A) 0.101
B) 0.924
C) 0.120
D) None of these
Correct Answer: A
Solution :
The half-cell reactions of the given cell are as At anode \[\frac{1}{2}{{H}_{2}}\xrightarrow{{}}{{H}^{+}}+{{e}^{-}};\,E_{1}^{o}=-0.00\,\,V\] At cathode \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}};\,E_{2}^{o}=0.771\,\,V\] ____________________________________ Net reaction \[F{{e}^{3+}}+\frac{1}{2}{{H}_{2}}\xrightarrow{{}}F{{e}^{2+}}+{{H}^{+}};E_{cell}^{o}\] \[=0.771\,\,V-0.00\,V=0.771\,\,V\] From Nernst equation, \[{{E}_{cell}}=E_{cell}^{o}=-\frac{0.0591}{n}\log \frac{[F{{e}^{2+}}]\,[{{H}^{+}}]}{[F{{e}^{3+}}]+pH_{2}^{1/2}}\] \[0.830=0.771-\frac{0.0591}{1}\log \left[ \frac{F{{e}^{2+}}}{F{{e}^{3+}}} \right]\] \[-\frac{0.059}{0.0591}=\log \frac{[F{{e}^{2+]}}}{[F{{e}^{3+}}]}\] \[\therefore \] \[\frac{[F{{e}^{2+]}}}{[F{{e}^{3+}}]}=\]antilog \[(-0.998)=0.101\]
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