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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 $[Fe^{3+}]$ to 0.50M keeping all other concentrations at unity,the emf of the cell will be changed by

$\begin{array}{1 1}-0.059V\\-0.0178V\\0.059V\\0.0178V\end{array} $

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1 Answer

Answer : 0.0178V
Substituting $[Fe^{3+}]=0.50M$ and all other species equal to 1M in Nernst equation,
$E=E^{\large\circ}-\large\frac{RT}{6F}$$\ln \large\frac{[Cr^{3+}]^2[Fe^{3+}]^6}{[Cr_2O_7^{2-}][H^+]^{14}[Fe^{2+}]^6}$
We get $E=E^{\large\circ}-\large\frac{RT}{6F}$$\ln (0.50)^6$
$\Rightarrow E^{\large\circ}-\big(\large\frac{6\times 0.059V}{6}\big)$$(-0.301)=E^{\large\circ}+0.0178$V
answered Jul 21, 2014 by sreemathi.v
 

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