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
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag
The cell reaction is Ni(s) + 2Ag+ (aq) → Ni2+(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,
Ni(s) + 2Ag+ (aq) → Ni2+(aq) + 2Ag(s)
which of the given species is the reducing agent?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Ni(s)
A reducing agent gives electrons to another species. In doing so it is itself oxidised, and its oxidation number increases. An oxidising agent takes electrons and is itself reduced.
Splitting the cell reaction into half reactions:
\( \displaystyle \ce{Ni(s) -> Ni^2+(aq) + 2e^-} \quad \text{(oxidation: 0 to +2)} \)
\( \displaystyle \ce{2Ag^+(aq) + 2e^- -> 2Ag(s)} \quad \text{(reduction: +1 to 0)} \)
Nickel loses electrons and reduces \( \ce{Ag^{+}} \) to Ag. So the reducing agent is Ni(s).
Why the other options are incorrect
Option (1) Ni2+(aq) — Incorrect. It is the product formed when the reducing agent is oxidised.
Option (3) Ag+(aq) — Incorrect. It gains electrons and is reduced, so it is the oxidising agent.
Option (4) Ag(s) — Incorrect. It is the product formed when the oxidising agent is reduced.