The e.m.f. of a Deniell cell at $298 \mathrm{~K}$ is $\mathrm{E}_1$ $\mathrm{Zn}\left|\begin{array}{c}…
The e.m.f. of a Deniell cell at $298 \mathrm{~K}$ is $\mathrm{E}_1$
$\mathrm{Zn}\left|\begin{array}{c}
\mathrm{ZnSO}_4 \\
(0.01 \mathrm{M})
\end{array}\right|\left|\begin{array}{c}
\mathrm{CuSO}_4 \\
(0.01 \mathrm{M})
\end{array}\right| \mathrm{Cu}$
When the concentration of $\mathrm{ZnSO}_4$ in 1.0 $\mathrm{M}$ and that of $\mathrm{CuSO}_4$ is $0.01 \mathrm{M}$, the e.m.f. changed to $\mathrm{E}_2$. What is the relationship between $E_1$ and $E_2$ ?
$\mathrm{E}_1 > \mathrm{E}_2$
$\mathrm{E}_1 < \mathrm{E}_2$
$\mathrm{E}_1=\mathrm{E}_2$
$\mathrm{E}_2=0 \neq \mathrm{E}_1$
Solution
Cell reaction can be represented as:
$\mathrm{Zn}+\mathrm{Cu}^{2+} \rightarrow \mathrm{Ca}+\mathrm{Zn}^{2+}$
Applying in both cases,
$\mathrm{E}^0=\frac{-0.0591}{2} \log \frac{\mathrm{Zn}^{2+}}{\mathrm{C} \mu^{2+}}$
Thus, $\mathrm{E}_1 > \mathrm{E}_2$