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A solution of $\mathrm{Fe}^{2+}$ is titrated potentiometrically using $\mathrm{Ce}^{4+}$ solution. When $80 \% \mathrm{Fe}^{2+}$ is titrated, the EMF of the system in $V$ is (Given, $E^{\circ} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$ and $\left.\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}\right)$
$(log 2=0.3, \log 3=0.5, \log 4=0.6)$
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$(log 2=0.3, \log 3=0.5, \log 4=0.6)$
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Verified Answer
The correct answer is:
0.734
$\because E=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Fe}^{3+}\right]}{\left[\mathrm{Fe}^{2+}\right]}$
Given, $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$
and, $\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}$
$\therefore$ For $80 \%$ of $\mathrm{Fe}^{2+}$
$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-}$
$\begin{aligned} \text { Initial } & =80 & & 20 \\ \text { Final } & =20 & & 80\end{aligned}$
$\because \quad E=0.77-0.059 \log \frac{80}{20}=0.77-0.05 \log 4$
$\therefore \quad E=0.77-0.0592 \log 2$
$=0.77-0.06=0.77-0.0354=0.734$
Given, $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$
and, $\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}$
$\therefore$ For $80 \%$ of $\mathrm{Fe}^{2+}$
$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-}$
$\begin{aligned} \text { Initial } & =80 & & 20 \\ \text { Final } & =20 & & 80\end{aligned}$
$\because \quad E=0.77-0.059 \log \frac{80}{20}=0.77-0.05 \log 4$
$\therefore \quad E=0.77-0.0592 \log 2$
$=0.77-0.06=0.77-0.0354=0.734$
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