Question
Write a balanced redox reaction for the combination of acidified KMnO4 and Iron tetraoxosulphate
Answers
Answered by
GPT-4o mini
To write the balanced redox reaction for the combination of acidified potassium permanganate (KMnO₄) and iron(II) sulfate (FeSO₄), we first need to identify the oxidation states of the elements involved and the changes that occur during the reaction.
In an acidified medium, the permanganate ion (\( \text{MnO}_4^- \)) is reduced, while the iron in iron(II) sulfate is oxidized. Here’s how to approach balancing the overall reaction:
1. **Identify the reduction half-reaction:**
The permanganate ion (\( \text{MnO}_4^- \)) is reduced to manganese(II) ion (\( \text{Mn}^{2+} \)).
\[
\text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O}
\]
2. **Identify the oxidation half-reaction:**
Iron(II) ion (\( \text{Fe}^{2+} \)) is oxidized to iron(III) ion (\( \text{Fe}^{3+} \)).
\[
\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-
\]
3. **Balance the number of electrons exchanged:**
Since 5 electrons are involved in the reduction of permanganate, we need to multiply the iron oxidation half-reaction by 5:
\[
5 \text{Fe}^{2+} \rightarrow 5 \text{Fe}^{3+} + 5 \text{e}^-
\]
4. **Combine the half-reactions:**
Now, combine the balanced half-reactions:
\[
\text{MnO}_4^- + 8 \text{H}^+ + 5 \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} + 5 \text{Fe}^{3+}
\]
5. **Final balanced redox reaction:**
The final balanced redox reaction is:
\[
\text{MnO}_4^- + 8 \text{H}^+ + 5 \text{FeSO}_4 \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} + 5 \text{Fe}_2(SO_4)_3
\]
This equation shows the stoichiometry of the oxidizing agent (KMnO₄) reacting with iron(II) sulfate in an acidic environment. The sulfate ions remain as spectator ions and are not included in the balanced reaction.
In an acidified medium, the permanganate ion (\( \text{MnO}_4^- \)) is reduced, while the iron in iron(II) sulfate is oxidized. Here’s how to approach balancing the overall reaction:
1. **Identify the reduction half-reaction:**
The permanganate ion (\( \text{MnO}_4^- \)) is reduced to manganese(II) ion (\( \text{Mn}^{2+} \)).
\[
\text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O}
\]
2. **Identify the oxidation half-reaction:**
Iron(II) ion (\( \text{Fe}^{2+} \)) is oxidized to iron(III) ion (\( \text{Fe}^{3+} \)).
\[
\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-
\]
3. **Balance the number of electrons exchanged:**
Since 5 electrons are involved in the reduction of permanganate, we need to multiply the iron oxidation half-reaction by 5:
\[
5 \text{Fe}^{2+} \rightarrow 5 \text{Fe}^{3+} + 5 \text{e}^-
\]
4. **Combine the half-reactions:**
Now, combine the balanced half-reactions:
\[
\text{MnO}_4^- + 8 \text{H}^+ + 5 \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} + 5 \text{Fe}^{3+}
\]
5. **Final balanced redox reaction:**
The final balanced redox reaction is:
\[
\text{MnO}_4^- + 8 \text{H}^+ + 5 \text{FeSO}_4 \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} + 5 \text{Fe}_2(SO_4)_3
\]
This equation shows the stoichiometry of the oxidizing agent (KMnO₄) reacting with iron(II) sulfate in an acidic environment. The sulfate ions remain as spectator ions and are not included in the balanced reaction.