Let’s analyze the given balanced redox reaction:
\[ \text{Fe}_2\text{O}_3(s) + 2\text{Al}(s) \rightarrow \text{Al}_2\text{O}_3(s) + 2\text{Fe}(l) \]
Answers:
l. What is the oxidation state of 2Al(s)?
- The oxidation state of elemental aluminum (Al) in \(2\text{Al}(s)\) is \(0\). Thus, for \(2\text{Al}(s)\), it is still \(0\).
m. What is the oxidation state of Fe in Fe2O3(s)?
- In \( \text{Fe}_2\text{O}_3 \), the oxidation state of Fe can be calculated as follows: Let the oxidation state of Fe be \( x \). The formula for iron(III) oxide is balanced as: \[ 2x + 3(-2) = 0 \] \[ 2x - 6 = 0 \] \[ 2x = 6 \] \[ x = +3 \] 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)?
- The oxidation state of oxygen (O) in \( \text{Fe}_2\text{O}_3 \) is \(-2\).
o. What is the oxidation state of 2Fe(l)?
- In elemental state \(2\text{Fe}(l)\), the oxidation state of iron is \(0\).
p. What is the oxidation state of Al in Al2O3(s)?
- In \( \text{Al}_2\text{O}_3 \): Let the oxidation state of Al be \( y \). The balancing gives: \[ 2y + 3(-2) = 0 \] \[ 2y - 6 = 0 \] \[ 2y = 6 \] \[ y = +3 \] 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)?
- The oxidation state of oxygen in \( \text{Al}_2\text{O}_3 \) is \(-2\).
r. What is the element that is oxidized?
- Aluminum (Al) is oxidized as it goes from an oxidation state of \(0\) in the reactant to \(+3\) in the product.
s. What is the element that is reduced?
- Iron (Fe) is reduced as it goes from an oxidation state of \(+3\) in the reactant \( \text{Fe}_2\text{O}_3 \) to \(0\) in the product \(2\text{Fe}(l)\).
t. What is the oxidation agent?
- The oxidizing agent is \( \text{Fe}_2\text{O}_3 \) because it contains iron that gains electrons (is reduced).
u. What is the reducing agent?
- The reducing agent is \( \text{Al} \) because it loses electrons (is oxidized).
v. How many electrons are transferred in the reaction as it is balanced?
- Each Al atom loses \(3\) electrons (0 to +3), and with \(2\) Al, that’s \(2 \times 3 = 6\) electrons lost. Conversely, \(2\) Fe atoms gain \(3\) electrons each from \( \text{Fe}_2\text{O}_3 \), so \(2 \times 3 = 6\) electrons are gained. Therefore, the total number of electrons transferred is \(6\).
This thoroughly answers all parts of your question regarding the redox reaction.
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