The correct option is B 1.5
C2H5OH+Cr2O2−7→CH3COOH+Cr3+
Oxidation state of C in C2H5OH=−2
Oxidation state of C in CH3COOH=0
Oxidation state of Cr in Cr2O2−7=+6
Oxidation state of Cr in Cr3+=+3
C2H5OH is undergoing oxidation and is the reducing agent
Cr2O2−7 is undergoing reduction and is oxidising agent.
For n-factor calculation,
nf=(|O.S.Product−O.S.Reactant|×number of atoms
nf of C2H5OH=4
nf of Cr2O2−7=6
Ratio of oxidation states is 2:3
Cross multiplying these with nf of each other.
we get,
3C2H5OH+2Cr2O2−7→CH3COOH+Cr3+
Balancing the elements other than oxygen and hydrogen on both sides,
3C2H5OH+2Cr2O2−7→3CH3COOH+4Cr3+
Adding the H2O to balance the oxygen,
3C2H5OH+2Cr2O2−7→3CH3COOH+4Cr3++11H2O
Adding H+ to balance hydrogen,
3C2H5OH+2Cr2O2−7+16H+→3CH3COOH+4Cr3++11H2O
Balance charge
total charge in reactant side=+12
total charge in product side=+12
3C2H5OH+2Cr2O2−7+16H+→3CH3COOH+4Cr3++11H2O)
This is the final balanced equation.
So, required ratio is nC2H5OHnCr2O2−7=32=1.5