Chemistry - Electrochemistry Question with Solution | TestHub

The half reactions are $\mathrm{Fe}(s)⟶{\mathrm{Fe}}^{2+}(\mathrm{aq})+2e^{-}\times 2$ $\begin{array}{r}{\mathrm{O}}_2(g)+4{\mathrm{H}}^{+}+4e^{-}\to 2{\mathrm{H}}_2\mathrm{O}+2\mathrm{Fe}(s)\\ \end{array}$

Net reaction

$$\begin{gathered} \text{displaylines}{\mathrm{O}}_2(g)+4{\mathrm{H}}^{+}\to 2{\mathrm{Fe}}^{2+}(aq)+2{\mathrm{H}}_2\mathrm{O}(l) \\ E=E^{\circ }-\frac{0.059}{4}\log \frac{{\left(10^{-3}\right)}^2}{{\left(10^{-3}\right)}^4(0.1)}=1.57\mathrm{V} \end{gathered}$$