Answer:
Al → Al³⁺ + 3 e⁻, is an oxidation process.
S²⁻ → S + 2 e⁻, is an oxidation process.
Cu → Cu²⁺ + 2 e⁻, is an oxidation process.
Explanation:
- The oxidation-reduction reaction contains a reductant and an oxidant (oxidizing agent).
- The oxidation process is the process in which electrons are lost and produce positively charged ions.
- The reduction process is the process in which electrons is gained and negatively charge ions are produced.
So,
- Al → Al³⁺ + 3 e⁻, is an oxidation process.
Al loses 3 electrons and produce Al³⁺c
- S²⁻ → S + 2 e⁻, is an oxidation process.
S²⁻ loses 2 electrons and produce S.
- Cu → Cu²⁺ + 2 e⁻, is an oxidation process.
Cu loses 2 electrons and produce Cu²⁺.
