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