Under which of the following conditions \(E\) value of the cell, for the cell reaction given is maximum?…
Under which of the following conditions \(E\) value of the cell, for the cell reaction given is maximum?
\(\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightleftharpoons \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q)\)
$\begin{aligned}
\frac{2.303 R T}{F} \text { at } 298 K &= 0.059 V, \\
E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ} &= -0.76 V, \\
E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ} &= +0.34 V
\end{aligned}$
\(C_1=0.1 \mathrm{M}, C_2=0.01 \mathrm{M}\)
\(C_1=0.01 \mathrm{M}, C_2=0.1 \mathrm{M}\)
\(C_1=0.1 \mathrm{M}, C_2=0.2 \mathrm{M}\)
\(C_1=0.2 \mathrm{M}, C_2=0.1 \mathrm{M}\)
Solution
From Nernst equation,
$\begin{aligned}
E & =E^{\circ}-\frac{2.303 R T}{n F} \log Q \\
\Rightarrow \quad E & =E^{\circ}-\frac{2.303 R T}{n F} \log \left(\frac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}}\right) \\
E^{\circ}_{\text {cell }} & =E^{\circ}_{C}-E^{\circ}_{A}=0.34-(-0.76) \mathrm{V}=1.1 \mathrm{~V} \\
E & =1.1-\frac{0.059}{n} \log \frac{C_{2}}{C_{1}} \quad\left(\begin{array}{l}
\mathrm{Zn}^{2+}=C_{2} \\
\mathrm{Cu}^{2+}=C_{1}
\end{array}\right)
\end{aligned}$
By analysing from the above equation, $E$ value of the cell will be maximum when, $\log \frac{C_{2}}{C_{1}}$ would come out to be minimum, when $\log \frac{C_{2}}{C_{1}}$ value would be minimum then, $\log \frac{0.01}{0.1}$=$\log 10^{-1}$ (minimum).
Thus, option (1) is correct.