Question

The correct coefficients of $MnO_4^ - $ , ${C_2}O_4^{2 - }$ and ${H^ + }$ are respectively :
\[MnO_4^ - + {C_2}O_4^{2 - } + {H^ + } \to C{O_2} + {H_2}O + M{n^{2 + }}\;\]
A.$2,5,16$
B.$16,5,2$
C.$5,16,2$
D.$2,16,5$

Answer 1

Hint:Firstly in the question we need to get the reaction we have to balance and then get the elements sorted. Then we have to apply the process in which we will balance all the elements except the oxygen and hydrogen. Then firstly we will find the change in oxidation state of carbon and manganese and then multiply the number in order to gain the balanced state. Then we will use water and protons in the end to balance the oxygen and hydrogen.

Complete answer:
The given reaction is not a balanced reaction as per the sorts. To get the coefficients right we have to make the reaction balanced at first.
Now,
We have to apply the reaction balancing step by step process:
Step 1 : in this step we have to balance the atoms other than the $H$ and $O$ .
To do that first we have to add coefficient $2$ in the $C{O_2}$ .
The $C$ is oxidized here and the oxidation number also increases from $ + 3$ to $ + 4$. Therefore the net increase will sum up to .
Step 2:
Now we can observe that the $Mn$ is reduced.
Therefore, the oxidation number of it will also decrease from $ + 7$ to $ + 2$ . Thus the net decrease in the oxidation number will be summed up to $ + 5$ .
Hence, the coefficient of $C$ will be multiplied by $5$ and the coefficient of $Mn$ is multiplied with $2$ in order to get the accurate combination.
Step 3:
Now, to balance the oxygen atoms we have to add $7{H_2}O$ molecules to the right and To balance $H$ atoms, $15$ protons are added to the left.
Hence, the balanced reaction is : \[2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 10C{O_2} + 8{H_2}O + 2M{n^{2 + }}\;\] .

Note:In a bond between two different elements, the bond's electrons are assigned to its main atomic contributor/higher electronegativity; in a bond between two atoms of the same element, the electrons are divided equally. This is because most electronegativity scales depend on the atom’s bonding state, which makes the assignment of the oxidation state a somewhat circular argument.