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

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.