In the reaction, Cr ₂ O ₇²⁻ + 4 H ₂ O ₂ + 2 H ^+ 2 CrO ₅ + H ₂ O The change in oxidation state of Cr is:

Step-by-step solution

In Cr ₂ O ₇²⁻ , the oxidation state of Cr is calculated as: 2x + 7(-2) = -2 2x = +12 x = +6 In CrO ₅ , the molecule has a butterfly structure containing two peroxo linkages ( -O-O- ) and one oxo linkage ( =O ). There are four oxygen atoms with an oxidation state of -1 and one oxygen atom with an oxidation state of -2 . The oxidation state of Cr in CrO ₅ is calculated as: y + 4(-1) + 1(-2) = 0 y = +6 Thus, the oxidation state of Cr remains +6 in both the reactant and the product. There is no change in the oxidation