Electrochemistry - Result Question 29
####31. In the electrochemical cell :- $[2003,2017]$ $Zn|ZnSO_4(0.01 M)||CuSO_4(1.0 M)| Cu$, the emf of this Daniel cell is $E_1$. When the concentration of $ZnSO_4$ is changed to $1.0 M$ and that of $CuSO_4$ 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, $\frac{R T}{F}=0.059$ )
(a) $E_1<E_2$
(b) $E_1>E_2$
(c) $E_2=0 \neq E_1$
(d) $E_1=E_2$
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Solution:
- (b) For cell,
$Zn|ZnSO_4(0.01 M) | CuSO_4(1.0 M)| Cu$
$E _{\text{cell }}=E _{\text{cell }}^{\circ}-\frac{2.303 R T}{n F} \frac{\log [Zn^{2+}]}{[Cu^{2+}]}$
$\therefore E_1=E _{\text{cell }}^{\circ}-\frac{2.303 R T}{2 \times F} \times \log \frac{(0.01)}{1}$
When concentrations are changed for $ZnSO_4$ and $CuSO_4$, we can write
$E_2=E _{\text{cell }}^{\circ}-\frac{2.303 R T}{2 F} \times \log \frac{1}{0.01}$
$\therefore \quad E_1>E_2$
