Question

Write a balanced half-reaction describing the oxidation of solid gold to aqueous gold (III) cations.​

Answer 1

The balanced half-reaction is: Au(s) -> Au³+(aq) + 3e¯. Solid gold is oxidized to form aqueous gold (III) cations by losing three electrons.

Step-by-step explanation:

The balanced half-reaction for the oxidation of solid gold to aqueous gold (III) cations is:

Au(s) -> Au³+(aq) + 3e¯

In this reaction, solid gold (Au) is oxidized to form aqueous gold (III) cations (Au³+). It loses three electrons to achieve a +3 charge. The electrons are represented on the product side of the reaction.

Answer 2

Answer: The half reaction for the oxidation of solid gold is written below

Step-by-step explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

`X\rightarrow X^(n+)+ne^-`

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

`X^(n+)+ne^-\rightarrow X`

The half reaction for the oxidation of gold to gold (III) ions follows:

`Au(s)\rightarrow Au^(3+)(aq.)+3e^-`

Hence, the half reaction for the oxidation of solid gold is written above.