<u>Answer:</u> The
of the reaction is 
<u>Explanation:</u>
For the given half reactions:
Oxidation half reaction: 
Reduction half reaction: 
Net reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
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 = 0.48 V
R = Gas constant = 8.314 J/K.mol
T = temperature of the reaction = ![25^oC=[273+25]=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5D%3D298K)
= equilibrium constant of the reaction = ?
Putting values in above equation, we get:

Hence, the
of the reaction is 