, hydrogen gas forms at the positive electrode (the carbon rod) as the cell works. This build-up of hydrogen bubbles is called polarisation. It insulates the electrode and increases the cell's internal resistance, so the current drops quickly.

A depolarising agent is an oxidising substance that removes this hydrogen by oxidising it to water before it can build up.

Parts of the Leclanché cell:

• Zinc case/plate: negative electrode
• Carbon rod: positive electrode (collects current)
• Ammonium chloride paste: electrolyte
• Manganese(IV) oxide (MnO₂): depolarising agent, packed around the carbon rod

Step-by-step solution:

1. Hydrogen forms at the carbon rod: 2H⁺ + 2e⁻ → H₂
2. Manganese(IV) oxide oxidises this hydrogen: 2MnO₂ + H₂ → Mn₂O₃ + H₂O
3. Since the hydrogen is removed, polarisation is prevented.

So the depolarising agent is manganese(IV) oxide.

• A. Carbon rod: it is the positive electrode, not the depolariser.
• B. Ammonium chloride: it is the electrolyte.
• D. Zinc plate: it is the negative electrode.

Exam tip: Remember "Manganese Makes hydrogen Melt away", so MnO₂ is the depolariser. Also, a Leclanché cell is a primary cell with zinc as negative, carbon as positive, NH₄Cl as electrolyte, and MnO₂ as depolariser.