A) \[{{10}^{0.32/0.0591}}\]
B) \[{{10}^{0.32/0.0295}}\]
C) \[{{10}^{0.26/0.0295}}\]
D) \[{{10}^{0.32/0.295}}\]
Correct Answer: B
Solution :
For celI \[Zn|Z{{n}^{2+}}(a=0.1\,M)||F{{e}^{2+}}(a=0.01M)|Fe\] The half-cell reactions are (i)\[Zn(S)\xrightarrow{{}}Z{{n}^{2+}}(aq)+2{{e}^{-}}\] (ii) \[F{{e}^{2+}}(aq)+2{{e}^{-}}\xrightarrow{{}}Fe(S)\] \[\overline{\underline{Zn(s)+F{{e}^{2+}}(aq)\xrightarrow{{}}Z{{n}^{2}}(aq)+Fe(s)}}\] On applying Nernst equation, \[E=E_{cell}^{o}-\frac{0.0591}{n}{{\log }_{10}}\frac{[Z{{n}^{2+}}]}{[F{{e}^{2+}}]}\] \[0.2905=E_{cell}^{o}-\frac{0.0591}{n}{{\log }_{10}}\frac{0.1}{0.01}\] \[0.2905=E_{cell}^{o}-0.0295\times {{\log }_{10}}10\] \[0.2905=E_{cell}^{o}-0.0295\times 1\] \[E_{cell}^{o}=0.2905+0.0295\] \[=0.32\,V\] At equilibrium, \[({{E}_{cell}}=0)\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}{{\log }_{10}}{{K}_{C}}\] \[\therefore \] \[0=E_{cell}^{o}-\frac{0.0591}{n}{{\log }_{10}}{{K}_{C}}\] or \[E_{cell}^{o}=\frac{0.0591}{2}{{\log }_{10}}{{K}_{C}}\] \[0.32=\frac{0.0591}{2}{{\log }_{10}}{{K}_{C}}\] or \[{{\log }_{10}}{{K}_{C}}=\frac{0.32}{0.02955}\] or \[{{K}_{C}}={{10}^{0.32/0.02955}}\]
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