Question
for following balanced redox reaction answer the following questions
Fe2O3(s) + 2Al(s) = Al2O3(s) + 2Fe(l)
l. What is the oxidation state of 2Al(s)?
m. What is the oxidation state of Fe in Fe2O3(s)?
n. What is the oxidation state of O in Fe2O3(s)?
o. What is the oxidation state 2Fe(l)
p. What is the oxidation state of Al in Al2O3(s)
q. What is the oxidation state of O in Al2O3(s)?
r. What is the element that is oxidized?
s. What is the element that is reduced?
t. What is the oxidation agent?
u. What is the reducing agent?
v. How many electrons are transferred in the reaction as it is balanced?
Fe2O3(s) + 2Al(s) = Al2O3(s) + 2Fe(l)
l. What is the oxidation state of 2Al(s)?
m. What is the oxidation state of Fe in Fe2O3(s)?
n. What is the oxidation state of O in Fe2O3(s)?
o. What is the oxidation state 2Fe(l)
p. What is the oxidation state of Al in Al2O3(s)
q. What is the oxidation state of O in Al2O3(s)?
r. What is the element that is oxidized?
s. What is the element that is reduced?
t. What is the oxidation agent?
u. What is the reducing agent?
v. How many electrons are transferred in the reaction as it is balanced?
Answers
GPT-4o mini
To analyze the balanced redox reaction \( \text{Fe}_2\text{O}_3(s) + 2\text{Al}(s) \rightarrow \text{Al}_2\text{O}_3(s) + 2\text{Fe}(l) \), let's address each question one by one:
l. **What is the oxidation state of 2Al(s)?**
- The oxidation state of pure elemental aluminum (Al) is **0**. Therefore, for \( 2\text{Al}(s) \), it remains **0**.
m. **What is the oxidation state of Fe in Fe2O3(s)?**
- In \( \text{Fe}_2\text{O}_3 \), there are two iron (Fe) atoms. Let the oxidation state of iron be \( x \). The oxidation state of oxygen is usually **-2**. The equation can be set up as follows:
\[
2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies 2x = 6 \implies x = +3
\]
- Therefore, the oxidation state of Fe in \( \text{Fe}_2\text{O}_3 \) is **+3**.
n. **What is the oxidation state of O in Fe2O3(s)?**
- As established previously, the oxidation state of oxygen (O) in \( \text{Fe}_2\text{O}_3 \) is **-2**.
o. **What is the oxidation state of 2Fe(l)?**
- The oxidation state of elemental iron (Fe) in its liquid state is also **0**. Therefore, for \( 2\text{Fe}(l) \), it remains **0**.
p. **What is the oxidation state of Al in Al2O3(s)?**
- In \( \text{Al}_2\text{O}_3 \), let the oxidation state of aluminum (Al) be \( y \). The equation can be set up as follows:
\[
2y + 3(-2) = 0 \implies 2y - 6 = 0 \implies 2y = 6 \implies y = +3
\]
- Therefore, the oxidation state of Al in \( \text{Al}_2\text{O}_3 \) is **+3**.
q. **What is the oxidation state of O in Al2O3(s)?**
- As in \( \text{Fe}_2\text{O}_3 \), the oxidation state of oxygen (O) in \( \text{Al}_2\text{O}_3 \) is **-2**.
r. **What is the element that is oxidized?**
- Oxidation is the loss of electrons. In this reaction, aluminum (Al) is oxidized from an oxidation state of 0 to +3. **Al** is the element that is oxidized.
s. **What is the element that is reduced?**
- Reduction is the gain of electrons. In this reaction, iron (Fe) is reduced from an oxidation state of +3 in \( \text{Fe}_2\text{O}_3 \) to 0 in elemental \( \text{Fe} \). **Fe** is the element that is reduced.
t. **What is the oxidizing agent?**
- The oxidizing agent is the species that is reduced. In this reaction, the oxidizing agent is **Fe2O3**.
u. **What is the reducing agent?**
- The reducing agent is the species that is oxidized. In this reaction, the reducing agent is **Al**.
v. **How many electrons are transferred in the reaction as it is balanced?**
- Each aluminum atom loses 3 electrons (oxidation from 0 to +3), and since there are 2 aluminum atoms, the total number of electrons lost is:
\[
2 \times 3 = 6 \text{ electrons}
\]
- Therefore, **6 electrons** are transferred in the reaction.
In summary:
- l: 0
- m: +3
- n: -2
- o: 0
- p: +3
- q: -2
- r: Al
- s: Fe
- t: Fe2O3
- u: Al
- v: 6 electrons
l. **What is the oxidation state of 2Al(s)?**
- The oxidation state of pure elemental aluminum (Al) is **0**. Therefore, for \( 2\text{Al}(s) \), it remains **0**.
m. **What is the oxidation state of Fe in Fe2O3(s)?**
- In \( \text{Fe}_2\text{O}_3 \), there are two iron (Fe) atoms. Let the oxidation state of iron be \( x \). The oxidation state of oxygen is usually **-2**. The equation can be set up as follows:
\[
2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies 2x = 6 \implies x = +3
\]
- Therefore, the oxidation state of Fe in \( \text{Fe}_2\text{O}_3 \) is **+3**.
n. **What is the oxidation state of O in Fe2O3(s)?**
- As established previously, the oxidation state of oxygen (O) in \( \text{Fe}_2\text{O}_3 \) is **-2**.
o. **What is the oxidation state of 2Fe(l)?**
- The oxidation state of elemental iron (Fe) in its liquid state is also **0**. Therefore, for \( 2\text{Fe}(l) \), it remains **0**.
p. **What is the oxidation state of Al in Al2O3(s)?**
- In \( \text{Al}_2\text{O}_3 \), let the oxidation state of aluminum (Al) be \( y \). The equation can be set up as follows:
\[
2y + 3(-2) = 0 \implies 2y - 6 = 0 \implies 2y = 6 \implies y = +3
\]
- Therefore, the oxidation state of Al in \( \text{Al}_2\text{O}_3 \) is **+3**.
q. **What is the oxidation state of O in Al2O3(s)?**
- As in \( \text{Fe}_2\text{O}_3 \), the oxidation state of oxygen (O) in \( \text{Al}_2\text{O}_3 \) is **-2**.
r. **What is the element that is oxidized?**
- Oxidation is the loss of electrons. In this reaction, aluminum (Al) is oxidized from an oxidation state of 0 to +3. **Al** is the element that is oxidized.
s. **What is the element that is reduced?**
- Reduction is the gain of electrons. In this reaction, iron (Fe) is reduced from an oxidation state of +3 in \( \text{Fe}_2\text{O}_3 \) to 0 in elemental \( \text{Fe} \). **Fe** is the element that is reduced.
t. **What is the oxidizing agent?**
- The oxidizing agent is the species that is reduced. In this reaction, the oxidizing agent is **Fe2O3**.
u. **What is the reducing agent?**
- The reducing agent is the species that is oxidized. In this reaction, the reducing agent is **Al**.
v. **How many electrons are transferred in the reaction as it is balanced?**
- Each aluminum atom loses 3 electrons (oxidation from 0 to +3), and since there are 2 aluminum atoms, the total number of electrons lost is:
\[
2 \times 3 = 6 \text{ electrons}
\]
- Therefore, **6 electrons** are transferred in the reaction.
In summary:
- l: 0
- m: +3
- n: -2
- o: 0
- p: +3
- q: -2
- r: Al
- s: Fe
- t: Fe2O3
- u: Al
- v: 6 electrons