Question

Use the half-reactions of the reaction Upper A u (upper O upper H) subscript 3 plus upper H upper I right arrow upper A U plus upper I subscript 2 plus upper H subscript 2 upper O. to answer the questions.
Reduction half-reaction:Upper A u superscript plus 3, plus 3 e superscript minus right arrow upper A u.
Oxidation half-reaction:2 upper i right arrow upper I subscript 2 plus 2 e superscript minus.

How many electrons does each gold atom gain?
How many electrons does each iodine atom lose?
What is the total number of electrons that are moved in the oxidation-reduction reaction?


Complete the final balanced equation based on the half-reactions.
Au(OH)3+
HI Right arrow.
Au +
I2 +
H2O

Answer 1

- Each gold atom gains 3 electrons (Au3+ + 3 e- -> Au).
- Each iodine atom (I−) loses 1 electron (I− -> I0 + e−).
- Total electrons transferred = 6 (LCM of 3 and 2).

Balanced equation (integer coefficients):
2 Au(OH)3 + 6 HI → 2 Au + 3 I2 + 6 H2O