Questions: Identify the species that is oxidized and the species that is reduced in each reaction. Write out two half reactions to show how many electrons are gained or lost by each species. a. Mg + Fe^2+ -> Mg^2+ + Fe b. Cu^2+ + Sn -> Sn^2+ + Cu c. 4 Na + O2 -> 2 Na2 O

Identify the species that is oxidized and the species that is reduced in each reaction. Write out two half reactions to show how many electrons are gained or lost by each species.
a. Mg + Fe^2+ -> Mg^2+ + Fe
b. Cu^2+ + Sn -> Sn^2+ + Cu
c. 4 Na + O2 -> 2 Na2 O

Transcript text: H 5.48 Identify the species that is oxidized and the species that is reduced in each reaction. Write out two half reactions to show how many electrons are gained or lost by each species. a. $\mathrm{Mg}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$ b. $\mathrm{Cu}^{2+}+\mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+}+\mathrm{Cu}$ c. $4 \mathrm{Na}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{O}$

Solution

Solution Steps

Step 1: Identify Oxidation States

First, we need to determine the oxidation states of each element in the reactants and products.

a. $\mathrm{Mg} + \mathrm{Fe}^{2+} \longrightarrow \mathrm{Mg}^{2+} + \mathrm{Fe}$

$\mathrm{Mg}$ (0) to $\mathrm{Mg}^{2+}$ (+2)
$\mathrm{Fe}^{2+}$ (+2) to $\mathrm{Fe}$ (0)

b. $\mathrm{Cu}^{2+} + \mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+} + \mathrm{Cu}$

$\mathrm{Cu}^{2+}$ (+2) to $\mathrm{Cu}$ (0)
$\mathrm{Sn}$ (0) to $\mathrm{Sn}^{2+}$ (+2)

c. $4 \mathrm{Na} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{O}$

$\mathrm{Na}$ (0) to $\mathrm{Na}^{+}$ (+1)
$\mathrm{O}_{2}$ (0) to $\mathrm{O}^{2-}$ (-2)

Step 2: Determine Oxidized and Reduced Species

Next, we identify which species are oxidized and which are reduced based on the change in oxidation states.

$\mathrm{Mg}$ is oxidized (0 to +2)
$\mathrm{Fe}^{2+}$ is reduced (+2 to 0)

$\mathrm{Sn}$ is oxidized (0 to +2)
$\mathrm{Cu}^{2+}$ is reduced (+2 to 0)

$\mathrm{Na}$ is oxidized (0 to +1)
$\mathrm{O}_{2}$ is reduced (0 to -2)

Step 3: Write Half-Reactions

Now, we write the half-reactions showing the electrons gained or lost.

Oxidation: $\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+} + 2e^{-}$
Reduction: $\mathrm{Fe}^{2+} + 2e^{-} \longrightarrow \mathrm{Fe}$

Oxidation: $\mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+} + 2e^{-}$
Reduction: $\mathrm{Cu}^{2+} + 2e^{-} \longrightarrow \mathrm{Cu}$

Oxidation: $4 \mathrm{Na} \longrightarrow 4 \mathrm{Na}^{+} + 4e^{-}$
Reduction: $\mathrm{O}_{2} + 4e^{-} \longrightarrow 2 \mathrm{O}^{2-}$

Final Answer

Oxidized species: $\mathrm{Mg}$
Reduced species: $\mathrm{Fe}^{2+}$
Half-reactions:
- Oxidation: $\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+} + 2e^{-}$
- Reduction: $\mathrm{Fe}^{2+} + 2e^{-} \longrightarrow \mathrm{Fe}$

\[ \boxed{\text{Oxidized: } \mathrm{Mg}, \text{ Reduced: } \mathrm{Fe}^{2+}} \]

Oxidized species: $\mathrm{Sn}$
Reduced species: $\mathrm{Cu}^{2+}$
Half-reactions:
- Oxidation: $\mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+} + 2e^{-}$
- Reduction: $\mathrm{Cu}^{2+} + 2e^{-} \longrightarrow \mathrm{Cu}$

\[ \boxed{\text{Oxidized: } \mathrm{Sn}, \text{ Reduced: } \mathrm{Cu}^{2+}} \]

Oxidized species: $\mathrm{Na}$
Reduced species: $\mathrm{O}_{2}$
Half-reactions:
- Oxidation: $4 \mathrm{Na} \longrightarrow 4 \mathrm{Na}^{+} + 4e^{-}$
- Reduction: $\mathrm{O}_{2} + 4e^{-} \longrightarrow 2 \mathrm{O}^{2-}$

\[ \boxed{\text{Oxidized: } \mathrm{Na}, \text{ Reduced: } \mathrm{O}_{2}} \]

Was this solution helpful?