Starting with correctly balanced half reactions, write the overall net ionic reaction in the following changes:
Chloride ion is oxidised to `Cl_(2)` by `underline(Mn)O_(4)^(Ө)` in acid solution.

Explanation

To solve the problem of balancing the redox reaction where chloride ion (Cl⁻) is oxidized to chlorine gas (Cl₂) by permanganate ion (MnO₄⁻) in acidic solution, we will follow these steps:

### Step 1: Write the oxidation half-reaction
The oxidation half-reaction involves the conversion of chloride ions to chlorine gas. The balanced half-reaction is:

\[ 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2 \text{e}^- \]

### Step 2: Write the reduction half-reaction
The reduction half-reaction involves the conversion of permanganate ion to manganese ions in acidic solution. The balanced half-reaction is:

\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 3: Equalize the number of electrons transferred
To combine the half-reactions, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. The oxidation half-reaction produces 2 electrons, while the reduction half-reaction consumes 5 electrons. To equalize, we can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:

- Multiply the oxidation half-reaction by 5:
\[ 10 \text{Cl}^- \rightarrow 5 \text{Cl}_2 + 10 \text{e}^- \]

- Multiply the reduction half-reaction by 2:
\[ 2 \text{MnO}_4^- + 16 \text{H}^+ + 10 \text{e}^- \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} \]

### Step 4: Combine the half-reactions
Now, we can add the two half-reactions together, ensuring that the electrons cancel out:

\[ 10 \text{Cl}^- + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow 5 \text{Cl}_2 + 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} \]

### Final balanced net ionic equation
The final balanced net ionic equation for the reaction is:

\[ 10 \text{Cl}^- + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow 5 \text{Cl}_2 + 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} \]