<u>Answer:</u> The electrode potential of the cell is 0.093 V
<u>Explanation:</u>
The given chemical cell follows:
<u>Oxidation half reaction:</u>
<u>Reduction half reaction:</u>
<u>Net cell 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 = +0.13 V
n = number of electrons exchanged = 2
Putting values in above equation, we get:
Hence, the electrode potential of the cell is 0.093 V