Question

Complete and balance the following chemical equations
${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + }}{{\text{I}}^{\text{ - }}} \to$

Answer 1

We can understand that this is a redox reaction. We will find the charge on Mn and I so that we can find out which is oxidized and which is reduced. The loss of electrons is called oxidation and gain of electrons is called reduction. That substance which gets oxidized is called a reducing agent and helps in reducing another substance. Moreover, the substance getting reduced is called oxidizing agent.

Complete step by step answer:
First, let us take ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ . We know that it is an ionic form of ${\text{KMn}}{{\text{O}}_{\text{4}}}$
Now, let us find the oxidation number of ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$
$x + 4( - 2) = - 1$
Therefore, we can find the charge on Mn,
x = - 1 + 4( - 2){\text{ }} \\\ x = + 7{\text{ }} \\\
So, we saw that Mn is in its highest oxidation state. Thus, it cannot donate or lose more electrons. Therefore, it cannot be oxidized.
But we know that Mn with $+ 7$ charge can be reduced to ${\text{Mn}}{{\text{O}}_{\text{2}}}$ with $+ 4$ charge.
${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }} \to {\text{Mn}}{{\text{O}}_{\text{2}}}$
As told before, when a substance gets reduced, it acts as an oxidizing agent and oxidizes the other molecule. So, now ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ oxidizes ${{\text{I}}^{\text{ - }}}$ to Iodine (${{\text{I}}_{\text{2}}}$ )
We saw that -1 charge of ${{\text{I}}^{\text{ - }}}$ is changed to Zero, thus loss of electrons takes place.
Similarly, since ${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}$ is a strong oxidizing agent, it oxidises ${{\text{H}}_{\text{2}}}{\text{O}} \to {\text{O}}{{\text{H}}^{\text{ - }}}$
Thus, the complete reaction will be
${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + }}{{\text{I}}^{\text{ - }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ + O}}{{\text{H}}^{\text{ - }}}$
Now, we will need to balance this chemical equation.
${\text{Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + }}{{\text{I}}^{\text{ - }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ + O}}{{\text{H}}^{\text{ - }}}$
Keep aside molecules with O and H, first let us balance all other elements.
First let us balance \ {\text{2M}}{{\text{n}}^{\text{7}}}{\text{ + 6}}{{\text{e}}^{\text{ - }}} \to {\text{2M}}{{\text{n}}^{\text{4}}} \\\ {\text{6}}{{\text{I}}^{\text{ - }}}{\text{ - 6}}{{\text{e}}^{\text{ - }}} \to {\text{3}}{{\text{I}}_{\text{2}}} \\\
On adding these 2 equations we get ${\text{2Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + 6}}{{\text{I}}^{\text{ - }}} \to {\text{2Mn}}{{\text{O}}_{\text{2}}}{\text{ + 3}}{{\text{I}}_{\text{2}}}$
Now we can balance H and O
${\text{2Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + 6}}{{\text{I}}^{\text{ - }}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2Mn}}{{\text{O}}_{\text{2}}}{\text{ + 3}}{{\text{I}}_{\text{2}}}{\text{ + 8O}}{{\text{H}}^{\text{ - }}}$
Thus, the answer is ${\text{2Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }}{\text{ + 6}}{{\text{I}}^{\text{ - }}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2Mn}}{{\text{O}}_{\text{2}}}{\text{ + 3}}{{\text{I}}_{\text{2}}}{\text{ + 8O}}{{\text{H}}^{\text{ - }}}$

Note:
The above mentioned equation is considered to be an example of redox reaction.
A redox reaction takes place when there will be a change of oxidation state of the atoms.
Moreover, the transfer of electrons takes place in between the species, so one species shows oxidation, and the other one reduces.
This is the reason that the above given reaction is a redox reaction.