Question

$E_{cell}^0$ for the reaction, $Fe+Z{n}^{2+}\rightleftharpoons F{e}^{2+}+Zn$ is $-0.32V$. The equilibrium concentration of $F{e}^{2+}$ when a piece of iron is placed in a $1MZ{n}^{2+}$ solution is:

• A. $1.4\times {10}^{-11}M$
• B. $1.5\times {10}^{-10}M$
• C. $3.5\times {10}^{-11}M$
• D. $3.5\times {10}^{-10}M$

Answer 1

The correct option is A $1.4\times {10}^{-11}M$

Since $E^o$ is $-ve$, only the reverse reaction is spontaneous.

$Z{n}^{2+}+Fe\rightleftharpoons Zn+F{e}^{2+}$; $E^0=-0.32\vee$

The expression for the standard emf of cell is,

$E_{cell}^0=\frac{0.0592}{n}logK=\frac{0.0592}{n}log\frac{\left[{Fe}^{2+}\right]}{\left[{Zn}^{2+}\right]}$

Substitute values in the above equation.

$-0.32V=\frac{0.0592}{2}log\frac{\left[{Fe}^{2+}\right]}{1}$

$log\left[{Fe}^{2+}\right]=-10.81$

$\left[{Fe}^{2+}\right]=1.55\times {10}^{-11}M$