Question:easy

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?
\(\mathrm{Ni(s) + 2Ag^{+}(aq) \to Ni^{2+}(aq) + 2Ag(s)}\)

Show Hint

Anode means oxidation, so look for the half reaction in which electrons are lost.
Updated On: Oct 1, 2026
  • \(\mathrm{Ni(s) \to Ni^{2+}(aq) + 2e^{-}}\)
  • \(\mathrm{Ni^{2+}(aq) + 2e^{-} \to Ni(s)}\)
  • \(\mathrm{Ag^{+}(aq) + e^{-} \to Ag(s)}\)
  • \(\mathrm{Ag(s) \to Ag^{+}(aq) + e^{-}}\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Use the cell notation:
In the cell $Ni(s)|Ni^{2+}(aq)||Ag^{+}(aq)|Ag$, the left electrode is the anode and the right one is the cathode. The left electrode is nickel.

Step 2: Write the anode process:
At the left side Ni(s) dips in $Ni^{2+}$. The reaction there is $Ni(s) \to Ni^{2+}(aq) + 2e^{-}$.

Step 3: Confirm with the overall reaction:
Add the cathode reaction $2Ag^{+} + 2e^{-} \to 2Ag$ to this. The electrons cancel and give the overall reaction.

Step 4: Eliminate the rest:
Options 2 and 3 are reductions, so they are cathode reactions. Option 4 is oxidation of silver, which runs opposite to the given reaction. Only option 1 remains.

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