Re: Electrolysis of aqueous KI

Dear Finola,

I’m very impressed by the clarity of your account. I was able to repeat your experiment in detail.

Potassium iodide solutions are colourless unless they have been oxidised to form a small amount of iodine. Our lab technicians reckon that this occurs more readily in sunlight, so KI is kept in brown glass bottles.

Iodine is soluble in iodide solutions because the brown/yellow triiodide ion (I₃^-) is formed.

The reaction proceeds exactly as described. The dense, brown triiodide solution is generated from iodide at the anode and sinks to the bottom of the u-tube. I’m fairly sure that the sharp boundary you observe is just the result of this dense liquid gently building up at the bottom of the u- tube.

Initially it seems that triiodide is discharged at the cathode because its colour disappears here. Soon, though, a gas, presumably hydrogen, is evolved and this causes stirring of the solution in the region of the electrode.

The pH around the cathode is high: perhaps 14 or so. Certainly it is very alkaline. This is due to the loss of H⁺ from H₂O, leaving OH^-. This would explain why the brown/yellow colour of triiodide is not seen here: in basic solution, triiodide disproportionates.

I tried a second experiment with universal indicator added. This confirmed the pHs and also that no dense colourless solution is generated at the cathode analogous to the dense brown solution generated at the anode.

The pH around the anode is low: pH 4 or so; but this is not significantly different from the pH of the potassium iodide solution I started with.

I obtained the numbers for equations a-e from: http://www.webelements.com/webelements/elements/text/I/redn.html and added the rest of the half-equations myself.

a I₂ + 2e^- = 2I^- +0.535 V

b I₃^- + 2e^- = 3I^- +0.54 V

i.e. iodine and triiodide are essentially identical in their electrochemistry

c 2HIO + 2H⁺ + 2e^- = I₂ + 2 H₂O 1.44 V (acidic solution)

d 2IO^- + 2H₂O + 2e^- = I₂ + 4OH^- 0.42 V (basic solution)

e IO₃^- + 2H₂O + 4e^- = IO^- + 4OH^- 0.15 V (basic solution)

Equation (a) minus equation (d) gives us:

f I₂ + 2OH^- = I^- + IO^- + H₂O 0.12 V

The positive cell potential indicates that under standard conditions the reaction is spontaneous i.e iodine disproportionates in basic solution. In acidic solution [equation (a) minus (c)] it is unfavourable.

Equations (a) plus (d) gives:

g IO^- + H₂O + 2e^- = 2OH^- + I^- 0.48 V

Subtracting equation (e) from this one (g) gives:

h 3 IO^- + = 2I^- + IO₃^- 0.33 V

i.e. iodate (I) will itself disproportionate into iodide and iodate (V)

Adding equations (f) and (h):

i 3I₂ + 6OH^- = 5I^- + IO₃^- + 3H₂O 0.45 V

So iodine disproportionates to iodate (V) and iodide in basic solution. This explains why the brown colour disappears when the solution becomes basic.

See also GF Liptrot p. 334 (1983) Modern Inorganic Chemistry

Ewen McLaughlin

Disproportionation: The transformation of a substance into two or more dissimilar substances usually by simultaneous oxidation and reduction. (Merriam Webster Online Dictionary.)