Question

Balance each redox reaction occurring in acidic aqueous solution. $$\begin{array}{l}{\text { a. } \mathrm{PbO}_{2}(s)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Pb}^{2+}(a q)+\mathrm{I}_{2}(s)} \\ {\text { b. } \mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)} \\ {\text { c. } \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Cl}^{-}(a q)}\end{array}$$

Short answer

a) PbO_2(s) + 2I^-(aq) + 4H^+ -> Pb^{2+}(aq) + I2(s) + 2H2O(l), b) 2MnO4^-(aq) + 5SO3^{2-}(aq) + 16H^+ -> 2Mn^{2+}(aq) + 5SO4^{2-}(aq) + 8H2O(l), c) S2O3^{2-}(aq) + Cl2(g) -> 2SO4^{2-}(aq) + 2Cl^-(aq).

Step-by-step solution

Balancing the reaction in acidic medium (a)

Write the half-reactions for the reduction and oxidation processes. Reduction: \( \mathrm{PbO}_2(s) + 4H^+ + 2e^- \rightarrow \mathrm{Pb}^{2+}(aq) + 2H_2O(l) \). Oxidation: \( 2\mathrm{I}^-(aq) \rightarrow \mathrm{I}_2(s) + 2e^- \). Balance the number of electrons in each half-reaction to equalize the loss and gain. The half-reactions are already balanced so they can be combined as is.

Balancing the reaction in acidic medium (b)

Write the half-reactions for reduction and oxidation processes. Reduction: \( \mathrm{MnO}_4^-(aq) + 8H^+ + 5e^- \rightarrow \mathrm{Mn}^{2+}(aq) + 4H_2O(l) \). Oxidation: \( \mathrm{SO}_3^{2-}(aq) \rightarrow \mathrm{SO}_4^{2-}(aq) + 2e^- \). Now, balance the number of electrons between the two half-reactions by multiplying the second half-reaction by 5 and the first by 2, then combine.

Balancing the reaction in acidic medium (c)

Write the half-reactions for the reduction and oxidation processes. Reduction: \( \mathrm{Cl}_2(g) + 2e^- \rightarrow 2\mathrm{Cl}^-(aq) \). Oxidation: \( \mathrm{S}_2O_3^{2-}(aq) \rightarrow \mathrm{SO}_4^{2-}(aq) + 2e^- \). As the number of electrons is the same, these half-reactions can be combined directly.

Combine the half-reactions

For each reaction, add the half-reactions together, making sure the electrons cancel out. a) \( \mathrm{PbO}_2(s) + 2\mathrm{I}^-(aq) + 4H^+ \rightarrow \mathrm{Pb}^{2+}(aq) + \mathrm{I}_2(s) + 2H_2O(l) \). b) \( 2\mathrm{MnO}_4^-(aq) + 5\mathrm{SO}_3^{2-}(aq) + 16H^+ \rightarrow 2\mathrm{Mn}^{2+}(aq) + 5\mathrm{SO}_4^{2-}(aq) + 8H_2O(l) \). c) \( \mathrm{S}_2O_3^{2-}(aq) + \mathrm{Cl}_2(g) \rightarrow 2\mathrm{SO}_4^{2-}(aq) + 2\mathrm{Cl}^-(aq) \) ensuring that the same number of atoms and charges appear on both sides of the equation.

Key concepts

Redox Reaction in Acidic Solution

Redox reactions in an acidic solution are a crucial component of chemical processes where electron transfer occurs in an environment abundant in hydrogen ions (\text{H}^+). In acidic solutions, certain elements gain electrons (reduction) while others lose electrons (oxidation). The acidic medium provides \text{H}^+ ions that can interact with the reactants or products to help balance the overall charge in the redox equation. For example, when balancing the reduction of lead dioxide (\text{PbO}_2), we see that hydrogen ions are necessary to balance out the charge as it gains electrons:
\[ \text{PbO}_2(s) + 4H^+ + 2e^- \rightarrow \text{Pb}^{2+}(aq) + 2H_2O(l) \]
It is essential to maintain the conservation of mass and charge by ensuring that the number of atoms and the total charge are balanced on both sides of the redox equation in an acidic solution.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions. This method involves separating the redox reaction into two half-reactions – one for oxidation and one for reduction. Each half-reaction is balanced individually for mass and charge. For example, in the oxidation half-reaction involving iodide ions, the electrons appear as products as iodide is oxidized to iodine:
\[ 2\text{I}^-(aq) \rightarrow \text{I}_2(s) + 2e^- \]
Once both half-reactions are balanced, they are then combined by equalizing the number of electrons transferred in each half-reaction to ensure that all electrons are accounted for and that the charges cancel out. This results in a balanced overall reaction, demonstrating the importance of electron balance in redox reactions.

Oxidation and Reduction Processes

Oxidation and reduction are two halves of the same process in redox reactions; one cannot occur without the other. Oxidation involves the loss of electrons, while reduction is the gain of electrons. During a redox reaction, the substance that loses electrons gets oxidized, and the substance that gains electrons gets reduced. These dual processes are often remembered with the mnemonic 'LEO says GER' (Loss of Electrons is Oxidation, Gain of Electrons is Reduction). An example drawn from our exercises involves the sulfate and permanganate ions:
\[ \text{Oxidation: } \text{SO}_3^{2-}(aq) \rightarrow \text{SO}_4^{2-}(aq) + 2e^- \]
\[ \text{Reduction: } \text{MnO}_4^-(aq) + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4H_2O(l) \]
The substance that undergoes oxidation, in this case, is the sulfite ion (SO_3^{2-}), which loses electrons to form sulfate (SO_4^{2-}), whereas the permanganate ion (MnO_4^{-}) undergoes reduction as it gains electrons.

Electron Balance in Redox

In a redox reaction, electron balance is achieved when the total number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. It is a fundamental aspect of the half-reaction method and ensures that the law of conservation of charge is upheld. This balancing act can sometimes require multiplying one or both half-reactions by appropriate coefficients. For instance, in the reaction between sulfite and permanganate ions, the oxidation half-reaction produces 2 electrons, whereas the reduction half-reaction requires 5 electrons:
\[ \text{Oxidation: } 5(\text{SO}_3^{2-}(aq) \rightarrow \text{SO}_4^{2-}(aq) + 2e^-) \]
\[ \text{Reduction: } 2(\text{MnO}_4^-(aq) + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4H_2O(l)) \]
By scaling the half-reactions so that they transfer the same number of electrons, they can be combined to give a correct overall equation, reflecting the precise electron balance required in a redox reaction.