Question

Balance these equations for redox reactions occurring in basic solution. (a) \(\mathrm{CrO}_{4}^{2-}+\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{3}^{2-}\) (b) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{N}_{2}(\mathrm{g})\) (c) \(\operatorname{Fe}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{MnO}_{4}^{-} \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{MnO}_{2}(\mathrm{s})\)

Short answer

The balanced equations are (a) \(\mathrm{CrO}_{4}^{2-} + 3\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s}) + 5\mathrm{OH}^{-}\) and \(\mathrm{S}_2\mathrm{O}_4^{2-} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{SO}_3^{2-}+\mathrm{H}_2\mathrm{O}\), (b) \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}+ \mathrm{N}_2\mathrm{H}_4 +2 \mathrm{OH}^{-}\rightarrow [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} + \mathrm{N}_2\mathrm{(g)}+ 4\mathrm{H}_2\mathrm{O}\), (c) \(\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_2\mathrm{O}\rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\), and (d) \(3\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{MnO}_{4}^{-}+\mathrm{OH}^{-} \rightarrow 3\mathrm{CH}_{3} \mathrm{COO}^{-} + \mathrm{MnO}_{2}(\mathrm{s}) + 4\mathrm{H}_2\mathrm{O}\).

Step-by-step solution

Balancing the first equation

First, the individual half reactions for the redox reaction are identified. The two half reactions for the equation are \(\mathrm{CrO}_{4}^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})\) and \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) to \(\mathrm{SO}_{3}^{2-}\). The half reactions are balanced separately by matching oxygen atoms with water molecules and hydrogen atoms with \(\mathrm{H}^{+}\) ions, and if necessary, matching the charges with electrons. Finally, the half equations are added together, cancelling out any same term from both sides of the reaction. So, the balanced equation is \(\mathrm{CrO}_{4}^{2-} + 3\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s}) + 5\mathrm{OH}^{-}\) and \(\mathrm{S}_2\mathrm{O}_4^{2-} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{SO}_3^{2-}+\mathrm{H}_2\mathrm{O}\).

Balancing the second equation

For the second equation, the half reactions are \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\) to \([\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}\) and \(\mathrm{N}_2\mathrm{H}_4\) to \(\mathrm{N}_2\mathrm{(g)}\). There is no redox principle in the first half reaction equation thereby, balancing this equation with water molecules, hydrogen ions, and electrons form it into: \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-} \rightarrow [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} + e^{-}\). Balancing the second half reaction equation results in: \(\mathrm{N}_2\mathrm{H}_4 + 2 \mathrm{OH}^{-} \rightarrow \mathrm{N}_2\mathrm{(g)}+ 4\mathrm{H}_2\mathrm{O}\). And finally, the overall equation is found by combining the half reactions to give: \([\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}+ \mathrm{N}_2\mathrm{H}_4 +2 \mathrm{OH}^{-}\rightarrow [\mathrm{Fe}(\mathrm{CN})_{6}]^{4-} + \mathrm{N}_2\mathrm{(g)}+ 4\mathrm{H}_2\mathrm{O}\).

Balancing the third equation

The half reactions for the third equation are \(\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})\) to \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) and \(\mathrm{O}_2(\mathrm{g})\) to \(\mathrm{H}_2\mathrm{O}\). The final balanced equation is obtained as: \(\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_2\mathrm{O}\rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\).

Balancing the fourth equation

For the fourth equation, the half reactions are \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) to \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) and \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{MnO}_{2}(\mathrm{s})\). The final balanced equation is obtained as: \(3\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{MnO}_{4}^{-}+\mathrm{OH}^{-} \rightarrow 3\mathrm{CH}_{3} \mathrm{COO}^{-} + \mathrm{MnO}_{2}(\mathrm{s}) + 4\mathrm{H}_2\mathrm{O}\).

Key concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry. It involves making sure that the number of atoms of each element is the same on both sides of the equation. This represents the law of conservation of mass, which states that matter cannot be created or destroyed. To balance an equation, we adjust the coefficients in front of compounds or elements in a chemical equation until the number of each type of atom is equal on both sides.

• Start by balancing atoms that appear in only one reactant and one product.
• Next, balance atoms that appear in two or more reactants or products.
• Finally, check to ensure all the coefficients are in the simplest ratio.

When balancing equations, it is crucial to ensure that both sides of the equation reflect the same number of each type of atom, thus showing how reactants recombine to form products without any change in the total mass.

Oxidation-Reduction

Oxidation-reduction, or redox reactions, are a type of chemical process where electrons are transferred between two substances. One substance undergoes oxidation, losing electrons, while the other undergoes reduction, gaining those electrons. Understanding this transfer is key to comprehending many chemical processes, from combustion to metabolism.
In any redox reaction:

• The substance that loses electrons is oxidized and is called the reducing agent.
• The substance that gains electrons is reduced and is called the oxidizing agent.

Understanding redox reactions involves identifying changes in oxidation states, tracing the flow of electrons from one substance to another, which often involves complex reactions but follows clear rules. Recognizing these can help clarify the direction and extent of the reaction, ensuring electrons are conserved in the process.

Half-Reaction Method

The half-reaction method is a systematic way to tackle redox reactions and ensure they are balanced correctly. It breaks down the redox process into two separate half-reactions: one for oxidation and another for reduction. Each half-reaction is independently balanced with respect to mass and charge, then combined to give the overall balanced equation. To use the half-reaction method effectively:

• Write the half-reactions for the oxidation and reduction processes.
• Balance each half-reaction for the number of atoms, ensuring this is done for all elements except hydrogen and oxygen.
• Balance the oxygen atoms by adding water molecules, and balance hydrogen by adding hydrogen ions (or hydroxide ions in basic solution).
• Balance the charge by adding electrons.
• Multiply the balanced half-reactions by appropriate factors if necessary to ensure that the electrons cancel when summed.
• Add the half-reactions back together to form the full redox equation and verify the mass and charge balance.

This method is particularly useful because it simplifies the complex process of balancing redox reactions by breaking it into manageable steps, allowing for a clear and orderly method to reach the balanced equation.