I would like to know how the products form in the following reaction: $$\ce{H2O2(aq) + 2KI(aq) + H2SO4(aq) -> I2(aq) + K2SO4(aq) + 2H2O(l)?}$$


First thing to notice is that in solution, $\ce{H2SO4}$ dissociates into $\ce{2H+}$ and $\ce{SO4^{2-}}$. Also, $\ce{KI}$ is actually $\ce{K+ + I-}$.

It is not really possible to draw a reaction mechanism for a redox reaction, but it can be split in half reactions for reduction and oxidation.

\begin{align} \ce{H2O2 +2H+ + 2e- &-> 2H2O}\tag{reduction}\\ \ce{2I- &-> I2 + 2e-}\tag{oxidation} \end{align}

After combining these half reactions, we get $$\ce{H2O2 + 2I- + 2H+ -> I2 + 2H2O}.$$

I addition to these participating ions, we also have $\ce{SO4^2-}$ and $\ce{K+}$ in solution, so called spectator ions. These would, if not dissolved, combine to create $\ce{K2SO4}$. This gives us the final equation:

$$\ce{H2O2(aq) + 2KI(aq) + H2SO4(aq) -> I2(aq) + K2SO4(aq) + 2H2O(l)}$$


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