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Question

The reaction of the permanganate ion with the oxalate ion in acidic solution 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
Steps for Balancing redox reactions:
  1. Identify the oxidation and reduction half .
  2. Find the oxidising and reducing agent.
  3. Find the n-factor of oxidising and reducing agent.
  4. Balance atom undergoing oxidation and reduction.
  5. Cross multiply the oxidising or reducing agent with simplified n-factor values
  6. Balance atoms other than oxygen and hydrogen.
  7. Balancing oxygen atoms
  8. Balancing hydrogen atoms
  9. Balance charge
For acidic medium:
As soon as we add x H2O units, we add 2x H+ ions on the opposite side.

nf=(|O.S.ProductO.S.Reactant|×number of atom
+7MnO4++3C2O24+2Mn2+++4CO2

+3C2O24+4CO2oxidation
nf=(|43|×2=2
+7MnO4+2Mn2+ reduction
nf=(|27|×1=5
MnO4 is oxidising agent
C2O24 is reducing agent.

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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