Electrochemistry Questions

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is ${\text{CH}}_3\text{OH}(l)+\frac{3}{2}{\text{O}}_2(g)\to {\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)$. At $298\text{K}$ standard Gibb’s energies of formation for ${\text{CH}}_3\text{OH}(l)$, ${\text{H}}_2\text{O}(l)$ and ${\text{CO}}_2(g)$ are $-166.2$, $-237.2$ and $-394.4{\text{kJ mol}}^{-1}$ respectively. If standard enthalpy of combustion of methanol is $-726{\text{kJ mol}}^{-1}$, efficiency of the fuel cell will be:

Given: $E_{{\text{Fe}}^{3+}/\text{Fe}}^{\circ }=-0.036\text{V}$, $E_{{\text{Fe}}^{2+}/\text{Fe}}^{\circ }=-0.439\text{V}$. The value of standard electrode potential for the change, ${\text{Fe}}^{3+}(aq)+e^{-}\to {\text{Fe}}^{2+}(aq)$ will be: