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Cr(OH)₃  undergo a oxidation reaction (increase in oxidation number) from +3 to +6, so Cr(OH)₃  is the reducing agent

Further explanation

Reduction oxidation reaction or abbreviated Redox is a chemical reaction in which there are changes in oxidation numbers

The general formula for determining oxidation numbers:

  • 1. Single element atomic oxidation = 0. For example Ar, Mg, Cu, Fe, N₂, O₂, etc. = 0, Group IA (Li, Na, K, Rb, Cs, and Fr)= +1 ,Group IIA (Be, Mg, Ca, Sr and Ba)= +2
  • H atoms in compounds = +1, except metal hydride compounds (Hydrogens that bind to groups IA or IIA) oxidation numbers H = -1, for example LiH, MgH₂, etc.
  • 2. O atoms in compounds = -2, except OF2 = + 2 and at peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.
  • 3. The oxidation number in the uncharged compound = 0,

On reaction :

2Cr(OH)₃+3OCl⁻+4OH⁻→2CrO₄²⁻+3Cl⁻+5H₂O

Oxidation number at Cr(OH)₃

Cr + 3. H + 3.O = 0

Cr + 3.1. + 3.-2 = 0

Cr + 3 -6 = 0

Cr = +3

Oxidation number at CrO₄²⁻

Cr + 4.O = -2

C + 4.-2 = -2

C = +6

The oxidation number Cr in Cr(OH)₃ is +3 and the oxidation number Cr in CrO₄²⁻ is +6, the change in oxidation number C from +3 to +6 is 3, so the Cr atom experiences an oxidation reaction (increase in oxidation number)

Oxidation number at OCl⁻

O + Cl = -1

-2 + Cl = -1

Cl = +1

Oxidation number at Cl⁻

Cl⁻ = =-1

The oxidation number Cl in OCl⁻ is +1 and the oxidation number Cl in Cl⁻ is -1, the change in Cl oxidation number from +1 to -1 is -2, so that the Cl atom experiences a reduction reaction (decreasing the oxidation number)

While O and H atoms do not experience changes in oxidation numbers (O = -2 and H = +1)

So the reducing agent which is experienced with an oxidation reaction (increase in oxidation number) is Cr(OH)₃

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Keywords: reduction, oxidation

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The substance that is a reducing agent is Cr(OH)3.

The term "reducing agent" implies a specie whose oxidation number increases from left to right in the reaction equation. So an the reducing agent must have a lower oxidation number on the left hand side of the reaction equation, but a higher oxidation number on the right hand side of the reaction equation.

We must look out for this as we study the equation;

2Cr(OH)3 + 3OCl^− + 4OH^− ------>  2CrO4^2- + 3Cl^− + 5H2O

It is easy to see from the equation that the oxidation number of chromium increased from +3 on the left hand side of the reaction equation to +6 on the right hand side of the reaction equation.

This means that Cr(OH)3 must be the reducing agent in this reaction.

Learn more; https://brainly.com/question/13978139

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