A) \[E{}^\circ \]
B) \[E{}^\circ +0.059\]
C) \[E{}^\circ +\frac{0.059}{2}\]
D) \[E{}^\circ -0.059\]
E) \[E{}^\circ -2\times 0.59\]
Correct Answer: C
Solution :
\[S{{n}^{2+}}\xrightarrow{{}}S{{n}^{4+}}+2{{e}^{-}}\] \[{{E}_{cell}}=E{}^\circ -\frac{0.059}{n}\log \frac{[S{{n}^{4+}}]}{[S{{n}^{2+}}]}\] \[=E{}^\circ -\frac{0.059}{2}\log \left[ \frac{0.01}{0.1} \right]\] \[=E{}^\circ -\frac{0.059}{2}\]You need to login to perform this action.
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