$A$ solution of $Fe^{2+}$ is titrated potentiometrically using $Ce^{4+}$ solution. When $80 \%$ of $Fe^{2+}$ is titrated,the $EMF$ of the system in $V$ is (Given,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \ V$ and $Fe^{2+} + Ce^{4+} \longrightarrow Fe^{3+} + Ce^{3+}$)
$(\log 2 = 0.3, \log 3 = 0.5, \log 4 = 0.6)$
$(\log 2 = 0.3, \log 3 = 0.5, \log 4 = 0.6)$
- A$0.806$
- B$0.532$
- C$0.734$
- D$0.756$
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For the given cell $Cu(s) | Cu^{2+}(C_1 \ M) || Cu^{2+}(C_2 \ M) | Cu(s)$,the change in Gibbs energy $(\Delta G)$ is negative,if:
The cell potential for the following cell notation is approximately
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$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)
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