Question

The emf of Daniell cell at $298K$ is $E_{1}Zn|ZnS{O}_4\left(0.01M\right)||CuS{O}_4\left(1.0M\right)|Cu$
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. $E_2<E_1$

Answer 1

The correct option is D $E_2<E_1$

Using the relation, $E_{cell}=E_{cell}^0-\frac{0.0591}{n}\log \frac{\left[anode\right]}{\left[cathode\right]}$

$=E_{cell}^0-\frac{0.0591}{n}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

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 $(E^0+0.0591)V$

$E_2=E^0-\frac{0.0591}{2}\log \frac{1}{0.01}$

$=E^0-\frac{0.0591}{2}\log {10}^2$

$=E^0-\frac{2\times 0.0591}{2}$ or $(E^0-0.0591)V$

Thus, $E_1>E_2$