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Question

What are the stoichiometric coefficients in the balanced redox reaction only for the species in the unbalanced reaction respectively.This reaction is used by the Breathalyzer to detect ethyl alcohol in a driver's breath?
C2H5OH+Cr2O27CH3COOH+Cr3+

A
Reactants: 3, 4; products: 3, 8
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B
Reactants: 2, 2; products: 2, 4
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C
Reactants: 3, 1; products: 3, 2
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D
Reactants: 3, 2; products: 3, 4
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Solution

The correct option is D Reactants: 3, 2; products: 3, 4
Steps for Balancing redox reactions:
  • Identify the oxidation and reduction halves.
  • Find the oxidising and reducing agent.
  • Find the n-factor of oxidising and reducing agents.
  • Cross multiply the oxidising and reducing agent with the simplified n-factor of each other.
  • Balance the atoms other than oxygen and hydrogen.
  • Balance oxygen atoms.
  • Balance hydrogen atoms.
  • Balance charge


formula used for the n-factor calculation,
nf=(|O.S.ProductO.S.Reactant|×number of atoms

Taking the given equation and following the above mentioned steps:

C2H5OH+Cr2O27CH3COOH+Cr3+

Oxidation state of C in C2H5OH=2
Oxidation state of C in CH3COOH=0
Oxidation state of Cr in Cr2O27=+6
Oxidation state of Cr in Cr3+=+3

Cleary, C2H5OH is undergoing oxidation and Cr2O27 is undergoing reduction.
using the formula of n-factor given above,

nf of C2H5OH=4
nf of Cr2O27=6
Ratio of oxidation states is 2:3.

Cross multiplying these with nf of each other.
we get,
3C2H5OH+2Cr2O27CH3COOH+Cr3+

Balancing the elements other than oxygen and hydrogen on both sides,
3C2H5OH+2Cr2O273CH3COOH+4Cr3+

Adding the H2O to balance the oxygen,

LHS of the equation contains 17 oxygens while RHS contains only 6 oxygens, so adding 11 H2O to RHS.
3C2H5OH+2Cr2O273CH3COOH+4Cr3++11H2O
adding H+ to balance hydrogen,

The LHS of the equation contains 18 hydrogens while RHS contains 34 hydrogens, so adding 16H+ to LHS.
3C2H5OH+2Cr2O27+16H+3CH3COOH+4Cr3++11H2O

Balance charge
total charge in reactant side=+12
total charge in product side=+12

3C2H5OH+2Cr2O27+16H+3CH3COOH+4Cr3++11H2O

This is the final balanced equation.
So, required coefficients, reactants: 3, 2; products: 3, 4


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