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A galvanic cell which makes use of the following reaction at 298K. $Cr_2O_7^{2-}+14H^++6Fe^{2+}\rightarrow 2Cr^{3+}+6Fe^{3+}+7H_2O$.Given : $E^{\large\circ}(Cr_2O_7^{2-},H^+,Cr^{3+}|Pt)=1.33V;E^{\large\circ}(Fe^{3+},Fe^{2+}|Pt)=0.77V$.The cell emf could be increased above the standard emf by

$\begin{array}{1 1}\text{increasing }[Cr^{3+}]\\\text{increasing }[Fe^{3+}]\\\text{decreasing }[Cr_2O_7^{2-}]\\\text{decreasing the }pH\end{array} $

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Answer : decreasing the pH
Nernst equation for the reaction $Cr_2O_7^{2-}+14H^++6Fe^{2+}\rightarrow 2Cr^{3+}+6Fe^{3+}+7H_2O$ is
$E=E^{\large\circ}-\large\frac{RT}{6F}$$\ln \large\frac{[Cr^{3+}]^2[Fe^{3+}]^6}{[Cr_2O_7^{2-}][H^+]^{14}[Fe^{2+}]^6}$
Decreasing pH (i.e.increasing $H^+$) will make the logarithm term more negative.Hence,$E > E^{\large\circ}$
answered Jul 21, 2014 by sreemathi.v
 

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