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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 : $E^{\large\circ}(Cr_2O_7^{2-},H^+,Cr^{3+}|Pt)=1.33V;E^{\large\circ}(Fe^{3+},Fe^{2+}|Pt)=0.77V$.Increasing $[H_3O^{+}]$ to 2.0M, keeping all other concentrations at unity,will change the cell emf by

$\begin{array}{1 1}0.0414V\\-0.0414V\\0.0207V\\-0.0207V\end{array} $

1 Answer

Answer : $0.0414V$
We have
$E=E^{\large\circ}-\large\frac{RT}{6F}$$\ln \large\frac{1}{(2)^{14}}$$=E^{\large\circ}+\big(\large\frac{14\times 0.059V}{6}\big)$$(0.301)=0.0414V$
answered Jul 21, 2014 by sreemathi.v
 

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