Question

Use the half-reactions of the reaction Au(OH)₃ + HI → Au+ I₂ +H₂O to answer the questions.

Reduction half-reaction: Au⁺³ +3e⁻ →Au
Oxidation half-reaction: 2l⁻ →I₂ + 2e⁻
1. How many electrons does each gold atom gain?

a) 1 electron
b) 2 electrons
c) 3 electrons
d) 6 electrons
2. How many electrons does each iodine atom lose?

a) 1 electron
b) 2 electrons
c) 3 electrons
d) 6 electrons
3. What is the total number of electrons that are moved in the oxidation-reduction reaction?

a) 2 electrons
b) 3 electrons
c) 5 electrons
d) 6 electrons
4. Complete the final balanced equation based on the half-reactions:
Au(OH)₃ + __HI -> __Au + __I₂ + __H₂O
a) 2Au(OH)₃ + 6HI -> 2Au + 3I₂ + 6H₂O
b) Au(OH)₃ + 3HI -> Au + I₂ + 3H₂O
c) Au(OH)₃ + 2HI -> Au + 2I₂ + 2H₂O
d) 3Au(OH)₃ + 2HI -> 3Au + 2I₂ + 6H₂O

Answer 1

Each gold atom gains 3 electrons in the reduction half-reaction, each iodine atom loses 2 electrons in the oxidation half-reaction, the total number of electrons moved in the oxidation-reduction reaction is 6 electrons, and the final balanced equation is Au(OH)₃ + 3HI → 2Au + 3I₂ + 3H₂O.

Step-by-step explanation:

1. Each gold atom gains 3 electrons in the reduction half-reaction: Au³⁺ + 3e⁻ → Au

2. Each iodine atom loses 2 electrons in the oxidation half-reaction: 2I⁻ → I₂ + 2e⁻

3. The total number of electrons moved in the oxidation-reduction reaction is 6 electrons.

4. The final balanced equation is: Au(OH)₃ + 3HI → 2Au + 3I₂ + 3H₂O