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Question: For the redox reaction: \(MnO_4^ - + {C_2}O_4^{2 - } + {H^ + } \to M{n^{2 + }} + C{O_2} + {H_2}O\) ...

For the redox reaction: MnO4−+C2O42−+H+→Mn2++CO2+H2OMnO_4^ - + {C_2}O_4^{2 - } + {H^ + } \to M{n^{2 + }} + C{O_2} + {H_2}O
The correct coefficients of the reactants MnO4−MnO_4^ - ,C2O42−{C_2}O_4^{2 - } and H+{H^ + } for the balanced reaction are ____, ____ and ____ respectively.
A.2, 5, 16
B.16, 5, 2
C.5, 16, 2
D.2, 16, 5

Explanation

Solution

A balanced chemical reaction is a reaction which has equal numbers of atoms of each element on both sides of an equation. In a redox reaction, it is important to balance the number of electrons transferred through the course of reaction. A balanced reaction follows both electrical neutrality and the law of conservation of mass.

Complete step by step answer:
From the law of conservation of mass, we know that matter can neither be created nor destroyed and nor can one element change into the other in a chemical reaction, thus the amount of each element must be the same throughout the entire reaction. For example, the number of atoms of any element in the reactants will be always equal to the number of atoms of that element in the products formed.
The given reaction is a redox reaction in which permanganate ion is the oxidizing agent and oxalate ion is the reducing agent. First we balance the number of atoms of each element except hydrogen and oxygen. We get,
MnO4−+C2O42−+H+→Mn2++2CO2+H2OMnO_4^ - + {C_2}O_4^{2 - } + {H^ + } \to M{n^{2 + }} + 2C{O_2} + {H_2}O
MnO4−→Mn2+MnO_4^ - \to M{n^{2 + }}
Manganese undergoes a change in oxidation number from +7→+2 + 7 \to + 2. It gains 5 electrons.
C2O42−→2CO2{C_2}O_4^{2 - } \to 2C{O_2}
Carbon undergoes a change in oxidation number from +2→+4 + 2 \to + 4. It loses 2 electrons.
Cross multiplying both the equations with electrons lost, we get,
2MnO4−+5C2O42−+H+→2Mn2++10CO2+H2O2MnO_4^ - + 5{C_2}O_4^{2 - } + {H^ + } \to 2M{n^{2 + }} + 10C{O_2} + {H_2}O
Now that the electrons are balanced, we balance the hydrogen and oxygen atoms keeping in mind to also neutralize the charge on both sides of the reaction. We get,
2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O

Thus, the correct answer is A.
Note:
The given equation in the question is a skeletal equation. A skeletal equation is an unbalanced chemical equation and does not follow the law of conservation of mass. It just denotes the reaction occurring but not the amount of the chemicals used or produced.