for the balanced equation when concentrated sodium hydroxide is added to chlorine in an aqueous mixture…

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for the balanced equation when concentrated sodium hydroxide is added to chlorine in an aqueous mixture. Under hot and concentrated alkaline conditions, chlorine disproportionates to form chloride ion

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The question asks for the balanced equation when concentrated sodium hydroxide is added to chlorine in an aqueous mixture. Under hot and concentrated alkaline conditions, chlorine disproportionates to form chloride ions ( $\text{Cl}^-$ ) and chlorate(V) ions ( $\text{ClO}_3^-$ ).

Write the unbalanced ionic half-reactions.
Oxidation: $\text{Cl}_2 \longrightarrow \text{ClO}_3^-$
Reduction: $\text{Cl}_2 \longrightarrow \text{Cl}^-$
Balance atoms other than oxygen and hydrogen.
Oxidation: $\text{Cl}_2 \longrightarrow 2\text{ClO}_3^-$
Reduction: $\text{Cl}_2 \longrightarrow 2\text{Cl}^-$
Balance oxygen atoms by adding $\text{H}_2\text{O}$ .
Oxidation: $\text{Cl}_2 + 6\text{H}_2\text{O} \longrightarrow 2\text{ClO}_3^-$
Reduction: $\text{Cl}_2 \longrightarrow 2\text{Cl}^-$
Balance hydrogen atoms by adding $\text{H}^+$ .
Oxidation: $\text{Cl}_2 + 6\text{H}_2\text{O} \longrightarrow 2\text{ClO}_3^- + 12\text{H}^+$
Reduction: $\text{Cl}_2 \longrightarrow 2\text{Cl}^-$
Balance charge by adding electrons ( $\text{e}^-$ ).
Oxidation: $\text{Cl}_2 + 6\text{H}_2\text{O} \longrightarrow 2\text{ClO}_3^- + 12\text{H}^+ + 10\text{e}^-$
Reduction: $\text{Cl}_2 + 2\text{e}^- \longrightarrow 2\text{Cl}^-$
Convert to basic medium by adding $\text{OH}^-$ to neutralize $\text{H}^+$ .
For oxidation: Add $12\text{OH}^-$ to both sides.
$\text{Cl}_2 + 6\text{H}_2\text{O} + 12\text{OH}^- \longrightarrow 2\text{ClO}_3^- + 12\text{H}^+ + 12\text{OH}^- + 10\text{e}^-$
$\text{Cl}_2 + 6\text{H}_2\text{O} + 12\text{OH}^- \longrightarrow 2\text{ClO}_3^- + 12\text{H}_2\text{O} + 10\text{e}^-$
Simplify $\text{H}_2\text{O}$ :
$\text{Cl}_2 + 12\text{OH}^- \longrightarrow 2\text{ClO}_3^- + 6\text{H}_2\text{O} + 10\text{e}^-$
The reduction half-reaction remains the same as it has no $\text{H}^+$ .
Equalize electrons by multiplying the reduction half-reaction by 5.
Oxidation: $\text{Cl}_2 + 12\text{OH}^- \longrightarrow 2\text{ClO}_3^- + 6\text{H}_2\text{O} + 10\text{e}^-$
Reduction: $5(\text{Cl}_2 + 2\text{e}^- \longrightarrow 2\text{Cl}^-) \implies 5\text{Cl}_2 + 10\text{e}^- \longrightarrow 10\text{Cl}^-$
Add the two half-reactions.
$(\text{Cl}_2 + 12\text{OH}^- \longrightarrow 2\text{ClO}_3^- + 6\text{H}_2\text{O} + 10\text{e}^-) + (5\text{Cl}_2 + 10\text{e}^- \longrightarrow 10\text{Cl}^-)$
$6Cl_2 + 12OH^- \longrightarrow

For the balanced equation when concentrated sodium hydroxide is added to chlorine in an aqueous mixture. Under hot and concentrated alkaline conditions, chlorine disproportionates to form chloride ion

Step-by-step solution

Write the unbalanced ionic half-reactions.

Balance atoms other than oxygen and hydrogen.

Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^+$ .

Balance charge by adding electrons ( $\text{e}^-$ ).

Convert to basic medium by adding $\text{OH}^-$ to neutralize $\text{H}^+$ .

Equalize electrons by multiplying the reduction half-reaction by 5.

Add the two half-reactions.

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