The electrical work done during the reaction at 298 K: 2Hg(l)+Cl2(g)→Hg2Cl2(s),isGiven that E∘Cl2Cl−=1.36V;E∘Hg2Cl2Hg,Cl−=0.27V;PCl2=1 atm,
210.37 kJ mol−1
105.185 kJ mol−1
420.74 kJ mol−1
110.37 kJ mol−1
−ΔG∘= electrical work done =nF E∘cell=2×96500×(1.36−0.27)J=210.37 kJ
Calculate the useful work done during the reaction in kJ/mol. Ag(S)+12 Cl2(g)→AgCl(s) Given that E0Cl2/Cl−=+1.36V and E0Ag+|AgCl|Cl−=+0.220V PCl2=1 atm and T = 298 K
p2 - 1 =