Question

Use the half-reaction method to balance each redox reaction occurring in acidic aqueous solution. (a) \(\mathrm{PbO}_{2}(s)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{I}_{2}(s)\) (b) \(\mathrm{SO}_{3}{ }^{2-}(a q)+\mathrm{MnO}_{4}{ }^{-}(a q) \longrightarrow \mathrm{SO}_{4}{ }^{2-}(a q)+\mathrm{Mn}^{2+}(a q)\) (c) \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{SO}_{4}{ }^{2-}(a q)+\mathrm{Cl}^{-}(a q)\)

Short answer

Balanced reactions: (a) \(2\mathrm{I}^{-} + \mathrm{PbO}_{2} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{I}_{2} + \mathrm{Pb}^{2+} + 4\mathrm{H}^{+}\). (b) \(5\mathrm{SO}_{3}^{2-} + 2\mathrm{MnO}_{4}^{-} + 16\mathrm{H}^{+} \rightarrow 5\mathrm{SO}_{4}^{2-} + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_{2}\mathrm{O}\). (c) \(2\mathrm{S}_{2}\mathrm{O}_{3}^{2-} + 6\mathrm{Cl}_{2} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{SO}_{4}^{2-} + 12\mathrm{Cl}^{-} + 12\mathrm{H}^{+}\).

Step-by-step solution

(a) - Write the two half-reactions

Divide the overall reaction into two half-reactions, one for oxidation and one for reduction. For (a): Oxidation: \(\mathrm{I}^{-} \rightarrow \mathrm{I}_{2}\). Reduction: \(\mathrm{PbO}_{2} \rightarrow \mathrm{Pb}^{2+}\).

(a) - Balance atoms other than O and H

Balance all atoms except oxygen and hydrogen in each half-reaction. Oxidation: \(2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2}\). Reduction remains unchanged as \(\mathrm{PbO}_{2} \rightarrow \mathrm{Pb}^{2+}\) since lead is already balanced.

(a) - Balance oxygen atoms

Balance oxygen atoms by adding water molecules. Reduction: \(\mathrm{PbO}_{2} + 2 \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Pb}^{2+} + 4\mathrm{H}^{+}\). Oxidation does not need balancing of oxygen.

(a) - Balance hydrogen atoms

Balance hydrogen atoms by adding \(\mathrm{H}^{+}\) ions. Reduction already shows balancing of hydrogen; Oxidation does not involve hydrogen.

(a) - Balance the charge with electrons

Balance the charge by adding electrons. Oxidation: \(2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2} + 2e^{-}\). Reduction: \(\mathrm{PbO}_{2} + 2 \mathrm{H}_{2}\mathrm{O} + 4e^{-} \rightarrow \mathrm{Pb}^{2+} + 4\mathrm{H}^{+}\).

(a) - Equalize the number of electrons

Make the number of electrons gained equal to the number of electrons lost by multiplying the half-reactions by appropriate coefficients. The coefficients are already equal.

(a) - Combine the half-reactions

Add the half-reactions together and simplify if necessary. \(2\mathrm{I}^{-} + \mathrm{PbO}_{2} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{I}_{2} + \mathrm{Pb}^{2+} + 4\mathrm{H}^{+}\). Cancel the 4 electrons on each side to get the balanced equation.

(b) - Write the two half-reactions

For (b), divide the overall reaction into two half-reactions. Oxidation: \(\mathrm{SO}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-}\). Reduction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\).

(b) - Balance atoms other than O and H

Balance all atoms except oxygen and hydrogen in each half-reaction. Oxidation and reduction reactions do not require balancing for sulfurs and manganese.

(b) - Balance oxygen atoms

Balance oxygen atoms by adding water molecules. Oxidation remains unchanged. Reduction: \(\mathrm{MnO}_{4}^{-} + 4\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{Mn}^{2+} + 8\mathrm{H}^{+}\).

(b) - Balance hydrogen atoms

Balance hydrogen atoms by adding \(\mathrm{H}^{+}\) ions. Oxidation does not require hydrogen balancing. Reduction is already balanced.

(b) - Balance the charge with electrons

Balance the charge by adding electrons. Oxidation: \(\mathrm{SO}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-} + 2e^{-}\). Reduction: \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\).

(b) - Equalize the number of electrons

Make the number of electrons gained equal to the number of electrons lost by multiplying the half-reactions by appropriate coefficients. Oxidation: Multiply by 5. Reduction: Multiply by 2.

(b) - Combine the half-reactions

Add the half-reactions together and simplify if necessary. \(5\mathrm{SO}_{3}^{2-} + 2\mathrm{MnO}_{4}^{-} + 16\mathrm{H}^{+} \rightarrow 5\mathrm{SO}_{4}^{2-} + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_{2}\mathrm{O}\). Cancel the 10 electrons on each side to obtain the balanced equation.

(c) - Write the two half-reactions

For (c), divide the overall reaction into two half-reactions. Oxidation: \(\mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow \mathrm{SO}_{4}^{2-}\). Reduction: \(\mathrm{Cl}_{2} \rightarrow \mathrm{Cl}^{-}\).

(c) - Balance atoms other than O and H

Balance all atoms except oxygen and hydrogen in each half-reaction. Oxidation: \(2\mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow 2\mathrm{SO}_{4}^{2-}\). Reduction: \(\mathrm{Cl}_{2} \rightarrow 2\mathrm{Cl}^{-}\).

(c) - Balance oxygen atoms

Balance oxygen atoms by adding water molecules. Oxidation: No additional water molecules needed since the number of oxygen atoms are already balanced.

(c) - Balance hydrogen atoms

Balance hydrogen atoms by adding \(\mathrm{H}^{+}\) ions if necessary. Oxidation does not require additional hydrogen atoms.

(c) - Balance the charge with electrons

Balance the charge by adding electrons. Oxidation: \((2)(2\mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow 2\mathrm{SO}_{4}^{2-}) + (2)(6e^{-} \rightarrow 0)\). Reduction: \(\mathrm{Cl}_{2} + 2e^{-} \rightarrow 2\mathrm{Cl}^{-}\).

(c) - Equalize the number of electrons

Make the number of electrons gained equal to the number of electrons lost by multiplying the half-reactions by appropriate coefficients. Oxidation is multiplied by 1, and reduction is multiplied by 6.

(c) - Combine the half-reactions

Add the half-reactions together and simplify if necessary. \(2\mathrm{S}_{2}\mathrm{O}_{3}^{2-} + 6\mathrm{Cl}_{2} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 2\mathrm{SO}_{4}^{2-} + 12\mathrm{Cl}^{-} + 12\mathrm{H}^{+}\). Cancel the 12 electrons on each side to get the balanced equation.

Key concepts

Understanding Redox Reactions

Redox reactions are a family of chemical reactions where reduction and oxidation occur simultaneously. They are essential to numerous biological and industrial processes and form the basis of operations such as batteries and corrosion. In a redox reaction, one substance transfers electrons to another substance. The loss of electrons from a molecule, atom, or ion is known as oxidation, while gaining electrons is called reduction. Remember the mnemonic 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain, which refers to the movement of electrons.During these reactions, the oxidation state of elements changes. It's an indicator of the degree of oxidation (loss of electrons) or reduction (gain of electrons) a compound undergoes in a reaction. For example, in the reaction provided, \[\[\begin{align*} \(\text{PbO}_2(s) + 2\text{I}^-(aq) \rightarrow \text{Pb}^{2+}(aq) + \text{I}_2(s)\) \end{align*}\]\]lead is reduced from an oxidation state of +4 in lead dioxide (PbO2) to +2 in the ion Pb2+, and iodine is oxidized from -1 in the iodide ion to 0 in iodine (I2).

Balancing Oxidation and Reduction in Equations

The key to understanding and mastering redox reactions is learning to balance the equation for the oxidation and reduction processes separately, using the half-reaction method. This approach simplifies the complex task of balancing redox reactions by dividing the overall process into two simpler equations. The steps follow a logical sequence: writing separate equations for oxidation and reduction (half-reactions), balancing the atoms and charges in each half-reaction, and then combining them back into the balanced overall chemical equation.When balancing these half-reactions, it is important to account for any differences in the elements and charges by adding electrons, water molecules, or protons (\[\[\begin{align*} \(\text{H}^+\) \end{align*}\]\]ions) as needed. Once the number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction, they can be added together to recover the overall equation. This balanced approach is demonstrated in the given exercises. For instance, in reaction \[\[\begin{align*} \(\text{SO}_{3}^{2-}(aq) + \text{MnO}_{4}^{-}(aq) \rightarrow \text{SO}_{4}^{2-}(aq) + \text{Mn}^{2+}(aq)\) \end{align*}\]\], sulfate is oxidized and permanganate is reduced, each undergoing its own balancing before being combined for the final equation.

Chemical Equation Balancing Strategies

Balancing chemical equations, including redox reactions, requires careful attention to detail, and there are several strategies that chemists use to ensure that the law of conservation of mass is satisfied. One such strategy is the half-reaction method used for balancing redox reactions, as seen in the provided exercises. This method involves the separation of the overall reaction into two half-reactions that can be independently balanced and then combined. Another approach, commonly employed for non-redox reactions, is to balance atoms of different elements, starting with the most complex molecule and working towards the simplest, adjusting coefficients as necessary to maintain balance.Crucially, whatever the method used, the total number of atoms of each element and the total charge must be the same on both sides of the balanced equation. With practice and an understanding of these principles, balancing chemical equations becomes more intuitive. Using the half-reaction method simplifies this process significantly for redox reactions and helps students to see the direct relationships between reactants and products, ensuring that all students can grasp this fundamental chemical process.