Redox titrations

Question

0.500 g of zinc powder was found to reduce an acidified solution of 25.50 mL of 0.200 M AO2+. What is the final oxidation number of metal A?

Answer 1

I'm assuming that the AO2+ is AO₂⁺ and not AO²⁺.

In AO₂⁺, the oxidation number of A would be 5+ in order that the species have a net charge of 1+.

moles Zn = 0.500 g x 1 mol Zn / 65.4 g = 0.007645 moles

Zn(s) ==> Zn²⁺ + 2e-

0.007645 mol Zn x 2 mol electrons / mol Zn = 0.01529 moles electrons

moles AO₂⁺ = 0.02550 L x 0.200 mol/L = 0.00510 moles AO₂⁺

0.01529 mol / 0.00510 = 2.998 ≈ 3

Thus, in AO₂⁺, the oxidation number of A is 5+. It was reduced by 3, so the final oxidation number of metal A would be 2+