$\mathrm{Zn}+\mathrm{Cu}^{2+}(a q) ightleftharpoons \mathrm{Cu}+\mathrm{Zn}^{2+}(a q)$ Reaction quotient is…

$\mathrm{Zn}+\mathrm{Cu}^{2+}(a q) ightleftharpoons \mathrm{Cu}+\mathrm{Zn}^{2+}(a q)$
Reaction quotient is $\mathrm{Q}=\frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}$


Variation of $E_{\text {cell }}$ with $\log Q$ is of the type with $\mathrm{OA}=1.10 \mathrm{~V} . \mathrm{E}_{\mathrm{cell}}$ will be $1.1591 \mathrm{~V}$ when :
  1. $\frac{\left[\mathrm{Cu}^{2+}ight]}{\left[\mathrm{Zn}^{2+}ight]}=0.01$
  2. $\frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}=0.01$
  3. $\frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}=0.1$
  4. $\frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}=1$

Solution

$\mathrm{E}_{\mathrm{cel}}=\mathrm{E}_{\text {cell }}^{o}-\frac{0.0591}{2} \log \left[\frac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}}ight]$
from graph $\mathrm{OA}=\mathrm{E}_{\mathrm{cell}}^{\mathrm{o}}=1.01 \mathrm{~V}$
$\mathrm{E}_{\mathrm{cell}}=1.159 \mathrm{~V}$
Substituting all values
$\frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}=10^{-2} \mathrm{~M}$
$=0.01$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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