Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App