QUESTION IMAGE
Question
oxygen is acting as an oxidizing agent in all of the following reactions except
a.
2 c(s) + o₂(g) → 2 co(g)
b.
s(s) + o₂(g) → so₂(g)
c.
2 f₂(g) + o₂(g) → 2 of₂(g)
d.
2 na(s) + o₂(g) → na₂o₂(s)
e.
2 mg(s) + o₂(g) → 2 mgo(s)
To determine in which reaction oxygen is not acting as an oxidizing agent, we analyze the oxidation states of oxygen in each reaction. An oxidizing agent is a substance that gets reduced (its oxidation state decreases).
Step 1: Analyze Reaction A
In \(2C(s) + O_2(g)
ightarrow 2CO(g)\):
- Oxidation state of \(O\) in \(O_2\) is \(0\).
- In \(CO\), oxidation state of \(O\) is \(-2\).
- Oxidation state of \(O\) decreases (from \(0\) to \(-2\)), so \(O_2\) is an oxidizing agent.
Step 2: Analyze Reaction B
In \(S(s) + O_2(g)
ightarrow SO_2(g)\):
- Oxidation state of \(O\) in \(O_2\) is \(0\).
- In \(SO_2\), oxidation state of \(O\) is \(-2\).
- Oxidation state of \(O\) decreases (from \(0\) to \(-2\)), so \(O_2\) is an oxidizing agent.
Step 3: Analyze Reaction C
In \(2F_2(g) + O_2(g)
ightarrow 2OF_2(g)\):
- Oxidation state of \(O\) in \(O_2\) is \(0\).
- In \(OF_2\), oxidation state of \(O\) is \(+2\) (since \(F\) has an oxidation state of \(-1\), and for \(OF_2\), let oxidation state of \(O\) be \(x\), then \(x + 2(-1) = 0 \implies x = +2\)).
- Oxidation state of \(O\) increases (from \(0\) to \(+2\)), so \(O_2\) is being oxidized (acting as a reducing agent), not an oxidizing agent.
Step 4: Analyze Reaction D
In \(2Na(s) + O_2(g)
ightarrow Na_2O_2(s)\):
- Oxidation state of \(O\) in \(O_2\) is \(0\).
- In \(Na_2O_2\) (peroxide), oxidation state of \(O\) is \(-1\).
- Oxidation state of \(O\) decreases (from \(0\) to \(-1\)), so \(O_2\) is an oxidizing agent.
Step 5: Analyze Reaction E
In \(2Mg(s) + O_2(g)
ightarrow 2MgO(s)\):
- Oxidation state of \(O\) in \(O_2\) is \(0\).
- In \(MgO\), oxidation state of \(O\) is \(-2\).
- Oxidation state of \(O\) decreases (from \(0\) to \(-2\)), so \(O_2\) is an oxidizing agent.
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C. \(2 F_2(g) + O_2(g)
ightarrow 2 OF_2(g)\)