Ask Questions, Get Answers

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

The $E_{cell}$ in which the reaction :$MnO_4^-+Fe^{+2}+H^+\rightarrow Mn^{+2}+Fe^{+3}+H_2O$ occurs is 0.59V at $25^{\large\circ}C$.The equilibrium constant for the reaction is

$\begin{array}{1 1}(a)\;50\\(b)\;10\\(c)\;10^{50}\\(d)\;10^5\end{array}$

Can you answer this question?

1 Answer

0 votes
For the given half-cell reaction
$-nFE^0_{cell}=-RTln K_{eq}$
$-5\times 96500\times 0.59=-2.303\times 8.314\times 298 \log K_{eq}$
$K_{eq}=7.8\times 10^{49}\approx 10^{50}$
Hence (c) is the correct answer.
answered Feb 17, 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