Problem 41 - General Chemistry: Principles and Modern Applications

Question

Balance these equations for disproportionation reactions. (a) \(\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}\) (basic solution) (b) \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}\) (acidic solution)

Step-by-Step Solution

Answer

The balanced equations for each reaction are: \n Reaction a: \n \(\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 5\mathrm{Cl}^{-} + \mathrm{ClO}_{3}^{-}\) \n Reaction b: \n \(3\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} + \mathrm{HSO}_{3}^{-}\)

Step 1

Assign the oxidation numbers for each species in the given reactions \n\n Reaction a: \n Chlorine in \(\mathrm{Cl}_{2}(\mathrm{g})\) has an oxidation number of 0. In \(\mathrm{Cl}^{-}\), it has an oxidation number of -1 and in \(\mathrm{ClO}_{3}^{-}\), it has an oxidation number of +5. \n\n Reaction b: \n Sulfur in \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) has an oxidation state of +4, in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), it has an oxidation number of +2 and in \(\mathrm{HSO}_{3}^{-}\) it has an oxidation number of +4.

Step 2

Create and balance half reactions for both oxidation and reduction \n\n Reaction a: \n Oxidation: \(\mathrm{Cl}\to\mathrm{ClO}_{3}^{-}+6e^{-}\) \n Reduction: \(\mathrm{Cl}+e^{-}\to\mathrm{Cl}^{-}\) \n\n Reaction b: \n Oxidation: \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} - 2e^{-}\to\mathrm{HSO}_{3}^{-}\) \n Reduction: \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\to\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+e^{-}\)

Step 3

Balance the number of electrons lost and gained in each half-reaction \n\n Reaction a: \n Multiply the reduction half-reaction by 6 and the oxidation half-reaction by 1 and then add them up: \n 6\(\mathrm{Cl} + e^{-} \to 6 \mathrm{Cl}^{-}\) \n \(\mathrm{Cl} \to \mathrm{ClO}_{3}^{-}+6e^{-}\) \n The final balanced equation for reaction a in basic solution is: \n \(\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 6 \mathrm{Cl}^{-} + \mathrm{ClO}_{3}^{-}\) \n\n Reaction b: \n Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1 and then add them up: \n 2 \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \to 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+2e^{-}\) \n \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} - 2e^{-} \to \mathrm{HSO}_{3}^{-} \) \n The final balanced equation for reaction b in acidic solution is: \n 3 \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} + \mathrm{HSO}_{3}^{-}\)

Key Concepts

Oxidation Numbers

Oxidation numbers are essential for understanding redox reactions like disproportionation. They represent the hypothetical charge an atom would have if electrons were transferred completely rather than shared. By assigning oxidation numbers, we can identify what is being oxidized and what is being reduced in a chemical reaction.

For example, in reaction (a) with chlorine gas,

• Chlorine (\(\mathrm{Cl}_{2}(\mathrm{g})\)) starts with an oxidation number of 0.
• Chloride ion (\(\mathrm{Cl}^{-}\)) takes on an oxidation number of -1, indicating it has gained an electron.
• The chlorate ion (\(\mathrm{ClO}_{3}^{-}\)) has an oxidation number of +5, showing it has lost electrons.

In reaction (b),

• Sulfur in \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) has an oxidation number of +4.
• In \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), sulfur has an oxidation number of +2, showing reduction.
• Meanwhile, in \(\mathrm{HSO}_{3}^{-}\), sulfur's number remains +4, meaning no change.

With these numbers, we clearly detect where oxidation and reduction occur.

Half Reactions

Half reactions simplify redox processes by splitting them into two parts: oxidation and reduction. They show electron movement, making balancing easier.

For chlorine (\(\mathrm{Cl}_{2}\)) in reaction (a):

• Oxidation involves chlorine turning into chlorate: \(\mathrm{Cl} o \mathrm{ClO}_{3}^{-} + 6e^{-}\). This half shows chlorine atoms losing electrons.
• Reduction is shown by: \(\mathrm{Cl} + e^{-} o \mathrm{Cl}^{-}\), where chlorine gains an electron to form chloride ions.

In reaction (b) with sulfur:

• Oxidation for sulfur involves \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} - 2e^{-} o \mathrm{HSO}_{3}^{-}\).
• Reduction yields \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} o \mathrm{S}_{2} \mathrm{O}_{3}^{2-} + e^{-}\), showing electron acceptance.

Breaking reactions into half reactions provides clarity and allows systematic balancing.

Electron Balancing

Balancing electrons is crucial in redox reactions to ensure the number of electrons lost equals those gained. This step ensures the overall charge remains balanced.

In reaction (a), we see:

• The reduction half-reaction is multiplied by 6: \(6(\mathrm{Cl} + e^{-} o \mathrm{Cl}^{-})\).
• Combine with oxidation: \(\mathrm{Cl} o \mathrm{ClO}_{3}^{-} + 6e^{-})\) for a balanced equation.

For reaction (b):

• Reduction is multiplied by 2: \(2(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} o \mathrm{S}_{2} \mathrm{O}_{3}^{2-} + 2e^{-})\).
• Combine with oxidation: \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-} - 2e^{-} o \mathrm{HSO}_{3}^{-})\).

This ensures electrons lost in oxidation are gained in reduction, balancing the reaction fully.