Reduction of iodate, IO3-(aq) + 5 H+(aq) + 4 e− → HIO(aq) + 2 H2O(aq), occurs with E0 = +1.13 V. What is the electrode potential, E, of the half reaction if the concentration/activity of all substances is 1.00 M?

The electrode potential, E, of the half reaction is +1.13 V

Explanation:

Applying Nernst Equation;

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;

Therefore, the electrode potential, E, of the half reaction is +1.13 V