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Q. What is the electrode potential of $Fe^{3+}/ Fe$ electrode in which concentration of $Fe^{3+}$ ions is$ 0.1\, M$. Given $E^\circ Fe^{3+}/Fe = + 0.771 \,V$

Electrochemistry

Solution:

$Fe^{3+} + 3e^- \rightarrow Fe$
$E = E^0 - \frac{0.059}{n}\,log \frac{[Fe]}{[Fe^{3+}]}$
$ = + 0.771 - \frac{0.059}{n} \,log \frac{1}{0.1} $
$ = + 0.75\,V$