Question

The emf of Daniell cell at $298K$ is $E_1$ $Zn\left|ZnS{O}_4(0.01M)\right|\left|CuS{O}_4(1.0M)\right|Cu$ When the concentration of $ZnS{O}_4$ is $1.0M$ and that of $CuS{O}_4$ is $0.01M$, 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 $I^{\text{st }}$
$\left[Z{n}^{2+}\right]=0.01M$
$\left[C{u}^{2+}\right]=1.0M$
$n=2$
$E_1=E_{\text{cell }}^{\circ }\frac{0.0591}{2}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
$E_1=E_{\text{cell }}^{\circ }-\frac{0.0591}{2}\log \frac{0.01}{1}$
$=E_{\text{cell }}^{\circ }+0.0591$
In case $I{I}^{nd}$
$\left[Z{n}^{2+}\right]=1.0M$
$\left[C{u}^{2+}\right]=0.01M$
$n=2$
$E_2=E_{\text{cell }}^{\circ }-\frac{0.0591}{2}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
$=E_{\text{cell }}^{\circ }-\frac{0.0591}{2}\log \frac{1}{0.01}$
$=E_{\text{cell }}^{\circ }-0.0591$
Thus, $E_1>E_2$