Question

Electrochemical cells
We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combinations of different half-cells. Each half-cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we do not require a salt bridge.
For the cell
\(\mathrm{Ni(s) \mid Ni^{2+}(aq) \parallel Ag^{+}(aq) \mid Ag}\)
The cell reaction is \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\).
Nernst equation relates the emf of the cell with standard emf and the concentration of reduced and oxidized species.

In the given cell,
\(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\)
which of the given species is the reducing agent?

• A. \(\mathrm{Ni^{2+}(aq)}\)
• B. \(\mathrm{Ni(s)}\)
• C. \(\mathrm{Ag^{+}(aq)}\)
• D. \(\mathrm{Ag(s)}\)

Hint:

The reducing agent loses electrons and is oxidised.

Answer 1

The Correct Option is B

Step 1: Write the two half reactions:
Oxidation: $Ni \to Ni^{2+} + 2e^{-}$. Reduction: $Ag^{+} + e^{-} \to Ag$.

Step 2: Use a quick rule:
The reducing agent is the one that appears on the reactant side of the oxidation half reaction. The oxidising agent is on the reactant side of the reduction half reaction.

Step 3: Pick the reactants:
The reactants are Ni(s) and $Ag^{+}(aq)$. Ni(s) is in the oxidation half, so it is the reducing agent. $Ag^{+}$ is the oxidising agent.

Step 4: Remove the products:
$Ni^{2+}$ and Ag(s) are products, so they cannot be agents acting in the forward reaction. Only option 2 is left.

Final Answer:
Ni(s) is oxidised from 0 to +2 and so it is the reducing agent.\[ \boxed{\mathrm{Ni(s)}} \]