How do you balance the redox reaction? #"KMnO"_4 + "H"_2"O"_2 + "H"_2"SO"_4 -> "MnSO"_4 + "K"_2"SO"_4 + "O"_2 + "H"_2"O"# ?
!! LONG ANSWER !!
Before you start balancing this equation, it's important to recognize that you're in acidic medium because the reaction take place in the presence of sulfuric acid.
This means that any protons that will remain unbalanced on the reactants' side will come from the sulfuric acid.
Moreover, not all the species are relevant to the reaction. More precisely, the potassium cation and the sulfate anion are spectator ions. The net ionic equation actually looks like this
So, start by assigning oxidation numbers to all the atoms that take part in the reaction
Notice that the oxidation number of mangenese changes from
On the other hand, the oxidation number of oxygen in hydrogen peroxide,
This means that oxygen is being oxidized. The two half-reactions will look like this
- reduction half-reaction
Here, each manganese atom gains 5 electrons.
Since you're in acidic medium, you can balance the oxygen atoms by adding water to the side that needs oxygen and the hydrogen atoms by adding protons,
The reduction half-reaction will thus be
- oxidation half-reaction
Here, each oxygen atom loses one electron, so two oxygen atoms will lose 2 electrons.
Once again, balance the oxygen and hydrogen accordingly
Now, in a redox reaction, the number of electrons gained in the reduction half-reaction must be equal to the number of electrons lost in the oxidation half-reaction.
This means that you need to multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5 to get a total of
Add these two half-reactions to get
Finally, reduce the number of protons to get
Now, all those protons are coming from the sulfuric acid, which means that the balanced chemical equations will be - add the spectator ions, too
SIDE NOTE This is actually the half-reaction method, not the oxidation numbers method. The oxidation numbers method does not use water and protons (for reactions that take place in acidic medium) to balance hydrogen and oxygen.
To balance this equation using the oxidation number method, figure out the number of electrons transferred, which is of course equal to 10, multiply the compounds so that these electrons are balanced, then balance the rest of the atoms by inspection.
So, you know that you need to multiply the reduction half-reaction by
Now balance the equation by balancing the number of sulfate anions present on both sides of the equation. Notice that you have one sulfate anion on the reactants' side and three on the products side, so multiply sulfuric acid by
Finally, balance hydrogen and oxygen by multiplying the water by
And there you have it.