Question

Given the balanced ionic equation:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Which equation represents the oxidation half reaction?
(1) Zn(s) + 2e– → Zn2+(aq)
(2) Zn(s) → Zn2+(aq) + 2e–
(3) Cu2+(aq) → Cu(s) + 2e–
(4) Cu2+(aq) + 2e– → Cu(s)

Answer 1

Answer: Option (2) is the correct answer.

Explanation:

Oxidation is the process in which there is loss of electrons or increase in oxidation number of the element.

Whereas reduction is the process in which there is gain of electrons or decrease in oxidation number of the element.

For example, $Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s)$

Oxidation-half reaction: $Zn(s) \rightarrow Zn^{2+} + 2e^{-}$

Reduction-half reaction: $Cu^{2+} + 2e^{-} \rightarrow Cu(s)$

Thus, we can conclude that the given equation Zn(s) → Zn2+(aq) + 2e– represents the oxidation half reaction.

Answer 2

The equation that represents the oxidation half reaction is Zn(s) → Zn2+(aq) + 2e–

What is oxidation reaction?

Oxidation reaction is the process in which there is loss of electrons or increase in oxidation number of the element.

From all the given half reaction, there is only one reaction with increase in oxidation number;

Zn(s) → Zn2+(aq) + 2e–

• Zn(s) - oxidation = 0
• Zn2+ - oxidation number = +2

Thus, there is increase in oxidation number of 0 to + 2.

Learn more about oxidation number here: https://brainly.com/question/27239694

#SPJ5