The rusting of iron takes place as follows. $1Fe+\frac{1}{2}{O}_2+2H^{\oplus }\to 1F{e}^{+2}+1H_{2}O;$ if $E^0$ for $O_2+4H^{\oplus }+4e^{-}\to 2H_{2}O$ is 1.23V and $E^0\left(Fe\left(F{e}^{+2}\right)\right)=0.44V$ then correct are is

$\left(1\right)O_2+4H^{\oplus }+4e^{-1}\to 2H_{2}O;E_1^0=1.23V$

$\left(2\right)Fe\stackrel{ }{\to }F{e}^{+2}+2e^{-};E_2^0=0.44V$

$\left(3\right)1Fe+\frac{1}{2}{O}_2+2H^{+}\stackrel{ }{\to }1F{e}^{+2}+1H_{2}O;E_3^0=E_{cell}^0=]$

 $\left(3\right)=(\left(2\right)\times 1)+(\left(1\right)\times \frac{1}{2})$

$(E_{cell}^0\times 2)=(0.44\times 2)+(1.23\times 2)$

$E_{cell}^0=1.67V$

But, $\Delta G^0=-nF{E}_{cell}^0=-2.303RT\log K_c$

Log $K_e=\left(\frac{-2\times 96500\times 1.67}{-2.303\times 8.314\times 298}\right)=56.4876$

$K_C={10}^{56}\times 2.85$