Chemistry, 20.09.2019 17:30 justaguy15

Which element is reduced in this reaction? 16h++2cr2o72−+c2h5oh→4cr3++11h2o+2c o2?

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The equation, for better presentation, is

6H⁺ + 2 Cr₂O₇²⁻ + C₂H₅OH → 4 Cr³⁺ + 11 H₂O + 2 CO₂

To tell which element is reduced, let's solve the oxidation numbers of derived compounds.

H⁺ becomes H₂O. The oxidation number of H⁺ is ⁺1, while the oxidation number of H in H₂O is:

2x -2 = 0
x = ⁺1

Hence, the oxidation number remained the same.

Next, Cr₂O₇²⁻ became Cr³⁺. The oxidation number of Cr in Cr³⁺ is ⁺3. The oxidation number of Cr in Cr₂O₇²⁻ isL
2x + 7(-2) = -2
x = 6

Thus, the oxidation from ⁺6 became ⁺3. The oxidation number reduced. So, the element reduced is Cr or Chromium.

Answer from: ineedhelp368

Cr atoms undergo a reduction reaction (decrease in oxidation number) from +6 to +3

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)= +2H 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 :

16H⁺ + 2Cr₂O₇²⁻ + C₂H₅OH → 4Cr³⁺+ 11H₂O + 2CO₂

Oxidation number at C₂H₅OH

2. C + 6. H + O = 0

2C + 6. + 1 + (-2) = 0

2C + 6 -2 = 0

2C + 4 = 0

2C = -4

C = -2

Oxidation number at CO₂

C + 2.O = 0

C + 2.-2 = 0

C = 4

The oxidation number C in C₂H₅OH is -2 and the oxidation number C in CO₂ is +4, the change in oxidation number C from -2 to +4 is 6, so the C atom experiences an oxidation reaction (increase in oxidation number)

Oxidation number at Cr₂O₇²⁻

2. Cr + 7. O = -2

2. Cr + 7.-2 = -2

2Cr -14 = -2

2Cr = 12

Cr = 6

Cr³⁺ = 3

The oxidation number Cr in Cr₂O₇²⁻ is +6 and the oxidation number Cr in Cr³⁺ is +3, the change in Cr oxidation number from +6 to +3 is -3, so that the Cr 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)

Learn more

an oxidation-reduction reaction

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a reducing agent during a redox reaction

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loses electrons in a chemical reaction

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Keywords: reduction, oxidation

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