What product(s) forms at the cathode in the electrolysis of an aqueous solution of Na3PO4?

Question

What product(s) forms at the cathode in the electrolysis of an aqueous solution of Na₃PO₄?

The correct answer is H₂ and OH^- can you explain why

Answer 1

There is a multi-step process involved.

To start, in water solution Na₃PO₄ partially dissociates to give 3 Na⁺ + H₂PO₄^- + 2 OH^- .

This insures that there will be H₂ PO₄^- available to be reduced at the cathode. The sodium ions are spectators and need not be considered further.

At the cathode, electrons get injected by the reaction

2 H₂PO₄^- + 2 e^- → H₂ + 2 HPO₄^2-

However, the HPO₄^2- does not hang around - it reacts with water

2 HPO₄^2- + 2 H₂O → 2 H₂PO₄^- + 2 OH^- ( the H₂PO₄^-1 gets regenerated !)

So the net reaction is simply 2 H₂O + 2e^- → H₂O + 2 OH^-

Thus the products produced at the cathode are H₂O and OH^-