Question

E value for the cell reaction
Cul Cu2+ (0.001 M) || Cu2+(0.1 M) I Cu is :-
(1) - PT In(0.01)
(3) + PT In(0.01)
(2) - PT In(0.1)
(9) RT 1n(0.1)
APMT Pro 2011​

Answer 1

E value for the cell reaction= $-\frac {RT} {F} ln \; {0.01}$

Explanation:

Cell Reaction

Cathode, $Cu^{2+}(0.1 M) + 2e^- \implies Cu$

Anode, $Cu \implies Cu^{2+} (0.001 M) + 2e^-$

Reaction became, $Cu^{2+}(0.1 M) \implies Cu^{2+}(0.001M)$

The Nernst equation for the cell potential is given below

$E = E^o - \frac{RT} {nF} ln \frac{[Oxidation]}{[reduction]}$

where $E^o = 0$ (concentration cell)

$= 0 - \frac{RT}{F}ln \frac{0.001}{0.1}$

$= -\frac {RT} {F} ln \frac{0.01}{1}$

Thus $-\frac {RT} {F} ln \; {0.01}$ is correct  answer.

Answer 2

Explanation:

so the correct option is -RT/F ln (0.1)