The expression of free energy is given by, [∆G0]=−nFE0 Thus ‌ For ‌n=1 −∆G10‌‌=nE10 ‌‌=E10 −∆G20‌‌=nE20 ‌‌=E20 ‌ For ‌n=2 −∆G30‌‌=nE30 ‌‌=2E30 ‌ For reaction ‌M2+(aq)+2e−→M(s) Therefore, −∆G30=(−)∆G10+(−)∆G20 2E30‌‌=(E10+E20) ‌‌=0.15+0.5 E30‌‌=0.325V