Assertion `(A):` For a Daniell cell `:`
`Zn|Zn^(2+)||Cu^(2+)|Cu` with `E_(cell)=1.1V`, the application of opposite potential greater than `1.1V` results into the flow of electron from cathod to anode. Reason : Zn is deposited at anode, and Cu is deposited at cathode.

Options

A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.
B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.
C. If `(A)` is correct, but `(R)` is incorrect.
D. If `(A)` is incorrect, `(R)` is correct.

Answer

b

Explanation

Both `(A)` and `(R)` are correct but `(R)` is not the correct explanation of `(A)`. Refer Figure.
Correct explanation `:` On applying external voltage greater than `1.1V` in a Daniell cell, the current flows in the reverse direction `i.e.,` from `Zn` to `Cu(` cathode to anode `)` and electrons flow from `Cu` to `Zn` .
`Zn` is deposited at `Zn` electrode and `Cu` dissolves at `Cu` electrode. The reaction is `:`
`Zn^(2+)+Cu rarr Zn+Cu^(2+)`