Question

For the redox reaction, $Mn{O_4}^ - + {C_2}{O_4}^{2 - } + {H^ + } \to M{n^{2 + }} + C{O_2} + {H_2}O$ , the correct coefficients of the reactants for the balanced equation are
A. $Mn{O_4}^ - = 2,{C_2}{O_4}^{2 - } = 16,{H^ + } = 5$
B. $Mn{O_4}^ - = 16,{C_2}{O_4}^{2 - } = 5,{H^ + } = 2$
C. $Mn{O_4}^ - = 5,{C_2}{O_4}^{2 - } = 16,{H^ + } = 2$
D. $Mn{O_4}^ - = 2,{C_2}{O_4}^{2 - } = 5,{H^ + } = 16$

Answer 1

The unbalanced given reaction is a redox reaction that is oxidation and reduction are occurring simultaneously in the reaction. Here the reduction of permanganate ion is happening that means permanganate ion is acting as an oxidizing agent in acidic solution.

Complete Step by step answer: n this reaction, $Mn{O_4}^ - + {C_2}{O_4}^{2 - } + {H^ + } \to M{n^{2 + }} + C{O_2} + {H_2}O$, on the left side manganese Mn exhibits 7 positive charge and on the right side it contains 2 positive charge. Thus manganese is getting reduced. Carbon on the left side exhibits 3 positive charges and on the right it has 4 positive charges, hence carbon is getting oxidized. So the reaction can be divided in two half reactions one is oxidation reaction and the other is reduction reaction.
In Reduction half reaction $M{n^{7 + }}$ in $Mn{O_4}^ -$ becomes $M{n^{2 + }}$ by taking on five electrons. Any oxygen however should become ${H_2}O$ as a byproduct, and water cannot form without hydrogen ion, hence ${H^ + }$ must be added to the left side. On balancing the reaction 4 water molecules should be formed. So the balanced reduction half reaction now become-
$Mn{O_4}^ - + 5e + 8{H^ + } \to M{n^{2 + }} + 4{H_2}O$
In the oxidation half reaction, ${C_2}{O_4}^ -$ becomes $C{O_2}$ by giving two electrons. So the balanced half oxidation reaction should be-
${C_2}{O_4}^ - \to C{O_2} + 2e$
On combining both half reactions, we have to balance the number of electrons, so multiply the reduction reaction by 2 and oxidation reaction by 5. And this will bring the total number of electrons to 10 on both the half reactions. By cancelling those 10 electrons from both reactions, the total full redox reaction will now become-
$2Mn{O_4}^ - + 16{H^ + } + 5{C_2}{O_4}^ - \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$

Hence the correct option will be D, $Mn{O_4}^ - = 2,{C_2}{O_4}^{2 - } = 5,{H^ + } = 16$

Note: The manganese reduction reaction includes the gain of 10 electrons, whereas the oxalate oxidation reaction involves the loss of 10 electrons. The electrons therefore cancel. Practically speaking, the five oxalate ions transfer a total of 10 electrons to 2 permanganate ions.