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: Read the cell notation convention.
In standard cell notation, the anode (oxidation) is written on the left of the salt bridge $||$ and the cathode (reduction) on the right.
Step 2: Identify the left electrode.
The left side is $\text{Ni}_{(s)}\,|\,\text{Ni}^{2+}$, so nickel metal is being oxidised: $\text{Ni} \rightarrow \text{Ni}^{2+} + 2e^-$.
Step 3: Identify the right electrode.
The right side is $\text{Ag}^+\,|\,\text{Ag}_{(s)}$, so silver ions are reduced: $\text{Ag}^+ + e^- \rightarrow \text{Ag}$.
Step 4: Recall the definition of a reducing agent.
A reducing agent donates electrons to something else and is itself oxidised in the process.
Step 5: Match the definition.
Solid nickel loses electrons (gets oxidised) and supplies them to reduce the silver ions, so nickel is the species being oxidised.
Step 6: Conclude.
The electron donor, and therefore the reducing agent, is metallic Ni.
\[ \boxed{\text{Ni is the reducing agent, option (3)}} \]