Cr\(_2\)O\(_7\)\(^{2-}\)(aq) + 14H\(^+\)(aq) + 6l\(^-\)(aq) → 2Cr\(^{3+}\)(aq) + 3l(g) + 7H\(_2\)O(I).
The change in the oxidation number of oxygen in the equation above is?
The change in oxidation number is from Cr\(_2\)O\(_7\)\(^{2-}\) to H\(_2\)O
The oxidation number of Oxygen in Cr\(_2\)O\(_7\)\(^{2-}\) is -2, while the oxidation number of Oxygen in H\(_2\)O is -2.
Change means final - initial = -2 - (-2)
= - 2 + 2 = 0
Therefore, the change in oxidation number of oxygen is 0
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