If a current of 965 A oxidises 0.1 mole of X 2 O 4 2 − in 100 seconds, what is the oxidation number of ' X ' in the product, assuming no change in the oxidation state of oxygen?

Question

If a current of 965 A oxidises 0.1 mole of X 2 ​ O 4 ​ 2 − in 100 seconds, what is the oxidation number of ' X ' in the product, assuming no change in the oxidation state of oxygen?

TestHub · Accepted Answer

Concepts: Faraday's laws, charge, moles, oxidation states.

Solution: Total charge $Q = I t = 965 \times 100 = 96500$ C. Moles of electrons $n_{e} = Q / F = 96500/96500 = 1$ mol. For 0.1 mol X₂O₄²⁻, 1 mol e⁻ implies 10 e⁻ per X₂O₄²⁻. Initial oxidation state of X in X₂O₄²⁻ is +3.

Final oxidation state of X is +3 + (10/2) = +8.

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