Question
In the redox reaction 2 AI (s) + 3 F(small 2) (g) -> AI^3+ (aq), which substance is being oxidized?
a. AI (s)
b. F (aq)
c. F(small 2)(g)
d. AI^3+ (aq)
a. AI (s)
b. F (aq)
c. F(small 2)(g)
d. AI^3+ (aq)
Answers
GPT-4o mini
In the given redox reaction:
\[ 2 \text{Al (s)} + 3 \text{F}_2 \text{(g)} \rightarrow 2 \text{Al}^{3+} \text{(aq)} + 6 \text{F}^- \text{(aq)} \]
we can determine which substance is being oxidized by looking at the change in oxidation states:
1. Aluminum (Al) starts with an oxidation state of 0 in its elemental form (Al(s)) and is oxidized to +3 (Al³⁺), indicating it loses electrons.
2. Fluorine (F₂) starts with an oxidation state of 0 as a diatomic molecule and is reduced when it gains electrons to become fluoride ions (F⁻).
Since oxidation involves the loss of electrons and an increase in oxidation state, aluminum (Al) is the substance being oxidized in this reaction.
Thus, the correct answer is:
a. Al (s)
\[ 2 \text{Al (s)} + 3 \text{F}_2 \text{(g)} \rightarrow 2 \text{Al}^{3+} \text{(aq)} + 6 \text{F}^- \text{(aq)} \]
we can determine which substance is being oxidized by looking at the change in oxidation states:
1. Aluminum (Al) starts with an oxidation state of 0 in its elemental form (Al(s)) and is oxidized to +3 (Al³⁺), indicating it loses electrons.
2. Fluorine (F₂) starts with an oxidation state of 0 as a diatomic molecule and is reduced when it gains electrons to become fluoride ions (F⁻).
Since oxidation involves the loss of electrons and an increase in oxidation state, aluminum (Al) is the substance being oxidized in this reaction.
Thus, the correct answer is:
a. Al (s)