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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 : $E^{\large\circ}(Cr_2O_7^{2-},H^+,Cr^{3+}|Pt)=1.33V;E^{\large\circ}(Fe^{3+},Fe^{2+}|Pt)=0.77V$.If $[Cr_2O_7^{2-}]=1.0M,[Cr^{3+}]=0.5M,[Fe^{2+}]=0.5M,[Fe^{3+}]=2.0M$ and $[H_3O^+]=2.0M$,then the cell emf at 298K would be

$\begin{array}{1 1}0.678V\\-0.678V\\0.572V\\0.548V\end{array} $

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Answer : $0.572V$
$E=E^{\large\circ}-\large\frac{RT}{6F}$$\ln \large\frac{[Cr^{3+}]^2[Fe^{3+}]^6}{[Cr_2O_7^{2-}][H^+]^{14}[Fe^{2+}]^6}$$=E^{\large\circ}-\big(\large\frac{0.059V}{6}\big)$$\log \large\frac{(0.5)^2(2.0)^6}{(1.0)(2.0)^{14}(0.5)^6}$
$\Rightarrow (1.33V-0.77V)-\big(\large\frac{0.059V}{6}\big)$$\log \big(\large\frac{1}{2^4}\big)=$$0.56V+4\times \big(\large\frac{0.059V}{6}\big)$$(0.301)$
$\Rightarrow 0.56V+0.012V=0.572$V
answered Jul 21, 2014 by sreemathi.v
 

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