Question

Under which of the following conditions $E$ value of the cell for the cell reaction given is maximum ?
$\ce{Zn_{(s)} + $\underset{{ C_1}}{\ce{ Cu^{2+}_{(aq)} }}$ <=> Cu_{(s)} + $\underset{{ C_2}}{\ce{Zn^{2+}_{(aq)} }}$ }$
$\left(\frac{2.303RT}{F} \; at \; 298 K = 0.059 V , E^{0}_{Zn^{2+}Zn} =-0.76V, E^{0}_{Cu^{2+}Cu} = +0.34 V\right)$

• A. $\ce{ C1 \,= \,0.1 M, C2\, =\, 0.01\,M }$
• B. $\ce{ C1\, =\, 0.01 M, C_2 = 0.1M }$
• C. $\ce{ C_1 =\, 0.1 M, C_2 =\, 0.2 M}$
• D. $\ce{ C1 \,= \,0.2M, C2 \,= \,0.1M }$

Answer 1

Answer: (A)

From Nernst equation,

$E =E^{\circ}-\frac{2.303 RT}{n F} \log \,Q $

$\Rightarrow E =E^{\circ}-\frac{2.303 R T}{n F} \log \left(\frac{ Zn ^{2+}}{ Cu ^{2+}}\right) $

$E_{ cell }^{\circ} =E_{C}^{\circ}-E_{A}^{\circ}=0.34-(-0.76) V =1.1 \,V $

$E= 1.1-\frac{0.059}{n} \log \frac{C_{2}}{C_{1}}$

$\begin{pmatrix} Zn ^{2+}=C_{2} \\ Cu ^{2+}=C_{1}\end{pmatrix}$

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).