Question

The e.m.f. of a Daniell cell at $298\, K$ is $E _{1}$.
$Zn / SO _{4}(0.01 M ) \| CuSO _{4}(1.0 M ) / Cu$
When the concentration of $ZnSO _{4}$ is $1.0\, M$ and that of $CuSO _{4}$ is $0.01\, M$, 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. $E_2 = 0 \neq E_1$
• D. $E_1 = E_2 $

Answer 1

Answer: (B)

Using the relation, $E _{\text {cell }}= E _{\text {cell }}^{0}-\frac{0.0591}{ n } \log \frac{\text { [anode] }}{\text { [cathode] }}$ $= E _{\text {cell }}^{0}-\frac{0.0591}{ n } \log \frac{\left[ Zn ^{2+}\right]}{\left[ Cu ^{2+}\right]}$
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, $E _{1}> E _{2}$