Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode):
Reduction half reaction (cathode):
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature =
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
= 3.5 M
=
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V