Problem 39 - General Chemistry: Principles and Modern Applications

Question

Balance these equations for redox reactions in basic solution. (a) \(\mathrm{MnO}_{2}(\mathrm{s})+\mathrm{ClO}_{3}^{-} \longrightarrow \mathrm{MnO}_{4}^{-}+\mathrm{Cl}^{-}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{OCl}^{-} \longrightarrow \mathrm{FeO}_{4}^{2-}+\mathrm{Cl}^{-}\) (c) \(\mathrm{ClO}_{2} \longrightarrow \mathrm{ClO}_{3}^{-}+\mathrm{Cl}\) (d) \(\mathrm{Ag}(\mathrm{s})+\mathrm{CrO}_{4}^{2-} \rightarrow \mathrm{Ag}^{+}+\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})\)

Step-by-Step Solution

Answer

The balanced chemical equations are: (a) \(4\, \mathrm{MnO}_2 + 4\, \mathrm{ClO}_3^- + 8\, \mathrm{OH}^- \rightarrow 4\, \mathrm{MnO}_4^- + 4\, \mathrm{Cl}^- + 2\, \mathrm{H}_2\mathrm{O}\), (b) \(2\, \mathrm{Fe(OH)}_3 + 5\, \mathrm{OCl}^- + 8\, \mathrm{H}_2\mathrm{O} \rightarrow 2\, \mathrm{FeO}_4^{2-} + 5\, \mathrm{Cl}^- + 8\, \mathrm{OH}^-\), (c) \(2\, \mathrm{ClO}_2 + 2\, \mathrm{OH}^- \rightarrow \mathrm{ClO}_3^- + 3\, \mathrm{Cl}^- + 2\, \mathrm{H}_2\mathrm{O}\), (d) \(\mathrm{Ag} + \mathrm{CrO}_4^{2-} + 1\, \mathrm{OH}^- \rightarrow \mathrm{Ag}^{+} + \mathrm{Cr(OH)}_3 + 4\, \mathrm{H}_2\mathrm{O}\)

Step 1 - Identify the oxidation states and which species are reduced and oxidized for each reaction

The oxidation state changes are for reaction (a) Mn (from +4 to +7), Cl (from +5 to -1), for (b) Fe (from +3 to +6), Cl (from +1 to -1), for (c) Cl (from +3 to +5 and from +3 to -1), and for (d) Ag (from 0 to +1), Cr (from +6 to +3).

Step 2 - Balance atoms other than Hydrogen and Oxygen

In (a) the Cl atoms are already balanced and so are the Mn. In (b) the Cl and Fe atoms are already balanced. In (c) there is one ClO2 molecule and one atom each of Chlorine and ClO3-. In (d) there is one atom of Ag on each side and one CrO4 on left and one Cr on right.

Step 3 - Balance Oxygen using H2O

Add, for (a) 4H2O to the right side, for (b) 4H2O to the left side, for (c) 2H2O to the right side, for (d) 4H2O to the right side.

Step 4 - Balance Hydrogen using OH-

Add, for (a) 8OH- to the left side, for (b) 8OH- to the right side, for (c) 2OH- to the left, for (d) 8OH- to the left.

Step 5 - Balance charges with e-

Balance the charges by adding, for (a) 2e- to the right side, for (b) 5e- to the left side, for (c) 2e- to the left of the first half reaction and 4e- to the right of the second half reaction, for (d) 1e- to the left side.

Step 6 - Combine half-reactions

Combine the half reactions by adding, for (a) the reduction and oxidation half-reactions, for (b) the reduction and oxidation half reactions, for (c) the reduction half reaction to the oxidation half reaction multiplied by 2, for (d) the reduction and oxidation half reactions.

Key Concepts

Oxidation and Reduction

In chemical reactions, oxidation and reduction are key processes. They involve the transfer of electrons between molecules, atoms, or ions.
Oxidation is when a substance loses electrons, increasing its oxidation state. Reduction is when it gains electrons, decreasing its oxidation state.

• An easy way to remember this is through the mnemonic: "OIL RIG"—Oxidation Is Loss, Reduction Is Gain (of electrons).

In a redox (reduction-oxidation) reaction, one substance is oxidized, losing electrons to another substance that is simultaneously reduced. For example, in reaction (a) Mn goes from an oxidation state of +4 to +7, meaning it loses electrons and is oxidized. Meanwhile, Cl changes from +5 to -1, indicating it gains electrons and is reduced. Understanding these processes is fundamental to balancing redox reactions, especially in determining which species are reducing agents (those that lose electrons) and oxidizing agents (those that gain electrons).
In redox reactions, these processes always occur together. If something is oxidized, something else must be reduced.

Oxidation States

Oxidation states, also known as oxidation numbers, are a helpful way to keep track of electrons in chemical reactions. They indicate the degree of oxidation of an atom in a chemical compound. Each atom is assigned a number, which can be positive, negative, or zero, based on specific rules and guidelines.

• Atoms in their elemental state have an oxidation state of 0. For example, \( \text{Ag} \) as a metal is 0 in reaction (d).
• For a monoatomic ion, the oxidation state is equal to its charge, such as \( \text{Cr}^{6+} \) having an oxidation state of +6 in \( \text{CrO}_4^{2-} \) in reaction (d).

By comparing the oxidation states before and after a reaction, you can determine which elements are oxidized and reduced. This is vital in redox reactions as it helps identify electron transfer processes. For instance, in reaction (b), Fe's change from +3 to +6 shows oxidation, while Cl's change from +1 to -1 shows reduction.Knowing how to assign oxidation states is crucial for balancing equations, and helps in understanding the electron flow within chemical reactions.

Half-Reactions

Half-reactions are a convenient way to separate the oxidation and reduction processes in a redox chemical reaction. Each half-reaction shows either the oxidation or the reduction process, making it easier to balance complex reactions.
Here is how they work:

• Start by identifying all the substances participating in the redox process and how their oxidation states change. The species gaining electrons undergo reduction, and the one losing electrons undergoes oxidation.
• Once identified, write a separate half-reaction for oxidation and another for reduction. For example, in reaction (a), the oxidation half reaction involves \( \text{MnO}_2 \rightarrow \text{MnO}_4^- \), and the reduction half-reaction involves \( \text{ClO}_3^- \rightarrow \text{Cl}^- \).

After separating the half-reactions, balance each one for mass (atoms) and charge (using electrons as needed). Then, multiply each half-reaction by appropriate coefficients so that electrons lost equal electrons gained, ensuring charge balance.Finally, add the two balanced half-reactions together, canceling out the electrons. This final complete reaction shows the redox process fully balanced both for mass and charge. Employing half-reactions not only helps in balancing but also offers clear insights into which species are oxidizing and reducing in a reaction.