Q. For the redox reaction, $Zn\ (s) + \ {Cu}^{2 +}\ (0.1\ M) \rightarrow$ $\ {Zn}^{2 +}\ (1\ M)\ + \ Cu\ (s)$ taking place in a cell, $E_{cell}^{{^\circ}}$ is 1.10 V. $E_{cell}^{{^\circ}}$ for the cell will be $\left( 2.303\ \frac{RT}{F} = 0.0591 \right)$
A) 2.14 V
B) 1.80 V
C) 1.07 V
D) 0.82 V
Explanation
(c)
$E_{cell}^{{^\circ}} = \ E_{cell}^{{^\circ}}\ - \ \frac{0.0591}{n}\log\mathcal{Q}$
${Cu}^{2 +}\ + \ Zn\ \rightarrow \ {Zn}^{2 +} + \ Cu$
0.1 M 1 M
$Q\ = \frac{\left\lbrack {Zn}^{2 +} \right\rbrack}{\left\lbrack {Cu}^{2 +} \right\rbrack} = \frac{1}{0.1}\ = \ 10$
$E_{cell}\ = \ 1.10\ –\ \frac{0.0591}{2}\log 10$
= 1.10 – 0.0295
= 1.0705 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.