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

Answer: (D)

$E_{c e l l}=E_{c e l l}^{\circ}+\frac{0.059}{n} \log \left[\frac{\text { product }}{\text { reactant }}\right]$
$S n^{2+} \rightarrow S n^{4+}+2 e^{-}$
$\therefore n=2$
Given $\left[S n^{2+}\right]=0.1\, M,\left[S n^{4+}\right]=0.01 \,M$
$E_{c e l l}=E_{c e l l}^{\circ}+\frac{0.059}{2} \log \left[\frac{S n^{4+}}{S n^{2+}}\right]$
$=E_{c e l l}^{\circ}+\frac{0.059}{2} \log \left[\frac{0.01}{0.1}\right]$
$=E_{c e l l}^{\circ}+\frac{0.059}{2} \log 0.1$
$=E_{c e l l}^{\circ}+\frac{0.059}{2} \times-1$
$=E_{c e l l}^{\circ}-\frac{0.059}{2}$