Cr2O72-(aq) + 14H+(aq) + 6l-(aq) → 2Cr3+(aq) + 3l(g) + 7H2O(I). The change in the oxidation number of oxygen in the equation above is?

Assessment: JAMB UTME - Chemistry - 1995 Subject: Chemistry

Question 1 Report

Cr2O72-(aq) + 14H+(aq) + 6l-(aq) → 2Cr3+(aq) + 3l(g) + 7H2O(I).

The change in the oxidation number of oxygen in the equation above is?

Answer Details
In the chemical equation provided, there are oxygen atoms in two different forms: O²⁻ and H₂O. The oxidation number of oxygen in O²⁻ is -2, while in H₂O it is -2 as well. Therefore, the change in oxidation number of oxygen is 0, because the oxidation number of oxygen is the same on both sides of the equation. Therefore, the correct answer is (a) 0.
Read lesson note on Oxidation And Reduction (JAMB)
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