Question

What is potential of platinum wire dipped into a solution of 0.1 M in $ S{{n}^{2+}} $ and 0.01 M in $ S{{n}^{4+}}? $

• A. $ {{E}_{0}} $
• B. $ {{E}_{0}}+0.059 $
• C. $ {{E}_{0}}+\frac{0.059}{2} $
• D. $ {{E}_{0}}=\frac{0.059}{2} $

Answer 1

The Correct Option is D

$ {{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} $