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
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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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 ($\text{ClO}_3^-$).
Step 1: Write the unbalanced ionic half-reactions.
Oxidation: $\text{Cl}_2 \longrightarrow \text{ClO}_3^-$
Reduction: $\text{Cl}_2 \longrightarrow \text{Cl}^-$
Step 2: Balance atoms other than oxygen and hydrogen.
Oxidation: $\text{Cl}_2 \longrightarrow 2\text{ClO}_3^-$
Reduction: $\text{Cl}_2 \longrightarrow 2\text{Cl}^-$
Step 3: 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: $\text{Cl}_2 \longrightarrow 2\text{Cl}^-$
Step 4: 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: $\text{Cl}_2 \longrightarrow 2\text{Cl}^-$
Step 5: 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: $\text{Cl}_2 + 2\text{e}^- \longrightarrow 2\text{Cl}^-$
Step 6: 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 $\text{H}^+$.
Step 7: 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: $5(\text{Cl}_2 + 2\text{e}^- \longrightarrow 2\text{Cl}^-) \implies 5\text{Cl}_2 + 10\text{e}^- \longrightarrow 10\text{Cl}^-$
Step 8: Add the two half-reactions.
$(\text{Cl}_2 + 12\text{OH}^- \longrightarrow 2\text{ClO}_3^- + 6\text{H}_2\text{O} + 10\text{e}^-) + (5\text{Cl}_2 + 10\text{e}^- \longrightarrow 10\text{Cl}^-)$
$6\text{Cl}_2 + 12\text{OH}^- \longrightarrow
