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

Each gold atom gains 3 electrons, each iodine atom loses 1 electron. The total number of electrons that are moved in the oxidation-reduction reaction is 6.

Let's consider the following unbalanced redox reaction.

Au(OH)₃ + HI ⇒ Au + I₂ + H₂O

The reduction half-reaction is:

Au⁺³ + 3 e⁻ ⇒ Au

As we can see each gold atom gains 3 electrons.

The oxidation half-reaction is:

2 I⁻ ⇒ I₂ + 2 e⁻

As we can see each iodine atom loses 1 electron.

We can balance the equation by multiplying the reduction by 2, the oxidation by 3, and adding them. The total number of electrons exchanged is 6.

2 (Au⁺³ + 3 e⁻ ⇒ Au)

3 (2 I⁻ ⇒ I₂ + 2 e⁻)

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2 Au⁺³ + 6 I⁻ ⇒ 2 Au + 3 I₂

The balanced redox reaction is:

2 Au(OH)₃ + 6 HI ⇒ 2 Au + 3 I₂ + 6 H₂O

For the reaction 2 Au(OH)₃ + 6 HI ⇒ 2 Au + 3 I₂ + 6 H₂O,

Each gold atom gains 3 electrons, each iodine atom loses 2 electrons. The total number of electrons that are moved in the oxidation-reduction reaction is 6.

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