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:
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Identify the reduction half-reaction: The permanganate ion (\( \text{MnO}_4^- \)) is reduced to manganese(II) ion (\( \text{Mn}^{2+} \)). \[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
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Identify the oxidation half-reaction: Iron(II) ion (\( \text{Fe}^{2+} \)) is oxidized to iron(III) ion (\( \text{Fe}^{3+} \)). \[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \]
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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: \[ 5 \text{Fe}^{2+} \rightarrow 5 \text{Fe}^{3+} + 5 \text{e}^- \]
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Combine the half-reactions: Now, combine the balanced half-reactions: \[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} + 5 \text{Fe}^{3+} \]
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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.