The EMF of the cell Zn | Zn2+ (0.01M) | | Fe2+ (0.001M) | Fe , at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction 10x. Find x.

The EMF of the cell  , at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction ${10}^{x}$. Find x.

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Solution:

For this cell, reaction is;

$\mathrm{Zn}+{\mathrm{Fe}}^{2+}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$

At equilibrium,

$\therefore {\mathrm{K}}_{\mathrm{eq}}={10}^{\frac{0.32}{0.0295}}$

Comparing the value of ${10}^{x}$,

$x=\frac{0.32}{0.0295}=10.847\approx 10.85$

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