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.

Which of the following is the anodic half cell reaction?
\(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\)

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

Hint:

Anode means oxidation, so look for the half reaction in which electrons are lost.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The anode is the electrode where oxidation takes place. Oxidation means loss of electrons, so electrons appear on the right side of an anodic half reaction.

Step 2: Key Formula or Approach:
Split \(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\) into two half reactions. Then pick the one in which electrons are lost.

Step 3: Check option (1):
Ni(s) becomes \(Ni^{2+}\) and releases 2 electrons. This is oxidation and it is also in the forward direction of the cell reaction. So option 1 is CORRECT.

Step 4: Check option (2):
This is \(Ni^{2+}\) gaining electrons. It is reduction, and it is the reverse of what happens in this cell. So option 2 is WRONG.

Step 5: Check option (3):
This is \(Ag^{+}\) gaining an electron. It is reduction, which takes place at the cathode. So option 3 is WRONG.

Step 6: Check option (4):
This is oxidation, but silver metal losing electrons is the reverse of the actual reaction. In this cell silver ions are reduced to silver metal. So option 4 is WRONG.

Final Answer:
The anodic half reaction is the oxidation of nickel, which is option 1.\[ \boxed{\mathrm{Ni(s) \to Ni^{2+}(aq) + 2e^{-}}} \]