Question

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\operatorname{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) c. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

Short answer

The balanced redox reactions in basic solutions are as follows: a. \(\operatorname{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q)+3\mathrm{OH}^- \rightarrow 2\operatorname{Cr}(\mathrm{OH})_{3}(s)\) b. \(5\mathrm{S}^{2-}(a q)+2\mathrm{MnO}_{4}^{-}(a q)+16\mathrm{OH}^- \rightarrow 2\mathrm{MnS}(s)+8\mathrm{H}_{2}\mathrm{O}(l)+5\mathrm{S}(s)\) c. \(3\mathrm{CN}^{-}(a q)+2\mathrm{MnO}_{4}^{-}(a q)+8\mathrm{OH}^- \rightarrow 3\mathrm{CNO}^{-}(a q)+2\mathrm{MnO}_{2}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\)

Step-by-step solution

Problem a

Balance the following reaction: \(\operatorname{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)\) To balance this reaction, we need to follow these steps: Step 1: Divide the reaction into half-reactions \[\text{Oxidation:} \quad \operatorname{Cr}(s) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)\] \[\text{Reduction:} \quad \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)\] Step 2: Balance the atoms involved in the redox process (excluding O and H) Oxidation: The Cr atoms are already balanced. Reduction: The Cr atoms are already balanced. Step 3: Balance the O atoms by adding H_2O Oxidation: Not necessary as there are no O atoms. Reduction: Add 3 H_2O to the right side. \[\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(l)\] Step 4: Balance the H atoms by adding OH^- Oxidation: Add 3 OH^- to the right side. \[\operatorname{Cr}(s) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{OH}^-\] Reduction: Add 6 OH^- to the left side. \[\mathrm{CrO}_{4}^{2-}(a q)+6\mathrm{OH}^- \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(l)\] Step 5: Balance charges by adding electrons (e^-) Oxidation: Add 3 e^- to the right side. \[\operatorname{Cr}(s) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{OH}^-+3\mathrm{e}^-\] Reduction: Add 3 e^- to the left side. \[\mathrm{CrO}_{4}^{2-}(a q)+6\mathrm{OH}^-+3\mathrm{e}^- \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(l)\] Step 6: Multiply each half-reaction by the necessary factor to make the number of electrons equal Oxidation: No need to multiply, as the number of electrons is already the same. Reduction: No need to multiply, as the number of electrons is already the same. Step 7: Add both half-reactions and cancel electrons and common species \[\operatorname{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q)+6\mathrm{OH}^- \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(l)+3\mathrm{OH}^-\] Simplify the reaction: \[\operatorname{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q)+3\mathrm{OH}^- \rightarrow 2\operatorname{Cr}(\mathrm{OH})_{3}(s)\] Now, the reaction is balanced.

Problem b

Balance the following reaction: \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) To balance this reaction, we need to follow these steps: Step 1: Divide the reaction into half-reactions \[\text{Oxidation:} \quad \mathrm{S}^{2-}(a q) \rightarrow \mathrm{S}(s)\] \[\text{Reduction:} \quad \mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnS}(s)\] Step 2: Balance the atoms involved in the redox process (excluding O and H) Oxidation: The S atoms are already balanced. Reduction: The Mn atoms are already balanced. Step 3: Balance the O atoms by adding H_2O Oxidation: Not necessary as there are no O atoms. Reduction: Add 4 H_2O to the right side. \[\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnS}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Step 4: Balance the H atoms by adding OH^- Oxidation: Not necessary as there are no H atoms. Reduction: Add 8 OH^- to the left side. \[\mathrm{MnO}_{4}^{-}(a q)+8\mathrm{OH}^- \rightarrow \mathrm{MnS}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Step 5: Balance charges by adding electrons (e^-) Oxidation: Add 2 e^- to the right side. \[\mathrm{S}^{2-}(a q) \rightarrow \mathrm{S}(s)+2\mathrm{e}^-\] Reduction: Add 5 e^- to the left side. \[\mathrm{MnO}_{4}^{-}(a q)+8\mathrm{OH}^-+5\mathrm{e}^- \rightarrow \mathrm{MnS}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Step 6: Multiply each half-reaction by the necessary factor to make the number of electrons equal Oxidation: Multiply by 5. \[\mathrm{5S}^{2-}(a q) \rightarrow 5\mathrm{S}(s)+10\mathrm{e}^-\] Reduction: Multiply by 2. \[\mathrm{2MnO}_{4}^{-}(a q)+16\mathrm{OH}^-+10\mathrm{e}^- \rightarrow 2\mathrm{MnS}(s)+8\mathrm{H}_{2}\mathrm{O}(l)\] Step 7: Add both half-reactions and cancel electrons and common species \[5\mathrm{S}^{2-}(a q)+2\mathrm{MnO}_{4}^{-}(a q)+16\mathrm{OH}^- \rightarrow 2\mathrm{MnS}(s)+8\mathrm{H}_{2}\mathrm{O}(l)+5\mathrm{S}(s)\] Now, the reaction is balanced.

Problem c

Balance the following reaction: \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\) To balance this reaction, we need to follow these steps: Step 1: Divide the reaction into half-reactions \[\text{Oxidation:} \quad \mathrm{CN}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)\] \[\text{Reduction:} \quad \mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)\] Step 2: Balance the atoms involved in the redox process (excluding O and H) Oxidation: The C and N atoms are already balanced. Reduction: The Mn atoms are already balanced. Step 3: Balance the O atoms by adding H_2O Oxidation: Not necessary as there are no O atoms. Reduction: Add 2 H_2O to the right side. \[\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+2\mathrm{H}_{2}\mathrm{O}(l)\] Step 4: Balance the H atoms by adding OH^- Oxidation: Not necessary as there are no H atoms. Reduction: Add 4 OH^- to the left side. \[\mathrm{MnO}_{4}^{-}(a q)+4\mathrm{OH}^- \rightarrow \mathrm{MnO}_{2}(s)+2\mathrm{H}_{2}\mathrm{O}(l)\] Step 5: Balance charges by adding electrons (e^-) Oxidation: Add 2 e^- to the right side. \[\mathrm{CN}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+2\mathrm{e}^-\] Reduction: Add 3 e^- to the left side. \[\mathrm{MnO}_{4}^{-}(a q)+4\mathrm{OH}^-+3\mathrm{e}^- \rightarrow \mathrm{MnO}_{2}(s)+2\mathrm{H}_{2}\mathrm{O}(l)\] Step 6: Multiply each half-reaction by the necessary factor to make the number of electrons equal Oxidation: Multiply by 3. \[\mathrm{3CN}^{-}(a q) \rightarrow 3\mathrm{CNO}^{-}(a q)+6\mathrm{e}^-\] Reduction: Multiply by 2. \[\mathrm{2MnO}_{4}^{-}(a q)+8\mathrm{OH}^-+6\mathrm{e}^- \rightarrow 2\mathrm{MnO}_{2}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Step 7: Add both half-reactions and cancel electrons and common species \[3\mathrm{CN}^{-}(a q)+2\mathrm{MnO}_{4}^{-}(a q)+8\mathrm{OH}^- \rightarrow 3\mathrm{CNO}^{-}(a q)+2\mathrm{MnO}_{2}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Now, the reaction is balanced.

Key concepts

Redox Reactions in Basic Solution

Understanding how to balance redox reactions in a basic solution is crucial for students studying chemistry. Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between two species. Balancing these reactions in basic solution differs slightly from balancing in acidic solutions mainly due to the presence of hydroxide ions (OH^-).

When balancing redox reactions in basic solution, the steps are similar to the acidic solution but with an additional step of neutralizing excess H⁺ with OH^- to form water (H₂O). This is because strong bases in solution exist essentially as hydroxide ions, which will react with any hydrogen ions that are present. Once H₂O molecules are added, some can be canceled out with OH^- ions on the other side to simplify the equation. By considering the stoichiometry of the reaction and being careful to balance both the mass and charge, students can successfully balance redox reactions in a basic environment.

Half-Reaction Method

The half-reaction method is an effective approach for balancing redox reactions. This technique separates the overall chemical equation into two separate half-reactions—one for oxidation and one for reduction. Each half-reaction is balanced separately for mass and charge.

In the oxidation half-reaction, an element loses electrons, and in the reduction half-reaction, another element gains electrons. After balancing these half-reactions for mass (elements) and then for charge by adding electrons, the two are then combined to give the balanced overall reaction. It's essential to multiply the half-reactions by integers to ensure that the same number of electrons is involved in both the oxidation and the reduction reactions before adding them together. This ensures that all electrons cancel out and that the final reaction is balanced in terms of both mass and charge.

Oxidation and Reduction Balancing

Oxidation and reduction balancing is a fundamental part of redox chemistry, which focuses on the changes in oxidation states of elements within the reacting species. Oxidation is the process where an element increases its oxidation state by losing electrons, while reduction sees a decrease in oxidation state by gaining electrons.

In balancing oxidation and reduction reactions, it's essential to identify the elements that are oxidized and reduced first, then balance their atoms and, importantly, their charges using the appropriate number of electrons. It is also vital to consider the side reactions that may occur in solution, particularly the self-reaction of water in basic solutions. Special attention should be given to ensuring that the electrons lost and gained are equal, which requires sometimes multiplying the half-reactions by appropriate factors.

Stoichiometry of Redox Reactions

Stoichiometry of redox reactions involves the quantitative relationship between the amount of reactants and products in a chemical reaction. This includes ensuring that atoms are conserved in the reaction and that the charges are balanced.

Often, balancing redox reactions requires an understanding of mole ratios, which dictate the proportions in which reactants react and products are formed. In the context of the exercise, it's important to note how the stoichiometry was utilized to multiply the half-reactions to ensure electrons are balanced. Furthermore, stoichiometric coefficients can be adjusted to account for the formation of water in the reduction half-reaction and its subsequent simplification by combining with hydroxide ions in basic solutions. Ultimately, correct stoichiometry ensures that the chemical equation is balanced in terms of both mass and charge.