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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. ResetHelp -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

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Answer:

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.
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The oxidation numbers of the given elements are:

Reactants= H2(+1), O2(-2), Cl(+7), O7(-2), O(-2), H(+1)

Products = Cl(+3), O2(-2), H2(+1), O(-2 ), O2(0)

Calculation of oxidation numbers

Oxidation numbers of an element are those total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.

To calculate the oxidation numbers of reactants;

H2O2= (+1)×2+(X) =0

= 2 + X = 0

X= -2

Cl2O7 = (X)×2 + (-2)7 = 0

= 2x +(-14) = 0

= 2x - 14 = 0

2x= 14

X = 14/2

X= +7

OH- = x + 1 = -1

X= -1-1

X= -2

To calculate the oxidation numbers of products;

ClO2– = X+(-2)×2= -1

= X - 4 = -1

X = 4-1

X = +3

H2O = +1, -2

O2 = 0

Learn more about oxidation here:

https://brainly.com/question/25886015

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