Respuesta :
The answer: C) each copper atom loses one (1) electron.
2Cu + S ⇒ Cu₂S
2Cu⁰ ⇒ 2Cu⁺ + 2e⁻ oxydation
2e⁻ + S⁰ ⇒ S⁻² reduction
In the reaction, copper is oxidized and it releases an electron and transfers it to the sulfur atom. In contrast to copper, sulfur is reduced and it receives two electrons.
2Cu + S ⇒ Cu₂S
2Cu⁰ ⇒ 2Cu⁺ + 2e⁻ oxydation
2e⁻ + S⁰ ⇒ S⁻² reduction
In the reaction, copper is oxidized and it releases an electron and transfers it to the sulfur atom. In contrast to copper, sulfur is reduced and it receives two electrons.
Answer: Option (c) is the correct answer.
Explanation:
The given reaction equation is as follows.
[tex]2Cu(s) + S(s) \rightarrow Cu_{2}S(s)[/tex]
The oxidation-reduction half reactions will be as follows.
Oxidation : [tex]Cu(s) \rightarrow Cu^{+} + 1e^{-}[/tex]
Reduction: [tex]S(s) + 2e^{-} \rightarrow S^{2-} [/tex]
By balancing the above half reactions, oxidation-half reaction will be as follows.
[tex]2Cu(s) \rightarrow 2Cu^{+} + 2e^{-}[/tex]
Whereas reduction-half reaction will remain the same.
Hence, it shows that each copper atom is losing one electron to combine with a sulfur atom.
Therefore, we can conclude that each copper atom loses one (1) electron.