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Question

Permanganate solution reacts with the oxalate ion in acidic medium and forms manganese(II) ion and carbon dioxide. What is the stoichiometric coefficient in front of manganese(II) in the balanced redox equation?
MnO4+C2O24Mn2++CO2

A
2
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B
1
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C
3
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D
5
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Solution

The correct option is A 2
The redox reaction is :
MnO4+C2O24Mn2++CO2
+7MnO4++3C2O24+2Mn2+++4CO2
MnO4 is oxidising agent
C2O24 is reducing agent.

nf=(|O.S.ProductO.S.Reactant|)×number of atom
+3C2O24+4CO2---oxidation
nf=(|43|×2)=2
+7MnO4+2Mn2+----reduction
nf=(|27|×1)=5
Balance atom undergoing oxidation and reduction.
MnO4+C2O24Mn2++2CO2
Cross mutiply the oxidising or reducing agent with simplified n-factor values.
2MnO4+5C2O242Mn2++10CO2

Balance atoms oxygen.

2MnO4+5C2O242Mn2++10CO2+8H2O

balance hydrogen atom
2MnO4+5C2O24+16H+2Mn2++10CO2+8H2O
Balance charge
charge in reactant side = +4
charge in product side = +4
so the balanced equation is 2MnO4+5C2O24+16H+2Mn2++10CO2+8H2O
the coefficient of Mn is 2.

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