Answer : The value of
and K is, -180 kJ/mol and 
Explanation :
The balanced cell reaction will be,

The half-cell reactions are:
Oxidation reaction (anode) : 
Reduction reaction (cathode) : 
Relationship between standard Gibbs free energy and standard electrode potential follows:

where,
= standard Gibbs free energy
F = Faraday constant = 96500 C
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = 0.93 V
Now put all the given values in the above formula, we get:


Now we have to calculate the value of 'K'.

where,
= standard Gibbs free energy = -180 kJ/mol
R = gas constant = 
T = temperature = 298 K
K = equilibrium constant = ?
Now put all the given values in the above formula 1, we get:


Therefore, the value of
and K is, -180 kJ/mol and 