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Question

Chemistry Question on Nernst Equation

What is potential of platinum wire dipped into a solution of 0.1 M in Sn2+S{{n}^{2+}} and 0.01 M in Sn4+?S{{n}^{4+}}?

A

E0{{E}_{0}}

B

E0+0.059{{E}_{0}}+0.059

C

E0+0.0592{{E}_{0}}+\frac{0.059}{2}

D

E0=0.0592{{E}_{0}}=\frac{0.059}{2}

Answer

E0=0.0592{{E}_{0}}=\frac{0.059}{2}

Explanation

Solution

Ecell=Ecello+0.059nlog⁡[productreactant]{{E}_{cell}}=E_{cell}^{o}+\frac{0.059}{n}\log \left[ \frac{\text{product}}{\text{reactant}} \right] Sn2+→Sn4++2e−S{{n}^{2+}}\xrightarrow{{}}S{{n}^{4+}}+2{{e}^{-}}
∴\therefore n=2n=2 Given [Sn2+]=0.1 M,[Sn4+]=0.01 M[S{{n}^{2+}}]=0.1\,M,[S{{n}^{4+}}]=0.01\,M
Ecell=Ecello+0.0592log⁡[Sn4+Sn2+]{{E}_{cell}}=E_{cell}^{o}+\frac{0.059}{2}\log \left[ \frac{S{{n}^{4+}}}{S{{n}^{2+}}} \right]
=Ecello+0.0592log⁡[0.010.1]=E_{cell}^{o}+\frac{0.059}{2}\log \left[ \frac{0.01}{0.1} \right]
=Ecello+0.0592log 0.1=E_{cell}^{o}+\frac{0.059}{2}\text{log}\,0.1
=Ecello+0.0592×−1=E_{cell}^{o}+\frac{0.059}{2}\times -1
=Ecello−0.0592=E_{cell}^{o}-\frac{0.059}{2}