# Explanation for the reaction of gold and iodine(iii) solution

The Experiment
A few days ago I saw a video of dissolving gold in the condition of $$\ce{ICl3/HCl(aq)}$$. In the video, $$\ce{ICl3(s)}$$ is added to a test tube, followed by adding conc. hydrochloric acid. The solid is dissolved in the solution and the solution turned yellow. No green color was shown.
A gold wire is then placed in the solution for a short period of time. Putting the wire on the starch indicator produces a color of blue.
As a comparison, a piece of titanium was placed in the solution and then indicated by starch. No blue was produced.
The Speculation
After adding $$\ce{HCl(aq)}$$ to $$\ce{ICl3(g)}$$, I can think of 3 types of reactions may occur.
1) Lewis acid-base reaction.$$\ce{ICl3(g) +Cl-(aq)<->ICl4-(aq)}$$ As I'm not sure whether $$\ce{ICl3}$$ is a strong Lewis acid and whether $$\ce{HICl4}$$ is a strong acid (I suggest it is), I don't know it may present as a molecule in the solution.
ii) Disproportionation. As no purple color was seen, which leads to no iodine was produced.
iii) Hydration. $$\ce{ICl4-(aq) +5H2O(l)<->HIO2(aq) +4Cl-(aq) +3H3O+(aq)}$$ This reaction is unlikely to occur as $$\ce{HIO2(aq)}$$ is unstable and will disproportionate but no disproportionation production could be observed.
The OP of the video claimed that the equation of the reaction of gold and the solution is $$\ce{2HICl4 +2Au->2HAuCl4 +I2}$$

Questions
a) Does the solution of iodine(iii) really contains much $$\ce{ICl4-}$$, and it is the cause of oxidation of the gold wire?
b) Does the equation claimed above really occur?
c) Is there any paper that investigates the electrode potential of $$\ce{ICl4-/I2}$$? Hence we can support the correctness of the equation thermodynamically.

• That seems to be OK, at least approximately. – Mithoron Oct 24 '19 at 18:51