Question

Consider the reaction Au(OH)₃ +HI → Au + I₂ + H₂O.

Label the half-reactions as oxidation (use ""O"") or reduction (use ""R"").
2l⁻ → I₂ + 2e⁻
Au⁺³ + 3e⁻→ Au .

A) 2l⁻ → I₂ + 2e⁻ (R)
Au⁺³ + 3e⁻ → Au (O)
B) 2l⁻ → I₂ + 2e⁻ (O)
Au⁺³ + 3e⁻ → Au (R)
C) I₂ + 2e⁻ → 2l⁻ (O)
Au⁺³ + 3e⁻ → Au (R)`
D) I₂ + 2e⁻ → 2l⁻ (R)
Au⁺³ + 3e⁻ → Au (O)

Answer 1

The half-reaction 2I⁻ → I₂ + 2e⁻ represents oxidation, labeled as "O" because electrons are lost; Au³⁺ + 3e⁻ → Au represents reduction, labeled as "R" because electrons are gained. Thus, the correct labels are "O" for oxidation and "R" for reduction.

Step-by-step explanation:

In order to determine whether a half-reaction represents oxidation or reduction, we need to analyze the movement of electrons. In an oxidation half-reaction, electrons are lost and appear as products. Conversely, in a reduction half-reaction, electrons are gained and appear as reactants. Let's examine the provided half-reactions:

• 2I− → I2 + 2e−
• Au3+ + 3e− → Au

For the first half-reaction, iodide ions (I−) are transformed into iodine (I2), and electrons are produced in the process. This indicates oxidation since electrons are being lost. The correct label for this half-reaction is "O" for oxidation.

For the second half-reaction, gold(III) ions (Au3+) are converting into metallic gold (Au) with the gain of electrons, indicating a reduction process. The correct label for this half-reaction is "R" for reduction.

Therefore, the correct answer is:

• 2I− → I2 + 2e− (O)
• Au3+ + 3e− → Au (R)