Chemistry - Electrochemistry Question with Solution | TestHub

The half cell reaction are at anode $2{\mathrm{O}}^{2-}+4e^{-}⟶{\mathrm{O}}_2]\times 3$ at cathode ${\mathrm{Al}}^{3+}⟶\mathrm{Al}+3e^{-}]4$ Net reaction $4\mathrm{Al}+6{\mathrm{O}}^{2-}⟶3{\mathrm{O}}_2+4\mathrm{Al}$ $\begin{array}{rl}& \text{or }\frac{4}{3}\mathrm{Al}+2{\mathrm{O}}^{2-}⟶{\mathrm{O}}_2+\frac{4}{3}\mathrm{Al}\\ & \therefore n=\frac{12}{3}=4\\ & \Delta G^{\circ }=-nF{E}^{\circ }\\ & \text{Here, }\Delta G^{\circ }=+960\mathrm{kJ}{\mathrm{mol}}^{-1}\\ & =960\times 1000\mathrm{J}{\mathrm{mol}}^{-1}\\ & n=4\\ & F=96500\text{coulomb }\\ & \therefore 960\times 1000=-4\times 96500\times E^{\circ }\\ & E^{\circ }=-\frac{960000}{4\times 96500}\\ & =-2.48\mathrm{V}\\ & \text{Potential difference }\approx 2.5\mathrm{V}\end{array}$