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? MnO−4+C2O2−4→Mn2++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 2Steps for Balancing redox reactions: Identify the oxidation and reduction half . Find the oxidising and reducing agent. Find the n-factor of oxidising and reducing agent. Balance atom undergoing oxidation and reduction. Cross multiply the oxidising or reducing agent with simplified n-factor values Balance atoms other than oxygen and hydrogen. Balancing oxygen atoms Balancing hydrogen atoms 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.Product−O.S.Reactant|×number of atom +7MnO−4++3C2O2−4→+2Mn2+++4CO2 +3C2O2−4→+4CO2oxidation nf=(|4−3|×2=2 +7MnO−4→+2Mn2+ reduction nf=(|2−7|×1=5 MnO−4 is oxidising agent C2O2−4 is reducing agent. Balance atom undergoing oxidation and reduction. MnO−4+C2O2−4→Mn2++2CO2 Cross mutiply the oxidising or reducing agent with simplified n-factor values. 2MnO−4+5C2O2−4→2Mn2++10CO2 Balance atoms oxygen. 2MnO−4+5C2O2−4→2Mn2++10CO2+8H2O balance hydrogen atom 2MnO−4+5C2O2−4+16H+→2Mn2++10CO2+8H2O Balance charge charge in reactant side = +4 charge in product side = +4 so the balanced equation is 2MnO−4+5C2O2−4+16H+→2Mn2++10CO2+8H2O the coefficient of Mn is 2.

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