Question

Balance the following in basic medium:
 $ F{e_3}{O_4} + MnO_4^ - \to F{e_2}{O_3} + Mn{O_2} $
(A) $ 6F{e_3}{O_4} + 2MnO_4^ - + {H_2}O \to 9F{e_2}{O_3} + 2Mn{O_2} + 2O{H^ - } $
(B) $ 6F{e_3}{O_4} + 4MnO_4^ - + {H_2}O \to 9F{e_2}{O_3} + 5Mn{O_2} + O{H^ - } $
(C) $ 6F{e_3}{O_4} + MnO_4^ - + {H_2}O \to 9F{e_2}{O_3} + 3Mn{O_2} + 2O{H^ - } $
(D) None of these

Answer 1

Hint: To answer this question, you must recall the steps for balancing a redox reaction. A reaction in which one species is oxidized while the other is reduced is known as a redox (reduction- oxidation reaction). In the basic medium, we use OH ions to equalize charge of the ions.

Complete step by step reaction
We are given the reaction, $ F{e_3}{O_4} + MnO_4^ - \to F{e_2}{O_3} + Mn{O_2} $
First we balance all the atoms other than hydrogen and oxygen in the reaction. We get,
 $ 2F{e_3}{O_4} + MnO_4^ - \to 3F{e_2}{O_3} + Mn{O_2} $
Next, we write both the half reactions. The reduction half reaction in which $ MnO_4^ - $ reduces to $ Mn{O_2} $ with a decrease in the oxidation state of manganese from $ + 7 $ to $ + 4 $ is given as:
 $ MnO_4^ - \to Mn{O_2} $ : Manganese loses three electrons
The Oxidation half reaction in which $ F{e_3}{O_4} $ is oxidized to $ F{e_2}{O_3} $ . The oxidation of iron changes from $ + 2.67 $ to $ + 3 $ . So the change in the oxidation state of iron is $ 0.33 $ . So we can say that 1 mole of $ F{e_3}{O_4} $ loses 0.33 electrons. So 6 moles of the iron will lose a total of two electrons. Thus, the oxidation reaction can be written as:
 $ 2F{e_3}{O_4} \to 3F{e_2}{O_3} $ : Iron loses two electrons
Cross multiplying the half cells with their electrons transferred and then adding them, we get,
 $ 6F{e_3}{O_4} + 2MnO_4^ - \to 9F{e_2}{O_3} + 2Mn{O_2} $
Now, the number of electrons transferred in the reaction is balanced for both the half- cell reactions. But the charge is not yet balanced. In basic medium, to balance the charges and number of oxygen atoms and hydrogen atoms, we add hydroxyl ions and water molecules accordingly. The left hand side of the reaction carries two negative charges. So we add 2 hydroxide ions on the right side and water molecules on the left side to balance the oxygen and hydrogen atoms.
 $ 6F{e_3}{O_4} + 2MnO_4^ - + {H_2}O \to 9F{e_2}{O_3} + 2Mn{O_2} + 2O{H^ - } $
Thus, the correct answer is A.

Note
After balancing the equation, we must always recheck the equation. For this, we verify that the equation has the same type and equal number of atoms on both sides of the equation. For acidic medium, we use hydrogen ions to equalize the charge of the ions.