Question
In the redox reaction 2 AI (s) + 3 F(small 2) (g) -> AI^3+ (aq), which substance is being reduced?
a. F(small 2) (g)
b. F (aq)
c. AI^3+ (aq)
d. AI (s)
a. F(small 2) (g)
b. F (aq)
c. AI^3+ (aq)
d. AI (s)
Answers
GPT-4o mini
To determine which substance is being reduced in the given redox reaction, we can analyze the oxidation states of the elements involved:
The reaction is:
\[
2 \text{Al (s)} + 3 \text{F}_2 \text{ (g)} \rightarrow \text{Al}^{3+} \text{ (aq)}
\]
1. Aluminum (Al) in its elemental form (Al (s)) has an oxidation state of 0. When it becomes \(\text{Al}^{3+}\), its oxidation state is +3. This is an increase in oxidation state, indicating that aluminum is being oxidized (losing electrons).
2. Fluorine (F_2) in its elemental form has an oxidation state of 0. When it reacts and engages in this redox process, the fluorine is transformed into fluoride ions (F⁻). In the ionic form, fluoride has an oxidation state of -1. This is a decrease in oxidation state, indicating that fluorine is being reduced (gaining electrons).
From this analysis, we see that the substance being reduced in this reaction is:
**a. F₂ (g)**
Fluorine (F₂) is reduced to fluoride ions (often represented as F⁻), which are not explicitly shown in the final equation but are implied due to the formation of Al³⁺ and the conservation of charge.
The reaction is:
\[
2 \text{Al (s)} + 3 \text{F}_2 \text{ (g)} \rightarrow \text{Al}^{3+} \text{ (aq)}
\]
1. Aluminum (Al) in its elemental form (Al (s)) has an oxidation state of 0. When it becomes \(\text{Al}^{3+}\), its oxidation state is +3. This is an increase in oxidation state, indicating that aluminum is being oxidized (losing electrons).
2. Fluorine (F_2) in its elemental form has an oxidation state of 0. When it reacts and engages in this redox process, the fluorine is transformed into fluoride ions (F⁻). In the ionic form, fluoride has an oxidation state of -1. This is a decrease in oxidation state, indicating that fluorine is being reduced (gaining electrons).
From this analysis, we see that the substance being reduced in this reaction is:
**a. F₂ (g)**
Fluorine (F₂) is reduced to fluoride ions (often represented as F⁻), which are not explicitly shown in the final equation but are implied due to the formation of Al³⁺ and the conservation of charge.