Q25 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Chemistry

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 $\left.\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}\right)$ $(log 2=0.3, \log 3=0.5, \log 4=0.6)$

  • A
    0.806
  • B
    0.532
  • Correct answer
    C
    0.734
  • D
    0.756

Solution

$\because E=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[\mathrm{Fe}^{3+}\right]}{\left[\mathrm{Fe}^{2+}\right]}$ 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 $\mathrm{Fe}^{2+}$ $\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-}$ $\begin{aligned} \text { Initial } & =80 & & 20 \\ \text { Final } & =20 & & 80\end{aligned}$ $\because \quad E=0.77-0.059 \log \frac{80}{20}=0.77-0.05 \log 4$ $\therefore \quad E=0.77-0.0592 \log 2$ $=0.77-0.06=0.77-0.0354=0.734$

Hence, the correct answer is option: 'C'.

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