What product(s) forms at the anode in the electrolysis of an aqueous solution of k3po4?

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⁺.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole