Question

In the cell represented as $\text{Ni}_{(s)} | \text{Ni}^{2+}_{(1\text{M})} || \text{Ag}^+_{(1\text{M})} | \text{Ag}_{(s)}$, the reducing agent is

• A. Ag
• B. $\text{Ag}^+$
• C. Ni
• D. $\text{Ni}^{2+}$

Hint:

Use the classic mnemonic LOAN: Left side is Oxidation at the Anode, and it involves a Negative electrode. Because oxidation happens on the left side, the starting reactant material on the far left (Ni) is always the species being oxidized, which automatically makes it the reducing agent!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem presents a standard line notation for a galvanic electrochemical cell. We must identify which chemical species serves as the reducing agent in the overall cell reaction.

Step 2: Key Formula or Approach:
1. According to standard IUPAC cell notation rules, the anode (where

oxidation takes place) is always written on the left side of the double vertical salt-bridge line ($||$). The cathode (where

reduction takes place) is written on the right side.
2. A

reducing agent is a substance that reduces another species by donating electrons, meaning the reducing agent itself undergoes oxidation during the process.

Step 3: Detailed Explanation:
Let's analyze the half-cell positions from the cell notation:

Left side (Anode): $\text{Ni}_{(s)} | \text{Ni}^{2+}_{(1\text{M})}$. This signifies that metallic nickel is being oxidized into nickel ions: $$\text{Ni}_{(s)} \rightarrow \text{Ni}^{2+}_{(aq)} + 2e^- \quad \text{(Oxidation Half-Reaction)}$

Right side (Cathode):\text{Ag}^+_{(1\text{M})} | \text{Ag}_{(s)}$. This signifies that silver ions are capturing electrons to become metallic silver: $$2Ag^+_(aq) + 2e^- 2Ag_(s) (Reduction Half-Reaction)$$ Since solid nickel (Ni) loses electrons and is oxidized, it acts as the supplier of electrons to reduce the silver ions. Therefore, solid nickel (Ni) is the reducing agent.

Step 4: Final Answer:
The reducing agent in the given cell is Ni, which corresponds to option (C).