Step 1: Recall the dual nature of hydrogen peroxide.
Hydrogen peroxide can act as an oxidizing agent or a reducing agent, depending on what it meets. Against a strong oxidizer, it behaves as a reducing agent.
Step 2: Identify the other reactant.
Potassium periodate $KIO_4$ has iodine in the high $+7$ state, so it is a strong oxidizing agent. So here hydrogen peroxide will act as the reducing agent.
Step 3: Write the redox reaction in alkaline medium.
The periodate ion is reduced and hydrogen peroxide is oxidized: \[ IO_4^{-} + H_2O_2 \rightarrow IO_3^{-} + H_2O + O_2 \]
Step 4: Track the iodine change.
Iodine goes from $+7$ in $IO_4^-$ down to $+5$ in $IO_3^-$. This drop in oxidation state is a reduction, confirming periodate is reduced.
Step 5: Track the oxygen change.
In hydrogen peroxide oxygen is at $-1$, and it rises to $0$ in $O_2$ gas. This increase is an oxidation, confirming peroxide is oxidized.
Step 6: State the products.
So the reaction gives potassium iodate, water, and oxygen gas.
\[ \boxed{KIO_3 + H_2O + O_2} \]