# Emf of the following cell

Question:

Emf of the following cell at $298 \mathrm{~K}$ in $\mathrm{V}$ is $\mathrm{x} \times 10^{-2} . \mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.1 \mathrm{M})\right|\left|\mathrm{Ag}^{+}(0.01 \mathrm{M})\right| \mathrm{Ag}$ The value of $x$ is__________ . (Rounded off to the nearest integer) [ Given$: E_{\mathrm{zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{~V} ; \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V} ; \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059$

Solution:

(147)

$\operatorname{Zn}(\mathrm{s})\left|\mathrm{Zn}^{+2}(0.1 \mathrm{M})\right| \mathrm{Ag}^{+}(0.01 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{s})$

$\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+} \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\mathrm{Zn}^{+2}$

$\mathrm{E}^{0}=0.80+0.76=1.56 ; \mathrm{Q}=\left\{\frac{\mathrm{Zn}^{2+}}{\left(\mathrm{Ag}^{+}\right)^{2}}\right\}$

$\mathrm{E}=\mathrm{E}^{0}-\frac{0.059}{\mathrm{n}} \log (\mathrm{Q})$

$\mathrm{E}=1.56-\frac{0.059}{2} \log \left[\frac{0.1}{(0.01)^{2}}\right]$

$\mathrm{E}=1.56-\frac{0.059}{2} \log \left[(10)^{3}\right]$

$\mathrm{E}=1.4715=147.15 \times 10^{-2}$ volt

$=x \times 10^{-2}$

$\mathrm{X}=147.15 \simeq 147 \mathrm{Ans}$