Statement 1: For the Daniell cell Zn | Zn2+ || Cu2+ | Cu with Ecel| = 1.1 V, the application of opposite potential greater than 1.1 V results into flow of electron from Cu to Zn. Statement 2: Zn is deposited at anode and Cu is deposited at cathode.

Question

Statement 1: For the Daniell cell Zn | Zn2+ || Cu2+ | Cu with Ecel| = 1.1 V, the application of opposite potential greater than 1.1 V results into flow of electron from Cu to Zn. Statement 2: Zn is deposited at anode and Cu is deposited at cathode. (A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct explanation of the statement 1. (B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct explanation of the statement 1. (C) If Statement 1 is true but Statement 2 is false. (D) If Statement 1 is false but Statement 2 is true

Tardigrade Admin · Accepted Answer

Zn is -ve electrode and Cu is +ve electrode in Daniel cell.

Q. Statement 1: For the Daniell cell $Z n ∣ Z n^{2 +} ∣∣ C u^{2 +} ∣$ Cu with $E_{ce l ∣} = 1.1 V,$ the application of opposite potential greater than 1.1 V results into flow of electron from Cu to Zn.
Statement 2: Zn is deposited at anode and Cu is deposited at cathode.

If both Statement 1 and Statement 2 are true and the statement 2 is the correct explanation of the statement 1.

If both Statement 1 and Statement 2 are true but statement 2 is not the correct explanation of the statement 1.

If Statement 1 is true but Statement 2 is false.

If Statement 1 is false but Statement 2 is true

$Z n$ is -ve electrode and $C u$ is +ve electrode in Daniel cell.

Q. Statement 1: For the Daniell cell Zn∣Zn2+∣∣Cu2+∣ Cu with Ecel∣​=1.1V, the application of opposite potential greater than 1.1 V results into flow of electron from Cu to Zn. Statement 2: Zn is deposited at anode and Cu is deposited at cathode.