Question

Under which of the following conditions $E$ value of the cell for the cell reaction given is maximum ?
$\mathrm{Zn}{}_{(\mathrm{s})}+\underset{C_1}{\mathrm{Cu}{}^{2+}{}_{(\mathrm{aq})}}\rightleftharpoons \mathrm{Cu}{}_{(\mathrm{s})}+\underset{C_2}{\mathrm{Zn}{}^{2+}{}_{(\mathrm{aq})}}$
$\left(\frac{2.303RT}{F}at298K=0.059V,E_{Z{n}^{2+}Zn}^0=-0.76V,E_{C{u}^{2+}Cu}^0=+0.34V\right)$

• A. $\mathrm{C}{}_1={}_0\cdot 1\mathrm{M},\mathrm{C}{}_2=0.01\mathrm{M}$
• B. $\mathrm{C}{}_1=0.01\mathrm{M},\mathrm{C}{}_2=0.1\mathrm{M}$
• C. $\mathrm{C}{}_1=0.1\mathrm{M},\mathrm{C}{}_2=0.2\mathrm{M}$
• D. $\mathrm{C}{}_1={}_0\cdot 2\mathrm{M},\mathrm{C}{}_2={}_0\cdot 1\mathrm{M}$

Answer 1

Answer: (A)

From Nernst equation,

$E=E^{\circ }-\frac{2.303RT}{nF}\log Q$

$\Rightarrow E=E^{\circ }-\frac{2.303RT}{nF}\log \left(\frac{Z{n}^{2+}}{C{u}^{2+}}\right)$

$E_{cell}^{\circ }=E_C^{\circ }-E_A^{\circ }=0.34-(-0.76)V=1.1V$

$E=1.1-\frac{0.059}{n}\log \frac{C_2}{C_1}$

$\left(\begin{array}{c}Z{n}^{2+}=C_2\\ C{u}^{2+}=C_1\end{array}\right)$

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).