What is the value of the equilibrium constant at 25 oc for the reaction between the pair: ag(s) and ni2+(aq) to give ni(s) and ag+(aq) use the reduction potential values for ag+(aq) of +0.80 v and for ni2+(aq) of -0.25 v give your answer using e-notation with no decimal places (e. g., 2 x 10-2 would be 2e-2; and 2.12 x 10-2 would also be 2e- do not include spaces, units, punctuation or anything else silly!

The of the reaction is .

Explanation:

Reduction at cathode :

Reduction potential of   to Ni=

Oxidation at anode:

Reduction potential of   to Ag=

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

and,

Equating these two equations, we get:

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

= standard electrode potential of the cell = -1.05 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction =

= equilibrium constant of the reaction = ?

Putting values in above equation, we get:

Hence, the of the reaction is

fe 55.85 g/mol and h2o 18.02 g/mol.