Question

Balance the following oxidation–reduction reactions that occur in basic solution. a. \(\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) 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 oxidation-reduction reactions in basic solution are: a. \(Cr(s) + CrO_{4}^{2-}(aq) + 8 H_2O \rightarrow 2\ Cr(OH)_{3}(s) + 6 OH^-(aq)\) b. \(2 MnO_4^-(aq) + S_2^{2-}(aq) + 16 H_2O \rightarrow 2 MnS(s) + S(s) + 16 OH^-(aq)\) c. \(3 CN^-(aq) + MnO_4^-(aq) + 2 H_2O \rightarrow 3 CNO^-(aq) + MnO_2(s) + 4 OH^-(aq)\)

Step-by-step solution

Identify the half-reactions

: First, we need to recognize the oxidation and reduction half-reactions. Oxidation half-reaction: \(\mathrm{Cr}(s) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) Reduction half-reaction: \(\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\)

Balance the half-reactions

: Next, we need to balance each half-reaction individually. Oxidation half-reaction: \(Cr \rightarrow Cr(OH)_3 \(+ 3 e^-) \\ The half-reaction is already balanced. Reduction half-reaction: \(CrO_4^{2-} + 8 H_2O \rightarrow Cr(OH)_3 + 6 OH^-(aq) + 3 e^-\)

Combine the balanced half-reactions

: Lastly, we need to sum the balanced half-reactions to form the balanced overall reaction: \(\mathrm{Cr}(s) + \mathrm{CrO}_{4}^{2-}(a q) + 8 H_2O \rightarrow 2\ \mathrm{Cr}(\mathrm{OH})_{3}(s) + 6 OH^-(aq)\) b. Balance the reaction: MnO4^-(aq) + S2-(aq) -> MnS(s) + S(s)

Identify the half-reactions

: Oxidation half-reaction: \(\mathrm{S}^{2-}(a q) \rightarrow \mathrm{S}(s)\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnS}(s)\)

Balance the half-reactions

: Oxidation half-reaction: \(S^{2-} \rightarrow S(s) + 2 e^-\) Reduction half-reaction: \(MnO_4^- + 8 H_2O + 5 e^- \rightarrow MnS + 8 OH^-\)

Combine the balanced half-reactions

: 2 MnO4^-(aq) + S2-(aq) + 16 H2O -> 2 MnS(s) + S(s) + 16 OH^-(aq) c. Balance the reaction: CN^-(aq) + MnO4^-(aq) -> CNO^-(aq) + MnO2(s)

Identify the half-reactions

: Oxidation half-reaction: \(\mathrm{CN}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)\)

Balance the half-reactions

: Oxidation half-reaction: \(CN^- \rightarrow CNO^- + e^-\) Reduction half-reaction: \(MnO_4^- + 2 H_2O + 3 e^- \rightarrow MnO_2 + 4 OH^-\)

Combine the balanced half-reactions

: 3 CN^-(aq) + MnO4^-(aq) + 2 H2O -> 3 CNO^-(aq) + MnO2(s) + 4 OH^-(aq)

Key concepts

Balancing Chemical Equations

Balancing chemical equations is like solving a puzzle where both sides must mirror each other in terms of quantity and type of elements. In these reactions, particularly oxidation-reduction reactions occurring in basic solutions, the process ensures that the number of each type of atom and charge is the same on both sides of the reaction.

Here's how you can achieve it in three simple steps:

• Start by identifying the two half-reactions within the equation. You need one for oxidation and another for reduction.
• Balance each half-reaction individually for atoms and charges. Use water, protons or hydroxide ions as needed to balance atoms other than oxygen and hydrogen. Then, balance the charges by adding electrons appropriately.
• Finally, combine the balanced half-reactions. Adjust to ensure that the electrons lost in oxidation match those gained in reduction, ensuring both mass and charge are balanced.

Every time you balance chemical equations, it feels fulfilling to see how the components align perfectly, just like fitting the right pieces in a jigsaw puzzle.

Basic Solution Chemistry

Basic solutions play a significant role in balancing redox reactions, potentially altering the approach compared to reactions in acidic solutions. When reactions occur in basic solutions, hydroxide ions ( OH⁻ ) are plentiful in the solution, influencing how we balance the half-reactions.

Here's how basic solutions affect the process:

• We often need to add water molecules ( H_2O ) to one side of the half-reaction to balance oxygen atoms.
• Hydroxide ions are added to counterbalance hydrogen ions ( H^+ ) formed during the balancing.
• This is crucial because every hydrogen atom in your equation should be neutralized to reflect the basic condition of the environment.

This method might seem a bit roundabout at first, but it's essential for mastering redox equations in basic solutions and getting the correct overall balanced equation. Remember, the environment of your reaction conditions how you approach balancing!

Half-Reactions

Half-reactions are a way to look at the changes happening in a redox reaction step by step, by dividing the oxidation and reduction processes. They help make the complex process of balancing reactions more approachable by separating it into manageable parts.

Here’s how to handle half-reactions effectively:

• Identify which substances are oxidized and reduced. Look for changes in oxidation states to spot which are donors (losing electrons) and acceptors (gaining electrons).
• Balance the number of atoms in each half-reaction. This includes adding water to balance oxygens, and using OH⁻ or H⁺ to balance hydrogens, depending on the solution's pH level.
• Fix the charge imbalance by adding electrons. Remember, electrons are what connect the separate half-reactions.

Once balanced separately, combining these half-reactions correctly will give you insight into the complete redox process. This method not only corrects stoichiometry but also reveals how electrons shuffle around, coloring the full picture of the chemical change.