Problem 32 - Chemistry Principles and Practice

Question

Balance each of the following redox reactions in acid solution. (a) \(\mathrm{Zn}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{N}_{2}(\mathrm{~g})\) (b) \(\mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_{2}(\mathrm{aq})\) (c) \(\mathrm{Ce}^{4+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{aq})+\mathrm{Ce}^{3+}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) \(5\mathrm{Zn} + 2\mathrm{NO}_{3}^{-} + 12\mathrm{H}^{+} \rightarrow 5\mathrm{Zn}^{2+} + \mathrm{N}_{2} + 6\mathrm{H}_{2}\mathrm{O}\). (b) \(\mathrm{IO}_{3}^{-} + 6\mathrm{I}^{-} + 6\mathrm{H}^{+} \rightarrow 3\mathrm{I}_{2} + 3\mathrm{H}_{2}\mathrm{O}\). (c) \(2\mathrm{Ce}^{4+} + 2\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2\mathrm{Ce}^{3+}\).

Step 1: Write half-reactions for Reaction (a)

For the reaction \( \mathrm{Zn} + \mathrm{NO}_{3}^{-} \rightarrow \mathrm{Zn}^{2+} + \mathrm{N}_{2} \), we need to split it into two half-reactions. Oxidation: \( \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^{-} \).Reduction: \( \mathrm{NO}_{3}^{-} + 10\mathrm{e}^{-} + 12\mathrm{H}^{+} \rightarrow \mathrm{N}_{2} + 6\mathrm{H}_{2}\mathrm{O} \).

Step 2: Balance electrons in Reaction (a)

Multiply the oxidation half-reaction by 5 to balance the electrons with the reduction half-reaction: Oxidation: \( 5(\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{e}^{-}) \Rightarrow 5\mathrm{Zn} \rightarrow 5\mathrm{Zn}^{2+} + 10\mathrm{e}^{-} \).The number of electrons in both half-reactions is now equal.

Step 3: Write balanced reaction for Reaction (a)

Combine the balanced half-reactions: \( 5\mathrm{Zn} + 2\mathrm{NO}_{3}^{-} + 12\mathrm{H}^{+} \rightarrow 5\mathrm{Zn}^{2+} + \mathrm{N}_{2} + 6\mathrm{H}_{2}\mathrm{O} \).This is the balanced equation for Reaction (a) in acid solution.

Step 4: Write half-reactions for Reaction (b)

For the reaction \( \mathrm{IO}_{3}^{-} + \mathrm{I}^{-} \rightarrow \mathrm{I}_{2} \), split it into two half-reactions:Oxidation: \( 2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2} + 2\mathrm{e}^{-} \).Reduction: \( \mathrm{IO}_{3}^{-} + 6\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow \mathrm{I}_{2} + 3\mathrm{H}_{2}\mathrm{O} \).

Step 5: Balance electrons in Reaction (b)

Multiply the oxidation half-reaction by 3 to balance the electrons with the reduction half-reaction:Oxidation: \( 6\mathrm{I}^{-} \rightarrow 3\mathrm{I}_{2} + 6\mathrm{e}^{-} \).Now the number of electrons in both half-reactions is equal.

Step 6: Write balanced reaction for Reaction (b)

Combine the balanced half-reactions:\( \mathrm{IO}_{3}^{-} + 6\mathrm{I}^{-} + 6\mathrm{H}^{+} \rightarrow 3\mathrm{I}_{2} + 3\mathrm{H}_{2}\mathrm{O} \).This is the balanced equation for Reaction (b) in acid solution.

Step 7: Write half-reactions for Reaction (c)

For the reaction \( \mathrm{Ce}^{4+} + \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + \mathrm{Ce}^{3+} \), split it into two half-reactions:Oxidation: \( 2\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2\mathrm{e}^{-} \).Reduction: \( \mathrm{Ce}^{4+} + \mathrm{e}^{-} \rightarrow \mathrm{Ce}^{3+} \).

Step 8: Balance electrons in Reaction (c)

Multiply the reduction half-reaction by 2 to balance the electrons with the oxidation half-reaction:Reduction: \( 2(\mathrm{Ce}^{4+} + \mathrm{e}^{-} \rightarrow \mathrm{Ce}^{3+}) \Rightarrow 2\mathrm{Ce}^{4+} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Ce}^{3+} \).Now the number of electrons in both half-reactions is equal.

Step 9: Write balanced reaction for Reaction (c)

Combine the balanced half-reactions:\( 2\mathrm{Ce}^{4+} + 2\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2\mathrm{Ce}^{3+} \).This is the balanced equation for Reaction (c) in acid solution.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial step when dealing with redox reactions. A chemical equation represents the chemical reaction occurring, with reactants transforming into products. Balancing ensures that the law of conservation of mass is upheld, where the number of atoms for each element in the reactants side equals the number in the products side.

In redox reactions, the task becomes a bit more complex because it involves balancing both atoms and charges. Considerations include:

• Ensuring the total charge before and after the reaction remains unchanged.
• Balancing the electrons exchanged during oxidation and reduction processes.
• Sometimes, adding appropriate numbers of water molecules (H\(_2\)O) and hydrogen ions (H\(^+\), in acid solutions) to balance hydrogen and oxygen atoms.

By systematically writing half-reactions and then combining them, you can effectively balance a redox equation.

Half-Reactions

The concept of half-reactions is fundamental in understanding redox reactions. Each redox equation can be broken down into two halves. One half represents oxidation, and the other represents reduction.

A half-reaction explicitly shows the movement of electrons, which is the essence of redox chemistry:

• Oxidation Half-Reaction: This process concerns the loss of electrons. For example, a neutral element may lose electrons to form a cation.
• Reduction Half-Reaction: This depicts the gaining of electrons, usually by an anion getting converted to a neutral molecule or another anion.

Understanding and writing half-reactions help clarify what is losing and gaining electrons, providing mechanical clarity to the chemical process. Once written, these half-reactions can be scaled by appropriate factors to ensure electrons lost in oxidation equal electrons gained in reduction, assisting in balancing the overall chemical equation.

Oxidation and Reduction

Oxidation and reduction, often referred to collectively as redox reactions, are processes where electrons are transferred between substances. This transfer of electrons results in changes in the oxidation states of the involved species.

Some key points to remember are:

• Oxidation: Involves the loss of electrons, which implies an increase in oxidation state. For instance, a metal turning into a cation (like Zn to Zn\(^{2+}\)).
• Reduction: Entails the gain of electrons, marking a decrease in oxidation state. An example is the conversion of nitrate ions (NO\(_3^-\)) to nitrogen gas (N\(_2\)).

It's crucial to identify which reactant is oxidized and which is reduced. The species that undergoes oxidation is the reducing agent, as it provides electrons for the reduction process of the other reactant, known as the oxidizing agent. This interplay maintains the charge balance and drives the chemical reaction towards completion.