Question

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:

FeI₂;

K₃PO₄.

Answer 1

(a). Electrolysis of FeI2 gives following reactions:

At cathode:

Fe²⁺   + 2e-    Fe(s) Eo = - 0.44 V

2H2O + 2e-     H2 + 2OH- Eo = - 0.827 V

Because it is much easier to reduce Fe²⁺ ions than water, the only product formed at the cathode is iron.

At anode:

2I-     I2 +    2e-    Eo = - 0.54 V

2H2O O2 + 4H+ + 4e- Eo = -1.23 V

The only product formed at anode is O2 gas.

(b). Electrolysis of K3PO4 gives following reactions:

At cathode:

K+ + e- K(s) Eo = - 2.93 V

2H2O + 2e-     H2 + 2OH- Eo = - 0.827 V

Because it is much easier to reduce water than K⁺ ions, the only product formed at the cathode is hydrogen gas.

At anode:

2H2O O2 + 4H+ + 4e- Eo = -1.23 V

The only product formed at anode is O2 gas.