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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 : $E^{\large\circ}(Cr_2O_7^{2-},H^+,Cr^{3+}|Pt)=1.33V;E^{\large\circ}(Fe^{3+},Fe^{2+}|Pt)=0.77V$.Reducing $[Cr^{3+}]$ to 0.50M,keeping all other concentrations at unity,the emf of the cell would be changed by

$\begin{array}{1 1}5.92mV\\2.96mV\\-5.92mV\\-2.96mV\end{array} $

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Answer : $5.92mV$
We have $E=E^{\large\circ}-\large\frac{RT}{6F}$$\ln (0.50)^2$
$\Rightarrow E^{\large\circ}-\big(\large\frac{2\times 0.059V}{6}\big)$$(-0.301)=E^{\large\circ}+0.0059V$
answered Jul 21, 2014 by sreemathi.v
 

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