Electrochemical Cell and Gibbs Energy of the Reaction
Calculate the emf of the cell in which the following reaction takes place : Ni ( s ) + 2Ag + ( 0.002 M ) → Ni 2 + ( 0.160 M ) + 2Ag ( s ). Given that E ø cell = 1.05 V .
Solution : By using Nernst equation
= 1.05 - 0.02955 log 4 × 10 4 = 1.05 - 0.02955 ( log 10000 + log 4 ) = 1.05 - 0.02955 ( 4 + 0.6021 ) = 0.914 V
Electrochemical Cell and Gibbs Energy of the Reaction
o Electrical work done ( 1 second ) = Electrical potential X Total charge passed o
o
o Passing the charges reversibly through the galvanic cell results in maximum work .
Reversible work done by galvanic cell = Decrease in Gibbs energy Let E = Emf of the cell nF = Amount of charge passed ΔrG = Gibbs energy of the reaction ΔrG = -nFEcell
For the reaction , Zn ( s ) + Cu 2 + ( aq ) -- > Zn 2 + ( aq ) + Cu ( s ) [ ΔrG = -2FEcell ] But when the equation becomes 2Zn ( s ) + 2Cu 2 + ( aq ) -- > 2Zn 2 + ( aq ) + 2Cu ( s ) [ ΔrG = -4FEcell ] Problem :