Electrolysis of Sodium Iodide and Water

This informal experiment is the electrolysis of sodium iodide and water. It would be advisable to check that current flows through the pencils, as the pencil lead may be broken. If you have access to graphite rods they can be used instead. In the second part of the experiment, there should be an obvious colour change when the phenolphthalein and $\text{NaOH}$ mix at the cathode.

If you have access to a laboratory, you are required to work with a concentrated, strong acid. Concentrated, strong acids can cause serious burns. Please be reminded that you have to be careful and wear the appropriate safety equipment when handling all chemicals, especially concentrated acids. The safety equipment includes gloves, safety glasses and protective clothing.

Optional Experiment: The electrolysis of sodium iodide and water

Aim

To study the electrolysis of water and sodium iodide.

Apparatus

$\text{4}$ pencils, copper wire attached to crocodile clips, $\text{9}$ $\text{V}$ battery, $\text{2}$ beakers, pencil sharpener, spatula, glass rod
distilled water, sodium iodide ($\text{NaI}$), phenolphthalein, $\text{1}$ $\text{mol.dm$^{-3}$}$ sulfuric acid ($\text{H}_{2}\text{SO}_{4}$) solution

Method

Label one beaker 1 and half fill it with distilled $\text{H}_{2}\text{O}$.
Sharpen both ends of two of the pencils. Strip away some of the wood to expose more of the graphite. (Graphite rods can be used instead of pencils if they are available).
Attach one end of the crocodile clips to the pencils and the other end to the battery.
Place the pencils in the beaker, making sure they are not touching each other. Observe what happens.
Pour approximately $\text{5}$ $\text{cm$^{3}$}$ of the $\text{H}_{2}\text{SO}_{4}$ solution into the beaker. Observe what happens.
Label a second beaker 2 and add $\text{5}$ spatula tips of $\text{NaI}$ to the beaker.
Half fill beaker $\text{2}$ with distilled $\text{H}_{2}\text{O}$ and stir with the glass rod until all the sodium iodide has dissolved.
Repeat steps $\text{2}$ - $\text{4}$ with the second set of pencils.
After a few minutes add $\text{3}$ - $\text{4}$ drops of the phenolphthalein to the beaker. Observe what happens.

Questions

In beaker 1:
1. What happened when the pencils were first put in the water?
2. What happened when the sulfuric acid was added to the beaker?
3. Why is the sulfuric acid necessary in this reaction?
4. What is happening at the negative electrode (the pencil attached to the negative terminal of the battery)?
5. What is happening at the positive electrode (the pencil attached to the positive terminal of the battery)?
6. Which electrode is the anode and which is the cathode?
In beaker 2:
1. What happened when the pencils were first put in the water?
2. What is happening at the negative electrode?
3. What is happening at the positive eletrode?
4. Which electrode is the anode and which is the cathode?
5. What happened when you added the phenolphthalein? Why did the change take place?

Results

In the electrolysis of water two $\text{H}^{+}$ ions each gain an electron (are reduced) and combine to form hydrogen gas ($\text{H}_{2}(\text{g})$):

$2\text{H}^{+}(\text{aq}) + 2\text{e}^{-}$ $\to$ $\text{H}_{2}(\text{g})$

Two $\text{O}^{2-}$ ions each lose two electrons (are oxidised) and combine to form oxygen gas ($\text{O}_{2}(\text{g})$):

$2\text{O}^{2-}(\text{aq})$ $\to$ $\text{O}_{2}(\text{g}) + 4\text{e}^{-}$
When the positive $\text{H}^{+}$ ions encounter the negative electrode they are reduced. Reduction is a gain of electrons at the cathode, therefore the negative electrode is the cathode.
When the negative $\text{O}^{2-}$ ions encounter the positive electrode they are oxidised. Oxidation is a loss of electrons at the anode, therefore the positive electrode is the anode.
Remember that pure water does not conduct electricity. So, an electrolyte (such as sulfuric acid) is necessary for the reaction to take place.
A salt dissolved in the water is also an electrolyte, so sulfuric acid is not necessary in the second beaker.
The electrolysis of water occurs in beaker 2, but there are other reactions taking place as well because of the presence of the $\text{Na}^{+}$ and $\text{I}^{-}$ ions.
When the negative $\text{I}^{-}$ ions encounter the positive electrode they are oxidised:

$2\text{I}^{-}(\text{aq})$ $\to$ $\text{I}_{2}(\text{s}) + 2\text{e}^{-}$

You should have observed solid iodine forming at the positive electrode.
When the positive $\text{Na}^{+}$ ions encounter the negative electrode they are reduced:

$\text{Na}^{+}(\text{aq}) + \text{e}^{-}$ $\to$ $\text{Na}(\text{s})$

$\text{Na}(\text{s})$ is very reactive with water:

$2\text{Na}(\text{s}) + 2\text{H}_{2}\text{O}(\text{l})$ $\to$ $2\text{NaOH}(\text{aq}) + \text{H}_{2}(\text{g})$.
Remember that phenolphthalein turns pink in the presence of a base. So when the phenolphthalein is added, the water around the negative electrode should become pink due to the $\text{NaOH}$.

Conclusion

The application of an electric current to water splits the water molecules and causes hydrogen and oxygen gas to form. This can only happen in the presence of an electrolyte (for example sulfuric acid). Sodium iodide dissolved in the water is also an electrolyte and enables the electrolysis of water. However, the cation and anion of the salt will also undergo a reaction. $\text{NaOH}$ will form at the cathode, while solid iodine will form at the anode.