Question

Assertion :For the Daniel cell, $Zn|Z{n}^{2+}||C{u}^{2+}|Cu$ with $E_{cell}=1.1V$, the application of opposite potential greater than $1.1V$ results into flow of electron from cathode to anode. Reason: $Zn$ is deposited at anode, and $Cu$ is deposited at cathode.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

In a Daniel cell, $Zn|Z{n}^{2+}\parallel C{u}^{2+}|Cu,E_{cell}=1.1V$

The oxidation half cell is

$Zn\to Z{n}^{2+}+2e^{-}$

Reduction half cell is:

$C{u}^{2+}+2e^{-}\to Cu$

So,

$Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$

Thus, here $Zn$ is oxidized and deposited at the anode, and $Cu$ is reduced and deposited at the cathode.

If the opposite potential is greater than $1.1V$, then electrons flow get reversed. Zn is deposited at the anode, and Cu is dissolved at the cathode.