For the cell $\mathrm{Zn}\left|{\mathrm{Zn}}^{2+}\left(1\mathrm{M}\right)\parallel {\mathrm{Sn}}^{2+}\left(1\mathrm{M}\right)\right|\mathrm{Sn},E_{\text{cell }}^{\circ }=0.6264\mathrm{V}$ The value of $K_{eq}^{\mathrm{\top }}$ for the reaction $\mathrm{Sn}+{\mathrm{Zn}}^{2+}\rightleftharpoons \mathrm{Zn}+{\mathrm{Sn}}^{2+}$ will be given by the expression

$E_{\text{cell }}°=0.6264 \mathrm{V}$

Cell reaction: ${\mathrm{Sn}}^{2+}+\mathrm{Zn}\to \mathrm{Sn}+{\mathrm{Zn}}^{2+}$

The given reaction is just opposite to the above reaction. Hence, the $E_{\text{cell }}°$ producing the given reaction would be -0.6264 V. Thus

$\log K_{\mathrm{eq}}°=\frac{nF{E}_{\mathrm{cell}}°}{2.303RT}=\frac{n{E}_{\text{cell }}°}{(2.303RT/F)}=\frac{2\times (-0.6264)}{0.059}=-21.23$