A variable, the opposite external potential (\(E_{ext}\)) is applied to the cell \(\mathrm{Zn \mid Zn^{2+}\ (1\ M) \parallel Cu^{2+}\ (1\ M) \mid Cu}\), of potential 1.1 V. When \(E_{ext} \lt 1.1\) V and \(E_{ext} \gt 1.1\) V, respectively electrons flow from
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- anode to cathode in both the cases.
- anode to cathode and cathode to anode.
- cathode to anode and anode to cathode.
- cathode to anode in both cases.
The Correct Option is B
Solution and Explanation
For \(E_{ext} \gt 1.1\) V the external source overcomes the cell and the reaction is reversed (electrolytic action): electrons flow from Cu to Zn, i.e. from the cathode to the anode of the original cell.
Answer: (2) anode to cathode and cathode to anode.
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