NAD +(aq)+2H+(aq)+2e is NAD(aq)+H+(aq)E°=-0.320v Oxaloacetate(aq)+2H+(aq)+2H+(aq)+2e is malate(aq) E°=-0.166v formular reaction E°cell=E electrode -Eanode If two half reaction, the spontaneeus reaction it will occur if less negative. potential reduction occurs at catode. that is E eclectrode=-0.166v, and Eanode= -0.320v To calculate the E°cell E°cell = -0.166v - (-0.320v) = 0.154v Electrical work done and equilibrium constant at suitable is related to each other as R. T.Ink =n.F.E°cell R,T,K,n, andF they stand for constant gas, absolute temperature, equilibrium, number of electrons and faraday constant respectively the value of n=2, T as 298K, F as 96500 then K U R.T.Ink =n.F.E°cell 8.314 J/mol. K×298K, F as 96500C×0.154V InK=29722/2477.57=162,173.544 K= E11.996
