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Assertion : For the Daniel cell, $\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}$ with $E_{\text {cell }}=1.1 \mathrm{~V}$, the application of opposite potential greater than $1.1 \mathrm{~V}$ results into flow of electron from cathode to anode.
Reason : $\mathrm{Zn}$ is deposited at anode and $\mathrm{Cu}$ is dissolved at cathode.
Options:
Reason : $\mathrm{Zn}$ is deposited at anode and $\mathrm{Cu}$ is dissolved at cathode.
Solution:
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Verified Answer
The correct answer is:
If both assertion and reason are true but reason is not the correct explanation of the assertion
If the opposing potential becomes slightly larger than that of the cell, the direction of current flow is reversed, and so is the cell reaction. Now zine ions are converted to zine at one electrode and $\mathrm{Cu}$ is converted into copper ion and the overall cell reaction becomes
$$
\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}
$$
$$
\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}
$$
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