Question

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

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