Consider the cell at $25^{\circ} \mathrm{C}$
$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq}),(1 \mathrm{M}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$
The fraction of total iron present as $\mathrm{Fe}^{3+}$ ion at the cell potential of $1.500 \mathrm{~V}$ is $\mathrm{x} \times 10^{-2}$. The value of $\mathrm{x}$ is __________ (Nearest integer)
$\left(\right.$ Given $\left.: \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=0.77 \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{~V}\right)$
$\mathrm{E}_{\text {cell }}^{0}=0.77-(0.76)$
$=1.53 \mathrm{~V}$
$1.50=1.53-\frac{0.06}{2} \log \left(\frac{\mathrm{Fe}^{2+}}{\mathrm{Fe}^{3+}}\right)^{2}$
$\log \left(\frac{\mathrm{Fe}^{2+}}{\mathrm{Fe}^{3+}}\right)=\frac{0.03}{0.06}=\frac{1}{2}$
$\frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=10^{1 / 2}=\sqrt{10}$
$\frac{\left[\mathrm{Fe}^{3+}\right]}{\left[\mathrm{Fe}^{2+}\right]}=\frac{1}{\sqrt{10}}$
$\frac{\left[\mathrm{Fe}^{3+}\right]}{\left[\mathrm{Fe}^{2+}\right]+\left[\mathrm{Fe}^{3+}\right]}=\frac{1}{1+\sqrt{10}}=\frac{1}{4.16}$
$=0.2402$
$=24 \times 10^{-2}$