Q32 / 160 · WBJEE 2015 · Chemistry
At temperature of $298 \mathrm{K}$, the emf of the following electrochemical cell. $\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+*}(0.1 \mathrm{M}) \| \mathrm{Zn}^{2 +}(0.1 \mathrm{M})\right| \mathrm{Zn}(\mathrm{s})$ will be (Given, $E^{\prime \prime}$ call $=-1.562 \mathrm{V}$ )
- Correct answerA$-1.532 \mathrm{V}$
- B$-1.503 V$
- C$1.532 \mathrm{V}$
- D$-3.06 V$
Solution
From the given cell, the cell reaction is
$2 \mathrm{Ag}(s)+Z \mathrm{n}^{2+}(01 \mathrm{M}) \longrightarrow$ $2 \mathrm{Ag}^{+}(0.1 \mathrm{M})+\mathrm{Zn}(s)$
The Nemst equation is,
$E_{\text {cel }}=E_{\text {cel }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Ag}^{+}\right]^{2}}{\left.\mid \mathrm{Zn}^{2+}\right]}$
or, $E_{\text {cal }}=(-1.562)-\frac{(0.0591)}{2} \log \frac{(0.1)^{2}}{(0.1)}$
$\begin{aligned} \text { Or, } E_{\text {cel }} &=(-1.562)-\frac{0.0591}{2} \log 10^{-1} \\ &=-1.562+\frac{0.0591}{2} \\ &=-1.562+0.02955 \\ &=-1.532 \mathrm{V} \end{aligned}$
Hence, the correct answer is option: 'A'.