To write a balanced ionic equation representing the oxidation of iodide ion (I⁻) by permanganate ion (MnO₄⁻) in basic solution to yield molecular iodine (I₂) and manganese(IV) oxide (MnO₂), we follow several steps. Let’s break it down:
Step 1: Write the half-reactions.
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Oxidation half-reaction (Iodide ions are oxidized to iodine): \[ 2 , \text{I}^- \rightarrow \text{I}_2 + 2 , \text{e}^- \]
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Reduction half-reaction (Permanganate is reduced to manganese(IV) oxide): \[ \text{MnO}_4^- + 8 , \text{e}^- + 4 , \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 8 , \text{OH}^- \]
Step 2: Balance the half-reactions.
The oxidation half-reaction is already balanced with respect to charge and mass.
The reduction half-reaction involves balancing oxygen and hydrogen with water and hydroxide ions. The half-reaction as given is already balanced.
Step 3: Equalize electron transfer.
The oxidation half-reaction produces 2 electrons, but the reduction half-reaction consumes 8 electrons. We need to equalize the number of electrons transferred by multiplying the oxidation half-reaction by 4:
\[ 4 , (2 , \text{I}^- \rightarrow \text{I}_2 + 2 , \text{e}^-) \]
This gives us: \[ 8 , \text{I}^- \rightarrow 4 , \text{I}_2 + 8 , \text{e}^- \]
Now the oxidation half-reaction becomes: \[ 8 , \text{I}^- \rightarrow 4 , \text{I}_2 + 8 , \text{e}^- \]
Step 4: Combine the balanced half-reactions.
Now we can combine the oxidation and reduction half-reactions: \[ 8 , \text{I}^- + \text{MnO}_4^- + 4 , \text{H}_2\text{O} \rightarrow 4 , \text{I}_2 + \text{MnO}_2 + 8 , \text{OH}^- \]
Step 5: Confirm that the equation is balanced.
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Atoms on the left:
- Iodine (I): 8 from \(8 , \text{I}^-\)
- Manganese (Mn): 1 from \(\text{MnO}_4^-\)
- Oxygen (O): 4 from \(\text{MnO}_4^-\) + 4 from \(4 , \text{H}_2\text{O}\) = 8
- Hydrogen (H): 8 from \(4 , \text{H}_2\text{O}\)
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Atoms on the right:
- Iodine (I): 8 from \(4 , \text{I}_2\)
- Manganese (Mn): 1 from \(\text{MnO}_2\)
- Oxygen (O): 2 from \(\text{MnO}_2\) + 8 from \(8 , \text{OH}^-\) = 10
- Hydrogen (H): 8 from \(8 , \text{OH}^-\)
The total number of each type of atom is balanced.
Final Balanced Ionic Equation:
\[ 8 , \text{I}^- + \text{MnO}_4^- + 4 , \text{H}_2\text{O} \rightarrow 4 , \text{I}_2 + \text{MnO}_2 + 8 , \text{OH}^- \]
This balanced equation shows the oxidation of iodide by permanganate in a basic solution yielding molecular iodine and manganese(IV) oxide.