Redox reaction with hydrogen peroxide with potassium iodide
With all redox reactions, we should first look at all oxidation states of our reactants and products.

$$\begin{array}{ccc}
\text{Element} &amp; \text{Oxidation state in reactants} &amp; \text{Oxidation state in products} \\
\hline
\ce{H} &amp; +1 &amp; +1 \\
\ce{O} &amp; -1\text{ in }\ce{H2O2}; -2\text{ in }\ce{H2SO4} &amp; -2 \\
\ce{K} &amp; +1 &amp; +1 \\
\ce{S} &amp; +6 &amp; +6 \\
\ce{I} &amp; -1 &amp; 0
\end{array}$$

Now, we have to look at what is being oxidized and reduced:

$$\ce{H_2O_2 + KI + H_2SO_4 -&gt; K_2SO_4 + I_2 + H_2O}$$

Hydrogen has the same oxidation state of 1+.

Potassium has the same oxidation state of 1+.

Iodine is oxidized from -1 to 0.

Oxygen is reduced from -1 to -2.

For redox reactions, we have to write out our half-reactions.

$$\ce{H2O2 -&gt; H_O}$$

$$\ce{KI + H2SO4 -&gt; K2SO4 + I2}$$

For our two half-reactions, we first have to make sure everything is balanced other than hydrogen and oxygen.

For our second half-reaction, we need two moles of potassium iodide in order to balance out the potassium and iodines.

$$\ce{H2O2 -&gt; H2O}$$

$$\ce{2KI + H2SO_4 -&gt; K2SO_4 + I2}$$

In order to balance oxygens, we add molecules of water.

$$\ce{H2O2 -&gt; 2H2O}$$

$$\ce{2KI + H2SO4 -&gt; K2SO4 + I2}$$

Since we are in acidic solution, we add protons to balance hydrogen.

$$\ce{H2O2 +2H+ -&gt; H2O}$$

$$\ce{2KI + H2SO4 -&gt; K2SO4 + I2 + 2H+}$$

We add electrons to balance charge:

$$\ce{H2O2 +2H+ -&gt; 2H2O + 2e-}$$

$$\ce{2KI + H2SO4 + 2e- -&gt; K2SO4 + I2 + 2H+}$$

Now we add 'em(the half-reactions) both up and cancel out necessary terms:

$$\ce{2KI + H2SO4 + H2O2 -&gt; K2SO4 + I2 + 2H2O}$$

This differs from your original equation since I think you forgot to balance it.

Hydrogen peroxide here is reduced to water in acidic solution.