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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 cathode of the cell might be

$\begin{array}{1 1}Cr\\Fe\\Pt\\\text{diamond}\end{array} $

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Answer : Pt
The cathodic half-cell will be $Cr_2O_7^{2-},Cr^{3+},H^+|Pt$.Hence,Pt is cathode.
answered Jul 21, 2014 by sreemathi.v
 

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