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Q. The oxidation states of the most electronegative element in the products of the reaction of $BaO_{2}$ with dilute $H_{2}SO_{4}$ are

Redox Reactions

Solution:

In $H_{2}O_{2}$ oxygen shows $= -1$ (peroxide) oxidation state and in $BaSO_{4}$ oxygen shows $= - 2$ oxidation state.