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

  • A
    $0.806$
  • B
    $0.532$
  • C
    $0.734$
  • D
    $0.756$

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