<u>Answer:</u> The cell voltage of the given cell is 2.01
<u>Explanation:</u>
The given chemical cell equation follows:
<u>Oxidation half reaction:</u> ( × 2)
<u>Reduction half reaction:</u> ( × 3)
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 = +2.00 V
R = Gas constant = 8.314 J/mol.K
T = temperature =
F = Faraday's constant = 96500
n = number of electrons exchanged = 6
Putting values in above equation, we get:
Hence, the cell voltage of the given cell is 2.01