Question

Write a balanced net ionic equation for the disproportionation reaction (a) of iodine to give iodate and iodide ions in basic solution. (b) of chlorine gas to chloride and perchlorate ions in basic solution.

Short answer

Question: Write balanced net ionic equations for the disproportionation reactions of (a) iodine to give iodate and iodide ions in basic solution, and (b) chlorine gas to form chloride and perchlorate ions in a basic solution. Answer: (a) 2 I₂ + 12 OH⁻ -> 2 IO₃⁻ + 5 I₂ + 6 H₂O (b) 2 Cl₂ + 20 OH⁻ -> 2 ClO₄⁻ + 10 Cl₂ + 10 H₂O

Step-by-step solution

Identify reactants and products

In this case, iodine (I\(_2\)) reacts to form iodate (IO\(_3^-\)) and iodide (I\(^-\)) ions in a basic solution.

Write the two half-reactions

We need to write two half-reactions for oxidation and reduction. Oxidation half-reaction: I\(_2\) -> IO\(_3^-\) Reduction half-reaction: I\(_2\) -> I\(^-\)

Balance the half-reactions

First, balance the Iodine atoms, and then balance the charges by adding electrons (e\(^-\)). Oxidation half-reaction: I\(_2\) -> 2 IO\(_3^-\) + 10e\(^-\) Reduction half-reaction: I\(_2\) + 2e\(^-\) -> 2 I\(^-\)

Balance final net ionic equation

Multiply the appropriate factor to each half-reaction to make electron transfer equal, then add both half-reactions and eliminate the electrons. Finally, use water (H\(_2\)O) and hydroxide ions (OH\(^-\)) to balance oxygen and hydrogen atoms. Balanced net ionic equation: 2 I\(_2\) + 12 OH\(^-\) -> 2 IO\(_3^-\) + 5 I\(_2\) + 6 H\(_2\)O (b) Disproportionation reaction of chlorine gas to chloride and perchlorate ions in basic solution.

Identify reactants and products

In this case, chlorine gas (Cl\(_2\)) reacts to form chloride (Cl\(^-\)) and perchlorate (ClO\(_4^-\)) ions in a basic solution.

Write the two half-reactions

We need to write two half-reactions for oxidation and reduction. Oxidation half-reaction: Cl\(_2\) -> ClO\(_4^-\) Reduction half-reaction: Cl\(_2\) -> Cl\(^-\)

Balance the half-reactions

First, balance the Chlorine atoms, and then balance the charges by adding electrons (e\(^-\)). Oxidation half-reaction: Cl\(_2\) -> 2 ClO\(_4^-\) + 10e\(^-\) Reduction half-reaction: Cl\(_2\) + 2e\(^-\) -> 2 Cl\(^-\)

Balance final net ionic equation

Multiply the appropriate factor to each half-reaction to make electron transfer equal, then add both half-reactions and eliminate the electrons. Finally, use water (H\(_2\)O) and hydroxide ions (OH\(^-\)) to balance oxygen and hydrogen atoms. Balanced net ionic equation: 2 Cl\(_2\) + 20 OH\(^-\) -> 2 ClO\(_4^-\) + 10 Cl\(_2\) + 10 H\(_2\)O