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Here are two oxidation-reduction systems with their standard → potentials : $O_2(g)+4H^+(aq)+4e^- $ $\leftrightarrows$ $2H_2O(l)$ $E^o_1$= $1.23 \;V$ $Ag^{+}(aq)+e^- $ $\leftrightarrows$ $Ag(s)$ $E^o_2$= $0.80 \;V$ Calculate the equilibrium constant at $ 25^oC $ (standard!) of the reaction $4Ag(s)$ -4$e^-+O_2(g)+4H^+(aq)$ +4$e^-$ $\leftrightarrows$ $4Ag^+(aq)+2H_2O(l)$
$E^o$ $ =$ $ E^o_1-E^o_2=0.43\; V$
$log\;K$ = $\frac{n}{0.059}E^o$ = $\frac{4}{0.059}0.43$ = $30$ $K$ = $10^{30}$