Question

The e.m.f. of a Daniell cell at $298K$ is $E_1$.
$Zn/S{O}_4(0.01M)\Vert CuS{O}_4(1.0M)/Cu$
When the concentration of $ZnS{O}_4$ is $1.0M$ and that of $CuS{O}_4$ is $0.01M$, the e.m.f. is changed to $E_2$. What is the relationship between $E_1$ and $E_2$ :

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

Answer 1

Answer: (B)

Using the relation, $E_{\text{cell }}=E_{\text{cell }}^0-\frac{0.0591}{n}\log \frac{\text{ [anode] }}{\text{ [cathode] }}$ $=E_{\text{cell }}^0-\frac{0.0591}{n}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
Substituting the given values in two cases.
$E_1=E^0-\frac{0.0591}{2}\log \frac{0.01}{1.0}$
$=E^0-\frac{0.0591}{2}\log 10^{-2}$
$=E^0+\frac{0.0591}{2}\times 2$ or $\left(E^0+0.0591\right)V$
$E_2=E^0-\frac{0.0591}{2}\log \frac{1}{0.01}$
$=E^0-\frac{0.0591}{2}\log 10^2$
$=E^0-\frac{2\times 0.0591}{2}$ or $\left(E^0-0.0591\right)V$
Thus, $E_1>E_2$