∆G1∘=−2×F×−1.18J (ii) Mn3++e−⟶Mn2+;E∘=1.51V ∆G2∘=−1×F×1.51J Aim : Mn3++3e−⟶Mn;∆G3∘=? (i) + (ii) gives the required results, i.e., ∆G3‌∘=∆G1‌∘+∆G2‌∘ =2.36F+(−1.51F)=0.85F ∴−nFEMn3+/Mn∘=0.85F or EMn3+/Mn∘=−‌