Problem 97 - Chemistry The Central Science In Si Units

Question

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)\) (b) \(\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)\) (acidic solution) (c) \(\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)\) (acidic solution) (d) \(\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution)

Step-by-Step Solution

Answer

(a) \(3\mathrm{Fe}^{2+} \to 2\mathrm{Fe}^{3+} + \mathrm{Fe}(s)\); (b) \(3\mathrm{Br}_{2}(l) + 6\mathrm{H}_{2}\mathrm{O} \to 5\mathrm{Br}^{-}(aq) + \mathrm{BrO}_{3}^{-}(aq) + 6\mathrm{H}^{+}(aq)\); (c) \(2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} \to \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{H}^{+} + 6\mathrm{Cr}(s)\); (d) \(3\mathrm{NO}(g) \to 2\mathrm{N}_{2}(g) + \mathrm{NO}_{3}^{-}(aq) + 2\mathrm{H}^{+}\)."

Step 1: Identify Oxidation and Reduction Half-Reactions (a)

**For part (a):**The reaction given is \[ \mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q) \]- **Reduction Half-Reaction**: \[ \mathrm{Fe}^{2+}(aq) + 2e^- \longrightarrow \mathrm{Fe}(s) \] Here, \( \mathrm{Fe}^{2+} \) gains electrons to form \( \mathrm{Fe} \).- **Oxidation Half-Reaction**: \[ \mathrm{Fe}^{2+}(aq) \longrightarrow \mathrm{Fe}^{3+}(aq) + e^- \] Here, \( \mathrm{Fe}^{2+} \) loses an electron to form \( \mathrm{Fe}^{3+} \).Combine these in a way so the electrons cancel each other.

Step 2: Balance the Half-Reactions for Electrons (a)

- Multiply the oxidation half-reaction by 2 to balance the electrons: \[ 2 \times (\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+} + e^-) \] which gives: \[ 2\mathrm{Fe}^{2+} \longrightarrow 2\mathrm{Fe}^{3+} + 2e^- \]- The reduction half-reaction is: \[ \mathrm{Fe}^{2+} + 2e^- \longrightarrow \mathrm{Fe} \]The combined balanced equation is:\[ 3\mathrm{Fe}^{2+} \longrightarrow 2\mathrm{Fe}^{3+} + \mathrm{Fe}(s) \]

Step 3: Identify Oxidation and Reduction Half-Reactions (b)

**For part (b):**The reaction given is\[ \mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(aq)+\mathrm{BrO}_{3}^{-}(aq) \]- **Reduction Half-Reaction**: \[ \mathrm{Br}_{2}(l) + 2e^- \longrightarrow 2\mathrm{Br}^{-}(aq) \]- **Oxidation Half-Reaction**: \[ \mathrm{Br}_{2}(l) \longrightarrow 2\mathrm{BrO}_{3}^{-}(aq) + 12e^- \] Then balance for electrons by multiplying the reduction by 6.

Step 4: Balance the Half-Reactions for Electrons and Atoms (b)

- Multiply the reduction half-reaction by 6: \[ 6\mathrm{Br}_{2}(l) + 12e^- \longrightarrow 12\mathrm{Br}^{-}(aq) \]- The oxidation half-reaction becomes: \[ \mathrm{Br}_{2}(l) \longrightarrow 2\mathrm{BrO}_{3}^{-}(aq) + 12e^- \]- Combine the balanced reactions:\[ 6\mathrm{Br}_{2}(l) + 12e^- \longrightarrow 12\mathrm{Br}_{-}(aq) \]\[ 2\mathrm{Br}_{2}(l) + 12e^- \longrightarrow 4\mathrm{BrO}_{3}^{-}(aq) \]Balanced equation:\[ 3\mathrm{Br}_{2}(l) + 6\mathrm{H}_{2}\mathrm{O} \longrightarrow 5\mathrm{Br}^{-}(aq) + \mathrm{BrO}_{3}^{-}(aq) + 6\mathrm{H}^{+}(aq) \]

Step 5: Identify Oxidation and Reduction Half-Reactions (c)

**For part (c):**The reaction given is\[ \mathrm{Cr}^{3+}(aq) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(aq)+\mathrm{Cr}(s) \]- **Reduction Half-Reaction**: \[ \mathrm{Cr}^{3+} + 3e^- \longrightarrow \mathrm{Cr}(s) \]- **Oxidation Half-Reaction**: \[ 2\mathrm{Cr}^{3+} \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6e^- \] Balance oxygen with water and hydrogen with hydrogen ions.

Step 6: Balance the Half-Reactions for Electrons and Atoms (c)

- Balance the oxidation half with water molecules: \[ 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{H}^{+} + 6e^- \]- Combine the reduction: \[ 6\mathrm{Cr}^{3+} + 3e^- \longrightarrow 3\mathrm{Cr}(s) \]Balanced equation:\[ 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{H}^{+} + 6\mathrm{Cr}(s) \]

Step 7: Identify Oxidation and Reduction Half-Reactions (d)

**For part (d):**The reaction given is\[ \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(aq) \]- **Reduction Half-Reaction**: \[ 2\mathrm{NO}(g) + 2e^- \longrightarrow \mathrm{N}_{2}(g) + \mathrm{O}^{2-}(aq) \]- **Oxidation Half-Reaction**: \[ \mathrm{NO}(g) \longrightarrow \mathrm{NO}_{3}^{-}(aq) + 2e^- \]

Step 8: Balance Full Reaction (d)

- Combine and balance both reactions: \[ 3\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g) + 2\mathrm{NO}_{3}^{-}(aq) + 2\mathrm{H}^{+} \] - Add water to balance oxygens where needed. - Electrons will cancel due to the same amount.

Key Concepts

Oxidation-Reduction Reactions

An oxidation-reduction reaction, also known as a redox reaction, is a chemical process where oxidation and reduction occur simultaneously. In these reactions, electrons are transferred between substances, which leads to changes in oxidation states of the elements involved.

Oxidation is defined as the loss of electrons, leading to an increase in oxidation state, while reduction is the gain of electrons, resulting in a decrease in oxidation state.

• In a redox reaction, the substance that loses electrons is called the reducing agent as it donates electrons to another substance.
• Conversely, the substance that gains electrons is known as the oxidizing agent since it accepts electrons from another species.

Understanding how electrons are transferred between these agents helps in recognizing how substances change during a reaction. The distinct roles of substances in redox reactions are crucial for processes like corrosion, combustion, and even in biological systems. Applying knowledge of redox reactions assists in determining both reactant and product species in chemical equations.

Half-Reaction Method

The half-reaction method is a systematic way to balance redox equations, particularly useful in complex reactions such as disproportionation. Disproportionation is a form of redox reaction where one substance is both oxidized and reduced.

In this method, you break down the overall equation into two separate equations focusing individually on oxidation and reduction processes.

• Each half-reaction is balanced separately first for mass, and then for charge by adding electrons (e")).
• Once both half-reactions are balanced, they are recombined ensuring the electrons cancel out, resulting in a fully balanced equation.
• This method allows for precise balancing, especially for electron transfer, which can be difficult by simple inspection.

The half-reaction method ensures correct stoichiometry and maintains electrical neutrality overall, making it essential for complex reactions involving multiple electrons or when the reaction occurs in stages.

Acidic Solution Balancing

Balancing redox reactions in acidic solutions involves additional steps compared to neutral environments. An acidic medium requires maintaining the acid-base balance using hydrogen ions (H\(^+\)) and water.

Here's how you can balance reactions in an acidic solution using the half-reaction method:

• Both half-reactions are first balanced by mass and charge as usual.
• Next, if oxygen atoms need balancing, water (H\(_2\)O) is added to the side lacking oxygen.
• After balancing oxygen, hydrogen ions (H\(^+\)) are added to balance the hydrogen atoms added via water.

Finally, the electrons are adjusted so they cancel out between the half-reactions when combined, completing the balanced equation.

This approach ensures the reaction is consistent with an acidic environment where you may have excess H\(^+\) ions. This is vital for reactions commonly occurring in nature and several industrial processes, including electrochemical cells and certain types of synthesis.