Pb + PbO2 + 2H2SO4 --> 2PbSO4 + 2H2O
ΔG = Σ(υ×ΔGprodusi) - Σ(υ×ΔGreactanti)
= [2ΔGapa + 2ΔGPbSO4] - [2ΔGH2SO4 + ΔGPbO2 + ΔGPb]
= [2(-56,7)+2(-193,9)] - [2(-176,5)+(-52,3)+0]
= -95,9 kJ/mol
F.e.m. = Ecatod-Eanod
ΔG = -zEF = -2xEx96500
=> E = 95,9x1000/2x96500
= 0,5 V
