<u>Answer:</u> The potential of the given cell is 0.856 V
<u>Explanation:</u>
The substance having highest positive potential will always get reduced.
Half reactions for the given cell follows:
<u>Oxidation half reaction:</u> ( × 2)
<u>Reduction half reaction:</u>
<u>Net reaction:</u>
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 the EMF of the cell, we use the Nernst equation, which is:
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +0.977 V
R = Gas constant = 8.314 J/K mol
T = temperature = 279 K
F = Faraday's constant = 96500 C
n = number of electrons exchanged = 2
Putting values in above equation, we get:
Hence, the potential of the given cell is 0.856 V