Identify the oxidation numbers for all the elements in the reactants and products for 4h2o2(aq)+cl2o7(g)+2oh−(aq)→2clo2−( aq)+5h2o(l)+4o2(g) identify the oxidation numbers by dragging the appropriate labels to their respective targets. -7-7 -5-5 -4-4 -3-3 -2-2 -1-1 00 +1+1 +2+2 +3+3 +4+4 +5+5 +7+7

H₂O₂: H (+1); O (-1)

Cl₂O₇: Cl (+7); O (-2)

OH⁻: O (-2); H (+1)

ClO₂⁻: Cl (+3); O (-2)

H₂O:  H (+1); O (-2)

O₂: O (0)

Explanation:

The oxidation number (o.n.) is the number of electrons an atom tends to gain or lose to form a chemical bond and gain stability.

In H₂O₂, H has the o. n. +1 (its most common o. n.) and O has the o.n. -1 in the peroxide ion O₂²⁻.In Cl₂O₇, Cl has the o.n. +7 (its highest o.n.) and O has the o.n. -2 (its most common o.n.)In OH⁻, O has the o.n -2 and H has the o.n. +1.In ClO₂⁻, Cl has the o.n. +3, and O has the o.n. -2. Cl has undergone a reduction.In H₂O, O has the o.n -2 and H has the o.n. +1.By definition, in O₂, O has the o.n. 0. It has undergone oxidation.