The emf of the cell $Zn\left|Z{n}^{+2}\left(0.01m\right)\right|\left|F{e}^{+2}\left(0.001M\right)\right|Fe$ at 298 K is 0.2905 volt. Then the value of equilibrium constant for the cell reaction is.

$Zn+F{e}^{+2}\stackrel{ }{\rightleftharpoons }Z{n}^{+2}+Fe(n=2)$
$E=E°-\frac{0.0591}{n}\log \left(1\right)$
$0.02905=E°-\frac{0.0591}{n}\log \frac{0.01}{0.001}$
$E°=0.02905+0.02905=0.32V$
But at equilibrium,$E=0$
$\therefore E°=\frac{0.0591}{2}\log K_{eq}$
$0.32=0.02945\log K_{eq}$