Question

For the redox reaction:
$MnO_4^ - + {C_2}O_4^{2 - } + {H^ + }\xrightarrow{{}}M{n^{2 + }} + C{O_2} + {H_2}O$
The correct coefficients of the reactants for the balanced equation are:

| $MnO_4^ -$| ${C_2}O_4^{2 - }$| ${H^ + }$
---|---|---|---
$\left( 1 \right)$| $16$| $5$| $2$
$\left( 2 \right)$| $2$| $5$| $16$
$\left( 3 \right)$| $2$| $16$| $5$
$\left( 4 \right)$| $5$| $16$| $2$

Answer 1

When we balance a given reaction certain factors have to be kept in mind, these factors include the nature of the reaction medium i.e. Acidic, Basic, Neutral in our question we have a reaction taking place in an acidic medium as ${H^ + }$is present in the left hand side of the equation. Also we must notice the ion pair present which here are $MnO_4^ - - M{n^{2 + }}$and ${C_2}O_4^{2 - } - C{O_2}$namely.

Complete step-by-step answer:
Let us start by separating the redox half reactions and handling them individually later on adding them up:
$MnO_4^ - \xrightarrow{{}}M{n^{2 + }}$
Step 1: Balance the central atom,
Here, $Mn$ is the central atom and is balanced on either sides
Step 2: Balance the charge on central metal atom on either sides,
Here on a closer look we see that$M{n^{7 + }} \to M{n^{2 + }}$i.e. a gain of 5 electrons occur so we now add $5{e^ - }$ on the left hand side.
Step 3: Balance the number of oxygen atoms on either side,
For this we add $4{H_2}O$on the right hand side this does the job of balancing the number of oxygen atoms.
Step 4: Balance the number of hydrogen atoms and the charge on either side of equation by adding proton
So, we add $8{H^ + }$on the left hand side
Thus, the final redox half reaction is:
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$………..$\left( 1 \right)$

(B)${C_2}O_4^{2 - } \to C{O_2}$
Step 1: Balance the central atom which here is Carbon
Multiply $2$on the right hand side to balance the central atom
Step 2: Balance the charge on central atom
Looking closely we can see that$2{C^{3 + }} \to 2{C^{4 + }}$i.e. loss of 2 electrons, so we add $2{e^ - }$ on the right hand side.
Step 3: balance the oxygen atoms
Here after step 1 the number of oxygen atoms was already balanced.
Step 4: Balance the number of hydrogen atom and overall charges on either side by adding protons
As there is no Hydrogen atom present and charges are also balance thoroughly so we don’t need to add proton,

So the redox half reaction is:
${C_2}O_4^{2 - } \to 2C{O_2} + 2{e^ - }$…………….$\left( 2 \right)$
Now to get the overall balanced reaction we just need to add both the redox half reactions in a manner that the electrons on either sides of the resultant cancels each other thus equation $\left( 1 \right)$is multiplied by $2$ and equation $\left( 2 \right)$is multiplied by $5$ to get the following:
$2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$

So, the option $\left( 2 \right)$ $2$ $5$ $16$ is the correct answer.

Note: Always follow the steps mentioned above in the order they are mentioned if you fail to do so then your balanced redox half equation would come out to be completely different and absurd when added together. Also since the medium was acidic so we used proton if the medium would have been basic we would then use hydroxide ions to balance redox half reactions.