Answer:
The electrode potential, E, of the half reaction is +1.13 V
Explanation:
Applying Nernst Equation;
[tex]E = E^o - \frac{0.0592}{n}LogQ[/tex]
Where;
E is the the electrode potential
E° is the standard electrode potential = +1.13 V
n is the number of electron transferred = 4
Q is the ratio of the concentration = 1
Substitute these values and solve for E;
[tex]E = E^o - \frac{0.0592}{n}LogQ\\\\E = 1.13 - \frac{0.0592}{4}Log[1]\\\\E =1.13 -0\\\\E =+1.13 V[/tex]
Therefore, the electrode potential, E, of the half reaction is +1.13 V
