A) \[{{E}_{0}}\]
B) \[{{E}_{0}}+0.059\]
C) \[{{E}_{0}}+\frac{0.059}{2}\]
D) \[{{E}_{0}}=\frac{0.059}{2}\]
Correct Answer: D
Solution :
Key Idea: \[{{E}_{cell}}=E_{cell}^{o}+\frac{0.059}{n}\log \left[ \frac{\text{product}}{\text{reactant}} \right]\] \[S{{n}^{2+}}\xrightarrow{{}}S{{n}^{4+}}+2{{e}^{-}}\] \[\therefore \] \[n=2\] Given \[[S{{n}^{2+}}]=0.1\,M,[S{{n}^{4+}}]=0.01\,M\] \[{{E}_{cell}}=E_{cell}^{o}+\frac{0.059}{2}\log \left[ \frac{S{{n}^{4+}}}{S{{n}^{2+}}} \right]\] \[=E_{cell}^{o}+\frac{0.059}{2}\log \left[ \frac{0.01}{0.1} \right]\] \[=E_{cell}^{o}+\frac{0.059}{2}\text{log}\,0.1\] \[=E_{cell}^{o}+\frac{0.059}{2}\times -1\] \[=E_{cell}^{o}-\frac{0.059}{2}\]
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