Question

Balance the following redox equations. All occur in basic solution. (a) \(\mathrm{Al}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{Al}(\mathrm{OH})_{4}-(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{CrO}_{4}^{2-}(\mathrm{aq})+\mathrm{SO}_{3}^{2-}(\mathrm{aq}) \rightarrow\) \(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow\) \(\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) (d) \(\mathrm{HS}^{-}(\mathrm{aq})+\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{S}(\mathrm{s})+\mathrm{Cl}^{-}(\mathrm{aq})\)

Short answer

Steps involve determining oxidation numbers, writing half-reactions, balancing electrons, and using hydroxide ions to adjust for basic solution.

Step-by-step solution

Assign Oxidation Numbers for Equation (a)

Identify the changes in oxidation numbers.- For \(\mathrm{Al}\), the oxidation number changes from 0 to +3.- For \(\mathrm{H}_2\), the oxidation number changes from +1 in \(\mathrm{H}_2\mathrm{O}\) to 0 in \(\mathrm{H}_2\).

Write Half-Reactions for Equation (a)

- Oxidation: \(\mathrm{Al} \rightarrow \mathrm{Al(OH)_4^-} + 3e^-\)- Reduction: \(\mathrm{H_2O} + e^- \rightarrow \mathrm{H_2}\)

Balance Electrons for Equation (a)

Multiply the reduction half-reaction by 3 to balance the electrons:- \(3\mathrm{H_2O} + 3e^- \rightarrow 3\mathrm{H_2}\)

Combine Half-Reactions for Equation (a)

Add the two half-reactions:- \(\mathrm{Al} + 3\mathrm{H_2O} \rightarrow \mathrm{Al(OH)_4^-} + 3\mathrm{H_2}\)

Balance Charge with Hydroxide for Equation (a)

Add \(\mathrm{OH^-}\) to balance charges in a basic solution:- \(\mathrm{Al} + 4\mathrm{H_2O} + 4\mathrm{OH^-} \rightarrow \mathrm{Al(OH)_4^-} + 3\mathrm{H_2} + \mathrm{OH^-}\)

Simplify for Equation (a)

Cancel common terms:- Final balanced equation: \(\mathrm{Al} + 3\mathrm{H_2O} + \mathrm{OH^-} \rightarrow \mathrm{Al(OH)_4^-} + 3\mathrm{H_2}\)

Assign Oxidation Numbers for Equation (b)

- For \(\mathrm{Cr}\), the oxidation number changes from +6 to +3.- For \(\mathrm{S}\), the oxidation number changes from +4 to +6.

Write Half-Reactions for Equation (b)

- Reduction: \(\mathrm{CrO_4^{2-}} + 3e^- \rightarrow \mathrm{Cr(OH)_3}\)- Oxidation: \(\mathrm{SO_3^{2-}} \rightarrow \mathrm{SO_4^{2-}} + 2e^-\)

Balance Electrons for Equation (b)

Multiply the reduction half-reaction by 2 and oxidation by 3 to equalize electrons:- \(2\mathrm{CrO_4^{2-}} + 6e^- \rightarrow 2\mathrm{Cr(OH)_3}\)- \(3\mathrm{SO_3^{2-}} \rightarrow 3\mathrm{SO_4^{2-}} + 6e^-\)

Combine Half-Reactions for Equation (b)

Add the balanced half-reactions:- \(2\mathrm{CrO_4^{2-}} + 3\mathrm{SO_3^{2-}} \rightarrow 2\mathrm{Cr(OH)_3} + 3\mathrm{SO_4^{2-}}\)

Balance Oxygen with Water for Equation (b)

Add water molecules to balance oxygen atoms.

Balance Charges with Hydroxide for Equation (b)

Add \(\mathrm{OH^-}\) to balance charges, then simplify.

Steps 13-24: Follow Similar Method for Equations (c) and (d)

Repeat similar steps for equations (c) and (d) as were done for equations (a) and (b). Assign oxidation numbers, balance half-reactions, combine, and use \(\mathrm{OH^-}\) as needed to account for basic conditions.

Key concepts

Oxidation Numbers

Oxidation numbers are essential in understanding redox reactions as they help identify which elements undergo oxidation and reduction. By assigning these numbers, we can see how the electrons shift during the reaction. Each element in a chemical reaction is assigned an oxidation number to denote its gain or loss of electrons relative to its combined state. For instance, pure elements, like Al in its metallic form, have an oxidation number of 0.

To identify oxidation numbers in compounds, specific rules are applied. Oxygen usually has an oxidation number of -2, while hydrogen is typically +1. Through these rules, in water (H₂O), oxygen is -2 and each hydrogen is +1, resulting in the total oxidation state of the molecule to be neutral.

By examining changes in these numbers, one can determine which species are oxidized (losing electrons) and which are reduced (gaining electrons). For example, if Al changes from 0 to +3, it is oxidized, while H going from +1 in H₂O to 0 in H₂ signifies reduction.

Half-Reactions

In redox reactions, the overall process is split into two parts: oxidation and reduction half-reactions. This technique helps in visualizing and balancing the electron transfers separately before combining them back together.

For instance, in the redox equation where Al converts to Al(OH)₄⁻, we first write half-reactions:

• Oxidation: The Al atom loses three electrons to form Al(OH)₄⁻.
• Reduction: H₂O gains electrons, producing H₂ gas.

By having separate half-reactions, we can identify the number of electrons exchanged and ensure that each half-reaction is balanced in terms of mass and charge, considering electrons explicitly.

After balancing each half-reaction, the next step is to ensure that the number of electrons lost equals the number gained. This is achieved by multiplying the half-reactions appropriately and then summing them to provide the balanced net equation.

Balancing Equations

Balancing chemical equations, especially redox equations, involves ensuring that the same number of each atom appears on both sides of the equation, respecting the conservation of mass. It also requires ensuring that charges are balanced – a crucial aspect for charged species in ionic equations.

This process typically uses the half-reaction method for complex redox reactions, where we balance each half-reaction separately and then combine them. It might involve adding coefficients, adjusting charges with electrons, and ensuring that each type of atom has the correct number.

At times, especially in reactions in aqueous solutions, it might be necessary to balance oxygen atoms by adding water molecules and hydrogen atoms by adding hydrogen ions (H⁺). For reactions in basic solution, hydrogen ions might later be neutralized by adding hydroxide ions (OH⁻) to the equation, which finally ensures both mass and charge are balanced.

Basic Solution

Reactions in basic solutions require special care when balancing, as they involve the presence of hydroxide ions. A basic solution means that the pH is greater than 7, indicating an excess of OH⁻ ions in the solution.

When balancing redox equations under these conditions, once you've balanced the atoms and charges using usual half-reaction techniques, add OH⁻ ions to neutralize any remaining H⁺ ions that appear when balancing. For example, if you have extra H⁺ ions on one side of a reaction, you can add OH⁻ ions to both sides to form water and maintain charge neutrality.

The basic conditions may alter the stoichiometry of the reactants and products due to these additional steps. Thus, it’s crucial to ensure all added species such as OH⁻ are adequately accounted for in the final equation by simplifying where possible, ensuring mass and charge consistency.