Question

How do you balance this redox reaction using the oxidation number method?
$F{e^{2 + }}\left( {aq} \right) + Mn{O_4}^-\left( {aq} \right) \to F{e^{3 + }}\left( {aq} \right) + M{n^{2 + }}\left( {aq} \right)$

Answer 1

In a redox reaction, at least one component gets oxidized, and at least one component gets decreased. Oxidation is the deficiency of electrons while decrease is the increase of electrons. A simple method to recall this is to think about the charges: a component's charge is diminished on the off chance that it picks up electrons. Redox responses for the most part happen in one of two conditions: acidic or fundamental. To adjust redox conditions, understanding oxidation states is essential.

Complete step by step answer:
Here is a redox response in corrosive arrangement that we will adjust:
$F{e^{2 + }}\left( {aq} \right) + Mn{O_4}^-\left( {aq} \right) \to F{e^{3 + }}\left( {aq} \right) + M{n^{2 + }}\left( {aq} \right)$)
Stage a Compose the unequal half conditions:
The oxidation half-cell reaction is,$F{e^{2 + }}\left( {aq} \right) \to F{e^{3 + }}\left( {aq} \right)$
The reduction half-cell reaction is,$Mn{O^{4 - }}\left( {aq} \right) \to M{n^{2 + }}\left( {aq} \right){\text{ }}$
Step B: Balance the half response.
Iron Half Reaction: To adjust this condition, we just need to add an electron to the correct side of the condition.
$F{e^{2 + }}\left( {aq} \right) \to {\text{ }}Fe3 + \left( {aq} \right){\text{ }} + {\text{ }}{e^ - }$
Permanganate Half Reaction: We add water to one side to adjust the four oxygens on the left.
$Mn{O_4}^ - \left( {aq} \right) \to M{n^{2 + }}\left( {aq} \right) + 4{H_2}O$
Next, to adjust the eight hydrogens in the water, we add 8H+ (aq) on the left.
$Mn{O_4}^ - \left( {aq} \right) + 8{H^ + }\left( {aq} \right) \to M{n^{2 + }}\left( {aq} \right) + 4{H_2}O$
Presently, taking a gander at the condition above, we can see the charges are unequal on the left versus the correct side. The general charge on the left is +7 and on the privilege is +2. To adjust the charge, we need to add five electrons to one side of the condition.
$Mn{O_4}^ - \left( {aq} \right) + 8{H^ + }\left( {aq} \right) + 5{e^ - } \to M{n^{2 + }}\left( {aq} \right) + 4{H_2}O$
Stage C: Duplicate the half conditions by proper numbers so they each contain a similar number of free electrons on each side. For this situation, we should increase the iron response by 5. Iron-permanganate-half-conditions
Step D: Add the half responses. The electrons on each side drop to give the general response:
$5F{e^{2 + }}\left( {aq} \right) + {\text{ }}Mn{O_4}^ - \left( {aq} \right) + 8{H^ + }\left( {aq} \right) \to {\text{ }}5F{e^{3 + }}\left( {aq} \right) + M{n^{2 + }}\left( {aq} \right) + 4{H_2}O$

Note: Now we can see the steps to balance the equation as follows,
Distinguish the oxidation number of each particle.
Decide the adjustment in oxidation number for every particle that changes.
Make the all out expansion in oxidation number equivalent to the absolute abatement in oxidation number.
Spot these numbers as coefficients before the equations containing those particles.
Balance all excess molecules other than O and H
Balance the number of oxygen atoms.
Balance the number of hydrogen atoms.
Watch that atom and charge balance.