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: