Question

The emf of Daniell cell at $298 K$ is $E_{1}$ $Zn \left| ZnSO _{4}(0.01 M )\right|\left| CuSO _{4}(1.0 M )\right| Cu$ When the concentration of $ZnSO _{4}$ is $1.0 M$ and that of $CuSO _{4}$ is $0.01 M$, the emf changed to $E_{2}$. What is the relation between $E_{1}$ and $E_{2}$ ?

• A. $E_1 = E_2$
• B. $E_2 = 0 \ne E_2$
• C. $E_1 > E_2$
• D. $E_1 < E_2$

Answer 1

Answer: (C)

In case $I ^{\text {st }}$
$\left[ Zn ^{2+}\right] =0.01 \,M $
$\left[ Cu ^{2+}\right] =1.0 \,M $
$n =2$
$E_{1} =E_{\text {cell }}^{\circ} \frac{0.0591}{2} \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]} $
$E_{1} =E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{0.01}{1} $
$=E_{\text {cell }}^{\circ}+0.0591$
In case $II ^{ nd }$
$\left[ Zn ^{2+}\right] =1.0\, M $
$\left[ Cu ^{2+}\right] =0.01 \,M $
$ n =2 $
$ E_{2} =E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]} $
$=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{1}{0.01} $
$=E_{\text {cell }}^{\circ}-0.0591$
Thus, $E_{1} >E_{2} $