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I think I've managed to write down the correct reaction:

$$\ce{H2O2 + KI + H2SO4 -> K2SO4 + I2 + H2O}$$

Now I need to balance this using the ion-electron method in an acidic environment.

Normally this is no problem, but I'm a bit confused on how to work with the reduction of $\ce{O}$ from $\ce{-I}$ to $\ce{-II}$ and I'm not sure how to write down the total ionic equation. I've tried the following:

$\ce{2H^{+1} + O_2^{-2} + K^{+1} + I^{-1} + 2H^{+1} + SO_4^{-2} -> 2K^{+1} +SO_4^{-2} + 2I^{} + 2H^{+1} + O^{-2}}$

But I'm not sure if that is correct, and if it is correct, I'm not sure what to do from here.

After this reaction and after diluting (with water?), I also need to know what will happen if I add a few drops of a starch-solution (I will actually have to perform this experiment so if it's merely something visible and not necessarily related to the reaction, then this part is less important.)

The final question is "what property of $\ce{H2O2}$ is demonstrated here?" and I suppose the answer to that is it's function as an oxidizer?

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  • $\begingroup$ Can you provide more information as to which starch indicator that you will be using? $\endgroup$ – Jun-Goo Kwak Mar 2 '14 at 21:00
  • $\begingroup$ That's the problem, my exercise just says to "dilute and add a few drops of a starch-solution" and doesn't explicitly refer to it after that. $\endgroup$ – Joshua Mar 3 '14 at 16:59
  • $\begingroup$ Hi, I now remember. If we look at the product side, we can see that we have made elemental iodine. We can't exactly see the iodine in solution, however, if we add any starch indicator, it will form a complex with iodine, and become very dark/blue solution, signifying that iodine was made in the reaction. $\endgroup$ – Jun-Goo Kwak Mar 3 '14 at 17:46
  • $\begingroup$ "We can't exactly see the iodine in solution" is not true. You can see molecular iodine pretty easily. chemistryexplained.com/photos/iodine-3326.jpg Starch just makes the initial colour change even more obvious. $\endgroup$ – orthocresol♦ Dec 10 '15 at 17:02
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With all redox reactions, we should first look at all oxidation states of our reactants and products.

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

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

$$\ce{H_2O_2 + KI + H_2SO_4 -> K_2SO_4 + I_2 + H_2O}$$

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.

$$\ce{H2O2 -> H_O}$$

$$\ce{KI + H2SO4 -> K2SO4 + I2}$$

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.

$$\ce{H2O2 -> H2O}$$

$$\ce{2KI + H2SO_4 -> K2SO_4 + I2}$$

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

$$\ce{H2O2 -> 2H2O}$$

$$\ce{2KI + H2SO4 -> K2SO4 + I2}$$

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

$$\ce{H2O2 +2H+ -> H2O}$$

$$\ce{2KI + H2SO4 -> K2SO4 + I2 + 2H+}$$

We add electrons to balance charge:

$$\ce{H2O2 +2H+ -> 2H2O + 2e-}$$

$$\ce{2KI + H2SO4 + 2e- -> K2SO4 + I2 + 2H+}$$

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

$$\ce{2KI + H2SO4 + H2O2 -> K2SO4 + I2 + 2H2O}$$

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

Hydrogen peroxide here is reduced to water in acidic solution.

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