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To determine which half-reactions correctly fit into the given overall reaction:
[tex]\[ 3 \text{MnO}_4^{-}(\text{aq}) + 24 \text{H}^{+}(\text{aq}) + 5 \text{Fe}(\text{s}) \rightarrow 3 \text{Mn}^{2+}(\text{aq}) + 5 \text{Fe}^{3+}(\text{aq}) + 12 \text{H}_2\text{O}(\text{l}) \][/tex]
we need to break it down and analyze which half-reactions, when combined, yield the overall reaction.
Firstly, we need to identify the reduction half-reactions for manganate and the oxidation half-reactions for iron and match them to the given overall equation.
### Analyzing the Options:
1. Option 1:
[tex]\[ \text{MnO}_4^{-}(\text{aq}) + 8 \text{H}^{+}(\text{aq}) + 5 \text{e}^{-} \rightarrow \text{Mn}^{2+}(\text{aq}) + 4 \text{H}_2\text{O}(\text{l}) \][/tex]
2. Option 2:
[tex]\[ 2 \text{MnO}_4^{-}(\text{aq}) + 12 \text{H}^{+}(\text{aq}) + 6 \text{e}^{-} \rightarrow 2 \text{Mn}^{2+}(\text{aq}) + 3 \text{H}_2\text{O}(\text{l}) \][/tex]
3. Option 3:
[tex]\[ \text{Fe}(\text{s}) \rightarrow \text{Fe}^{3+}(\text{aq}) + 3 \text{e}^{-} \][/tex]
4. Option 4:
[tex]\[ \text{Fe}(\text{s}) \rightarrow \text{Fe}^{2+}(\text{aq}) + 2 \text{e}^{-} \][/tex]
5. Option 5:
[tex]\[ \text{Fe}^{2+}(\text{s}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^{-} \][/tex]
### Determining the Correct Half-Reactions:
Given the overall reaction:
[tex]\[ 3 \text{MnO}_4^{-}(\text{aq}) + 24 \text{H}^{+}(\text{aq}) + 5 \text{Fe}(\text{s}) \rightarrow 3 \text{Mn}^{2+}(\text{aq}) + 5 \text{Fe}^{3+}(\text{aq}) + 12 \text{H}_2\text{O}(\text{l}) \][/tex]
Step 1: Look at the oxidation half-reaction for Fe.
From the overall reaction, Fe solid is converted to [tex]$\text{Fe}^{3+}$[/tex]:
[tex]\[ 5 \text{Fe}(\text{s}) \rightarrow 5 \text{Fe}^{3+}(\text{aq}) \][/tex]
This suggests that iron is being oxidized directly to [tex]$\text{Fe}^{3+}$[/tex]. The half-reaction that matches this is:
[tex]\[ \text{Fe}(\text{s}) \rightarrow \text{Fe}^{3+}(\text{aq}) + 3 \text{e}^{-} \][/tex]
which is Option 3.
Step 2: Look at the reduction half-reaction for [tex]$\text{MnO}_4^{-}$[/tex].
The overall charge balance and the number of electrons must match. The reduction half-reaction must correctly balance the Mn species in the overall reaction:
Using the overall reaction stoichiometry:
[tex]\[ 3 \text{MnO}_4^{-}(\text{aq}) \rightarrow 3 \text{Mn}^{2+}(\text{aq}) \][/tex]
Matching this with the given options, Option 1 adequately suits this because it involves [tex]$\text{MnO}_4^{-}$[/tex] reducing to [tex]$\text{Mn}^{2+}$[/tex] and balances the electrons provided by the oxidation half-reaction (combined stoichiometrically by a factor that meets the overall electron balance):
[tex]\[ \text{MnO}_4^{-}(\text{aq}) + 8 \text{H}^{+}(\text{aq}) + 5 \text{e}^{-} \rightarrow \text{Mn}^{2+}(\text{aq}) + 4 \text{H}_2\text{O}(\text{l}) \][/tex]
Thus, after considering electron matching, stoichiometry, and correct species conversions, the correct two options that contribute to the overall reaction are:
[tex]\[ \boxed{3 \text{ and } 5} \][/tex]
[tex]\[ 3 \text{MnO}_4^{-}(\text{aq}) + 24 \text{H}^{+}(\text{aq}) + 5 \text{Fe}(\text{s}) \rightarrow 3 \text{Mn}^{2+}(\text{aq}) + 5 \text{Fe}^{3+}(\text{aq}) + 12 \text{H}_2\text{O}(\text{l}) \][/tex]
we need to break it down and analyze which half-reactions, when combined, yield the overall reaction.
Firstly, we need to identify the reduction half-reactions for manganate and the oxidation half-reactions for iron and match them to the given overall equation.
### Analyzing the Options:
1. Option 1:
[tex]\[ \text{MnO}_4^{-}(\text{aq}) + 8 \text{H}^{+}(\text{aq}) + 5 \text{e}^{-} \rightarrow \text{Mn}^{2+}(\text{aq}) + 4 \text{H}_2\text{O}(\text{l}) \][/tex]
2. Option 2:
[tex]\[ 2 \text{MnO}_4^{-}(\text{aq}) + 12 \text{H}^{+}(\text{aq}) + 6 \text{e}^{-} \rightarrow 2 \text{Mn}^{2+}(\text{aq}) + 3 \text{H}_2\text{O}(\text{l}) \][/tex]
3. Option 3:
[tex]\[ \text{Fe}(\text{s}) \rightarrow \text{Fe}^{3+}(\text{aq}) + 3 \text{e}^{-} \][/tex]
4. Option 4:
[tex]\[ \text{Fe}(\text{s}) \rightarrow \text{Fe}^{2+}(\text{aq}) + 2 \text{e}^{-} \][/tex]
5. Option 5:
[tex]\[ \text{Fe}^{2+}(\text{s}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^{-} \][/tex]
### Determining the Correct Half-Reactions:
Given the overall reaction:
[tex]\[ 3 \text{MnO}_4^{-}(\text{aq}) + 24 \text{H}^{+}(\text{aq}) + 5 \text{Fe}(\text{s}) \rightarrow 3 \text{Mn}^{2+}(\text{aq}) + 5 \text{Fe}^{3+}(\text{aq}) + 12 \text{H}_2\text{O}(\text{l}) \][/tex]
Step 1: Look at the oxidation half-reaction for Fe.
From the overall reaction, Fe solid is converted to [tex]$\text{Fe}^{3+}$[/tex]:
[tex]\[ 5 \text{Fe}(\text{s}) \rightarrow 5 \text{Fe}^{3+}(\text{aq}) \][/tex]
This suggests that iron is being oxidized directly to [tex]$\text{Fe}^{3+}$[/tex]. The half-reaction that matches this is:
[tex]\[ \text{Fe}(\text{s}) \rightarrow \text{Fe}^{3+}(\text{aq}) + 3 \text{e}^{-} \][/tex]
which is Option 3.
Step 2: Look at the reduction half-reaction for [tex]$\text{MnO}_4^{-}$[/tex].
The overall charge balance and the number of electrons must match. The reduction half-reaction must correctly balance the Mn species in the overall reaction:
Using the overall reaction stoichiometry:
[tex]\[ 3 \text{MnO}_4^{-}(\text{aq}) \rightarrow 3 \text{Mn}^{2+}(\text{aq}) \][/tex]
Matching this with the given options, Option 1 adequately suits this because it involves [tex]$\text{MnO}_4^{-}$[/tex] reducing to [tex]$\text{Mn}^{2+}$[/tex] and balances the electrons provided by the oxidation half-reaction (combined stoichiometrically by a factor that meets the overall electron balance):
[tex]\[ \text{MnO}_4^{-}(\text{aq}) + 8 \text{H}^{+}(\text{aq}) + 5 \text{e}^{-} \rightarrow \text{Mn}^{2+}(\text{aq}) + 4 \text{H}_2\text{O}(\text{l}) \][/tex]
Thus, after considering electron matching, stoichiometry, and correct species conversions, the correct two options that contribute to the overall reaction are:
[tex]\[ \boxed{3 \text{ and } 5} \][/tex]
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