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: Understanding the Concept:
A reducing agent is the species that gives away electrons. In doing so it is itself oxidised, and its oxidation number goes up.

Step 2: Key Formula or Approach:
Look at each species in \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\) and see which one loses electrons and which one gains them.

Step 3: Check option (1):
\(Ni^{2+}\) is a product. Its oxidation number is +2, and it has already lost its electrons. It is the oxidised form, not the reducing agent. So option 1 is WRONG.

Step 4: Check option (2):
Ni(s) changes from 0 to +2. It loses 2 electrons and is oxidised, so it makes the other species reduce. So Ni(s) is the reducing agent, and option 2 is CORRECT.

Step 5: Check option (3):
\(Ag^{+}\) goes from +1 to 0. It gains an electron and is reduced, so it is the oxidising agent. So option 3 is WRONG.

Step 6: Check option (4):
Ag(s) is the product formed by reduction. It is not oxidised in this reaction, so it is not the reducing agent. So option 4 is WRONG.

Final Answer:
Nickel metal loses electrons, so Ni(s) is the reducing agent. This is option 2.\[ \boxed{\mathrm{Ni(s)}} \]