Question

Write a balanced ionic equation to represent the oxidation of iodide ion by permanganate ion in basic solution to yield molecular iodine and manganese (Iv) oxide.
Show all necessary steps

Answer 1

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.

1. **Oxidation half-reaction** (Iodide ions are oxidized to iodine):
\[
2 \, \text{I}^- \rightarrow \text{I}_2 + 2 \, \text{e}^-
\]

2. **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.

- **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 \(4 \, \text{H}_2\text{O}\)

- **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.