Question

In the electrochemical cell:
$Zn|ZnS{O}_4\left(0.01M\right)||CuS{O}_4\left(1.0M\right)Cu$, the emf of this Daniell cell is $E_1$. When the concentration of $ZnS{O}_4$ is changed to 1.0 M and that of $CuS{O}_4$ changed to 0.01 M, the emf changes to $E_2$. From the followings, which one is the relationship between $E_1$ and $E_2$? (Given, RT/F = 0.059)

• A. $E_1<E_2$
• B. $E_1>E_2$
• C. $E_2=-E_1$
• D. $E_1=E_2$

Answer 1

The correct option is B $E_1>E_2$

$E_{cell}=E_{cell}^0-\frac{0.059}{n}log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
$E_1=E^0-\frac{0.059}{2}log\frac{0.01}{0}$
$E_1=E^{\circ }-\frac{0.059}{2}(-2)=E^{\circ }+0.059$
$E_2=E^{\circ }-\frac{0.059}{2}log\frac{1}{0.01}=E^{\circ }-0.059$
Hence, $E_1>E_2$