Question

Relation between emf and equlibrium constant in reaction

Answer 1

LET THE REACTION BE

A+B---->C+D

E=E°-0.059/N*log([A][B]/[C][D])

at equilibrium E=0

E°=0.059/N*log([A][B]/[C][D])

K=[C][D]/[A][B]

So,

E°=O.O59/N*log(1/K)

N= no of electrons exchanged

Answer 2

The value of Ecell for the voltaic cell below:

Pt(s)|Fe2+(0.1M),Fe3+(0.2M)||Ag+(0.1M)|Ag(s)(2)

SOLUTION

To use the Nernst equation, we need to establish  Eocell  and the reaction to which the cell diagram corresponds so that the form of the reaction quotient (Q) can be revealed. Once we have determined the form of the Nernst equation, we can insert the concentration of the species.

Solve:

Eocell=Eocathode−Eoanode(3)

= EoAg/Ag - EoFe3+/Fe2+

= 0.800V-0.771V = 0.029V

Now to determine Ecell for the reaction

Fe2+(0.1M) + Ag+(1.0M) → Fe3+(0.20M) + Ag(s)

Use the Nernst equation

Ecell= 0.029V -(0.0592V/1)log [Fe3+]/[Fe2+][Ag]

=0.029V - 0.0592V*log [0.2]/[0.1]*[1.0]

=0.011V