The e.m.f. of a Daniell cell at $298 \mathrm{~K}$ is $\mathrm{E}_{1}$. Zn…

The e.m.f. of a Daniell cell at $298 \mathrm{~K}$ is $\mathrm{E}_{1}$. Zn $\left|\begin{array}{c}\mathrm{ZnSO}_{4} \\ (0.01 \mathrm{M})\end{array}ight|\left|\begin{array}{c}\mathrm{CuSO}_{4} \\ (1.0 \mathrm{M})\end{array}ight| \mathrm{Cu}$
When the concentration of $\mathrm{ZnSO}_{4}$ is $1.0 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ is $0.01 \mathrm{M}$, the e.m.f. changed to $E_{2}$. What is the relationship between $E_{1}$ and $E_{2}$ ?
  1. $\quad E_{2}=0 \neq E_{1}$
  2. $E_{1}>E_{2}$
  3. $E_{1} < E_{2}$
  4. $\quad E_{1}=E_{2}$

Solution

Cell reaction is, $\mathrm{Zn}+\mathrm{Cu}^{2+} ightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ell \mathrm{n} \frac{\left[\mathrm{Zn}^{2+}ight]}{\left[\mathrm{Cu}^{2+}ight]}$

Greater the factor $\left[\frac{\left(\mathrm{Zn}^{2+}ight)}{\left(\mathrm{Cu}^{2+}ight)}ight]$, less is the EMF Hence $E_{1}>E_{2}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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