Respuesta :

The SRP(standard hydrogen electrode) of 2H⁺ + 2e⁻→H₂ is 0
and SRP of K⁺ + e⁻→K is lower than 0.
It means that if there are potassium ion and hydogen ion, it is easy for H⁺ to be reduced than oxidized. (H⁺ is reduced before K⁺ is reduced.)
By process '4H₂O + 4e⁻ → 2H₂ + 4OH⁻
H₂ is formed
PO₄³⁻ is much bigger than H₂O so the charge density is much lower than the O²⁻ side of H₂O.
PO₄³⁻ can't be formed.
By process '2H₂O→O₂ + 4H⁺ + 4e⁻'
O₂ is formed
dsdrajlin

Answer:

O₂ and H⁺

Explanation:

In electrolysis, the anode is the electrode where the oxidation occurs. Neither K⁺ nor PO₄³⁻ can oxidize because K and P are in their highest oxidation states. Then, oxygen in water will oxidize according to the following half-reaction:

Oxidation: 2 H₂O(l) → O₂(g) + 4 H⁺(aq) + 4 e⁻

As we can see, the products at the anode are O₂ and H⁺.

Otras preguntas