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# Complete and balance the following chemical equation.$F{{e}^{2+}}+Mn{{O}_{4}}^{-}+{{H}^{+}}\to$$F{{e}^{2+}}+Mn{{O}_{4}}^{-}+{{H}^{+}}\to$

Hint: Solve the reducing half reaction and oxidizing half reaction separately and then add both reactions to get a balanced reaction.

In the given reaction we know that $Mn{{O}_{4}}^{-}$will be reduced to as $M{{n}^{2+}}$is in its highest oxidation state of 7 and the medium is acidic.
On the other side $F{{e}^{2+}}$will get oxidized to $F{{e}^{3+}}$.
We will now write the separate half reactions.
Oxidizing half reaction:

$F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}$
Reducing half reaction:

$Mn{{O}_{4}}^{-}+5{{e}^{-}}+8{{H}^{+}}\to M{{n}^{2+}}+4{{H}_{2}}O$
As we can see 5 electrons are needed to reduce one mole of $Mn{{O}_{4}}^{-}$ we multiply the oxidizing half reaction by 5 to cancel out the electrons on both the sides. We then add both the half reactions and once again check if all atoms are accounted for.

$Mn{{O}_{4}}^{-}+5F{{e}^{2+}}+8{{H}^{+}}\to M{{n}^{2+}}+5F{{e}^{3+}}+4{{H}_{2}}O$

Therefore, this is the complete and balanced chemical equation.
Additional Information:The above question could be solved by directly substituting the products and then writing the required coefficients however this does not hold good for all reactions as sometimes the electron may not be balanced, in the case of comproportionation or disproportionation reaction. That is the reason why the half reaction method is reliable as we can see the electron transfer directly and can balance them first.

Note: In some chemical reactions the same reactant undergoes oxidation as well as reduction. In such cases write the reactant in oxidation half reaction as well as in reducing half reaction and then add the two half reactions to get a final balanced chemical reaction.