Question

How would I balance ${H_2}{O_2} \to {O_2}$ as an oxidation reaction?

Answer 1

To solve this question, first we will mention and explain both types of reaction such as reduction half equation and oxidation half equation. And then combine both of them to satisfy our given reaction. And finally balance the final reaction properly.

Complete step by step solution:
$+ IH - - IO - - IO - + IH$
As always, the oxidation number is the charge left on the atom of interest when the two electrons that constitute the atom-atom bond are broken with the charge assigned to the most electronegative atom... And when we do this for ${H_2}O$ we get. $2 \times H( + I) + O( - II)$ ; when we do this for $\;{F_2}O\;$ we gets $2 \times F( - I) + O( + II)$ for $HN{O_3}$ we gets $3 \times O( - II) + H( + I) + N( + V)$ . Now of course this is a FORMALISM, a useful fiction, but this is a very useful fiction when we use the oxidation states to balance chemical reactions.
But when we do this for an element-element bond, the two atoms are of equal electronegativity...and so $HO - OH\;$ gives NEUTRAL $2 \times OH$ ...i.e. a formal $O( - I)$ oxidation state.
And so reduction half equation-
$Mn{O^{ - 4}} + 8H + + 5{e^ - } \to M{n_2} + 4{H_2}O(l)$
Oxidation half equation-
${H_2}{O_2} \to {O_2} + 2H + + 2{e^ - }$
After adding both the upper equations, we have:
$2Mn{O^{ - 4}} + 16{H^ + } + 10{e^ - } + 5{H_2}{O_2} \to 2M{n_2}^ + + 8{H_2}O(l) + 5{O_2} + 10{H^ + } + 10{e^ - }$
And after cancellation we gets:
$2Mn{O^{ - 4}} + 6{H^ + } + 5{H_2}{O_2} \to 2Mn{2^ + } + 8{H_2}O(l) + 5{O_2} \uparrow$

Note:
In the oxidation number change method the underlying principle is that the gain in the oxidation number (number of electrons) in one reactant must be equal to the loss in the oxidation number of the other reactant. Step 1. Write down the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known.