Which half-reaction correctly represents oxidation? A Fe(s) B U D Fe +2 Fe(s) → Fe +2 Fe +2 + (aq) → Fe(s) (aq) + 2e¯ (aq) + 2e¯ +2e → Fe +2 + 2e → (aq) Fe (s)

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**Oxidation Half-Reaction Identification**

**Question:**
Which half-reaction correctly represents oxidation?
   
**Options:**
   
A. \( \text{Fe}_{(s)} \rightarrow \text{Fe}^{+2}_{(aq)} + 2e^- \)

B. \( \text{Fe}^{+2}_{(aq)} \rightarrow \text{Fe}_{(s)} + 2e^- \)

C. \( \text{Fe}_{(s)} + 2e^- \rightarrow \text{Fe}^{+2}_{(aq)} \)

D. \( \text{Fe}^{+2}_{(aq)} + 2e^- \rightarrow \text{Fe}_{(s)} \)

**Explanation:**

Oxidation is a process in which an element loses electrons. This can be represented in a half-reaction where the reactant gives up electrons to form a product. Here are the provided options along with their corresponding reactions:

- **Option A:**
  \( \text{Fe}_{(s)} \rightarrow \text{Fe}^{+2}_{(aq)} + 2e^- \)  
  This reaction shows that solid iron (Fe) loses two electrons to form a ferrous ion (\( \text{Fe}^{+2} \)) in aqueous solution. This is an example of an oxidation reaction since iron loses electrons.

- **Option B:**
  \( \text{Fe}^{+2}_{(aq)} \rightarrow \text{Fe}_{(s)} + 2e^- \)  
  This reaction shows a ferrous ion (\( \text{Fe}^{+2} \)) in aqueous solution losing electrons to form solid iron (Fe). This represents reduction, where the iron ion gains electrons to revert to its metallic form.

- **Option C:**
  \( \text{Fe}_{(s)} + 2e^- \rightarrow \text{Fe}^{+2}_{(aq)} \)  
  This reaction incorrectly places electrons on the reactant side for an oxidation process, which is not possible since electrons are lost (not gained) during oxidation.

- **Option D:**
  \( \text{Fe}^{+2}_{(aq)} + 2e^- \rightarrow \text{Fe}_{(s)} \)  
  This reaction shows a reduction process where a ferrous ion (\( \text{Fe}^{+2} \))
Transcribed Image Text:**Oxidation Half-Reaction Identification** **Question:** Which half-reaction correctly represents oxidation? **Options:** A. \( \text{Fe}_{(s)} \rightarrow \text{Fe}^{+2}_{(aq)} + 2e^- \) B. \( \text{Fe}^{+2}_{(aq)} \rightarrow \text{Fe}_{(s)} + 2e^- \) C. \( \text{Fe}_{(s)} + 2e^- \rightarrow \text{Fe}^{+2}_{(aq)} \) D. \( \text{Fe}^{+2}_{(aq)} + 2e^- \rightarrow \text{Fe}_{(s)} \) **Explanation:** Oxidation is a process in which an element loses electrons. This can be represented in a half-reaction where the reactant gives up electrons to form a product. Here are the provided options along with their corresponding reactions: - **Option A:** \( \text{Fe}_{(s)} \rightarrow \text{Fe}^{+2}_{(aq)} + 2e^- \) This reaction shows that solid iron (Fe) loses two electrons to form a ferrous ion (\( \text{Fe}^{+2} \)) in aqueous solution. This is an example of an oxidation reaction since iron loses electrons. - **Option B:** \( \text{Fe}^{+2}_{(aq)} \rightarrow \text{Fe}_{(s)} + 2e^- \) This reaction shows a ferrous ion (\( \text{Fe}^{+2} \)) in aqueous solution losing electrons to form solid iron (Fe). This represents reduction, where the iron ion gains electrons to revert to its metallic form. - **Option C:** \( \text{Fe}_{(s)} + 2e^- \rightarrow \text{Fe}^{+2}_{(aq)} \) This reaction incorrectly places electrons on the reactant side for an oxidation process, which is not possible since electrons are lost (not gained) during oxidation. - **Option D:** \( \text{Fe}^{+2}_{(aq)} + 2e^- \rightarrow \text{Fe}_{(s)} \) This reaction shows a reduction process where a ferrous ion (\( \text{Fe}^{+2} \))
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