Question

Question #efb20

Answer 1

That actually depends on how dilute the solution is.

Explanation:

The anode is where oxidation takes place, meaning that the species discharged at the anode will be the one that's more readily oxidized.

If you look at the standard electrode potentials for the hydroxide and iodide ions, you'll see that

`O_(2(g)) + 2H_2O_((l)) + 4e^(-) rightleftharpoons 4OH_((aq))^(-)`, `" "E^@ = "+0.401 V"`

`I_(2(s)) + 2e^(-) rightleftharpoons 2I_((aq))^(-)`, `" "E^@ = "+0.54 V"`

Now, remember what oxidation means. When something is being oxidized, it loses electrons. In your case, this implies that the easier it is for an ion to lose electrons, the easier it will be oxidized, and thus discharged at the anode.

Take a look at the two equilibrium reactions. Since both `E^@` values are positive, both equilibria will lie to the right,

In order for the ions to be oxidized, you need the equilibrium to shift left. This will happer easier for the ion that has the less positive `E^@` value, which in your case is the hydroxide ion.

So, if the solution is dilute enough, i.e. if you have comparable concentrations of `OH^(-)` and `I^(-)` ions, then the hydroxide ion will be discharged at the anode and oxygen gas, `O_2`, will be produced.

If, however, the solution is concentrated, i.e. you have more `I^(-)` than `OH^(-)` ions present, then iodide will be oxidized and a dark brown solution, which is `I_(2(aq))`, will be formed at the anode.

I recommend this excellent video demonstrating the electrolysis of potassium iodide, `KI`: