Question

Balance the following redox equations. (a) \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}_{2}(g)\) (acidic) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) (basic)

Short answer

Question: Balance the following redox reactions: (a) Cu(s) + NO3^-(aq) → Cu^2+(aq) + NO2(g) (acidic) (b) Cr(OH)3(s) + ClO^-(aq) → CrO4^2-(aq) + Cl^-(aq) (basic) Answer: (a) Cu(s) + 2NO3^-(aq) + 4H+(aq) → Cu^2+(aq) + 2NO2(g) + 2H2O(l) (b) 2Cr(OH)3(s) + 4ClO^-(aq) → 2CrO4^2-(aq) + 4Cl^-(aq) + 2H2O(l)

Step-by-step solution

Identify the oxidation and reduction half-reactions

Oxidation: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)\) Reduction: \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)\)

Balance the atoms, except for oxygen and hydrogen

Oxidation: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)\) (already balanced) Reduction: \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)\) (already balanced)

Balance the oxygen atoms by adding water molecules

Oxidation: No need to balance oxygen atoms Reduction: \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g) + H_2O (l)\)

Balance hydrogen atoms by adding protons

Oxidation: No need to balance hydrogen atoms Reduction: \(\mathrm{NO}_{3}^{-}(a q) + 2H^+(aq) \longrightarrow \mathrm{NO}_{2}(g) + H_2O (l)\)

Balance charges by adding electrons

Oxidation: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q) + 2e^-\) Reduction: \(\mathrm{NO}_{3}^{-}(a q) + 2H^+(aq) + e^- \longrightarrow \mathrm{NO}_{2}(g) + H_2O (l)\)

Make the number of electrons equal and combine half-reactions

\(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q) + 2e^-\) \(\mathrm{2NO}_{3}^{-}(a q) + 4H^+(aq) + 2e^- \longrightarrow 2\mathrm{NO}_{2}(g) + 2H_2O (l)\) Combine: \(\mathrm{Cu}(s) + \mathrm{2NO}_{3}^{-}(a q) + 4H^+(aq) \longrightarrow \mathrm{Cu}^{2+}(a q) + 2\mathrm{NO}_{2}(g) + 2H_2O (l)\) (b) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) (basic)

Identify the oxidation and reduction half-reactions

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) Reduction: \(\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)\)

Balance the atoms, except for oxygen and hydrogen

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (already balanced) Reduction: \(\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)\) (already balanced)

Balance the oxygen atoms by adding water molecules

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) + 3H_2O (l) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) Reduction: \(\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q) + H_2O(l)\)

Balance hydrogen atoms by adding hydroxide ions

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) + 3H_2O (l) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) + 6OH^-(aq)\) Reduction: \(\mathrm{ClO}^{-}(a q) + 2OH^-(aq) \longrightarrow \mathrm{Cl}^{-}(a q) + H_2O(l)\)

Balance charges by adding electrons

Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) + 3H_2O (l) + 2e^- \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) + 6OH^-(aq)\) Reduction: \(\mathrm{ClO}^{-}(a q) + 2OH^-(aq) + e^- \longrightarrow \mathrm{Cl}^{-}(a q) + H_2O(l)\)

Make the number of electrons equal and combine half-reactions

\(\mathrm{2Cr}(\mathrm{OH})_{3}(s) + 6H_2O (l) + 4e^- \longrightarrow \mathrm{2CrO}_{4}^{2-}(a q) + 12OH^-(aq)\) \(\mathrm{4ClO}^{-}(a q) + 8OH^-(aq) + 4e^- \longrightarrow \mathrm{4Cl}^{-}(a q) + 4H_2O(l)\) Combine: \(\mathrm{2Cr}(\mathrm{OH})_{3}(s) + 4\mathrm{ClO}^{-}(a q) \longrightarrow 2\mathrm{CrO}_{4}^{2-}(a q) + 4\mathrm{Cl}^{-}(a q) + 2H_2O (l)\)

Key concepts

Half-Reaction Method

The half-reaction method is an essential tool in chemistry for balancing redox (reduction-oxidation) reactions. Redox reactions involve the transfer of electrons between different chemical species. Using half-reactions can simplify the balancing process:

• In a redox reaction, two half-reactions are identified: one for oxidation and one for reduction.
• In each half-reaction, we separately consider the oxidation and reduction processes.
• The oxidation half-reaction shows electrons being lost, while the reduction half-reaction shows electrons being gained.

For example, in our exercise:

• Oxidation: The transformation from Cu(s) to Cu²⁺(aq).
• Reduction: The transformation from NO₃^-(aq) to NO₂(g).

By focusing individually on these processes, we can systematically balance each half-reaction concerning atoms and charge, making the final combination straightforward.

Oxidation and Reduction

Understanding oxidation and reduction is pivotal in chemistry, especially when discussing redox reactions. These processes involve electron transfer and are two halves of a redox reaction.

• Oxidation occurs when a substance loses electrons, leading to an increase in oxidation state. For instance, Cu in our first example becomes Cu²⁺, losing two electrons.
• Reduction, on the other hand, is the gain of electrons, resulting in a decrease in oxidation state. In the same reaction, NO₃^- gains electrons to become NO₂.

To track these changes, we often use oxidation numbers and compare them before and after the reaction. Overall, one substance's oxidation accompanies another's reduction, highlighting the interconnected nature of these processes.

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom is present on both sides of a reaction, adhering to the law of conservation of mass. In the context of redox reactions, balancing also involves ensuring charge balance.

• Start by writing the unbalanced equation and then separate it into half-reactions as mentioned earlier.
• Balance all atoms except hydrogen and oxygen first.
• Balance oxygen atoms by adding water (H₂O) molecules.
• Next, balance hydrogen atoms by adding protons (H⁺) or hydroxide ions (OH^-) depending on the solution’s acidity or basicity.
• Finally, balance the charge by adding electrons (e^-).

Each step brings us closer to a balanced chemical equation, ensuring both the mass and charge are aligned, as seen in our examples, whether the environment is acidic or basic.