Question

Complete and balance the following equations, and identify the oxidizing and reducing agents: $$ \begin{array}{l}{\text { (a) } \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)} \\ \quad {\text { (acidic solution) }} \\ {\text { (b) } \mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+} \\ \quad {\mathrm{HCOOH}(a q) \text { (acidic solution) }}\end{array} \\ {\text {(c) } \mathrm{I}_{2}(s)+\mathrm{OCl}^{-}(a q) \longrightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)} \\ {\text { (acidic solution) }} \\ {\text { (d) } \mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)} \\ {(\text { acidic solution })} \\ {\text { (e) } \operatorname{MnO}_{4}^{-}(a q)+\operatorname{Br}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{3}^{-}(a q)} \\ {\text { (basic solution) }} \\ {\text { (f) } \mathrm{Pb}(\mathrm{OH})_{4}^{2-}(a q)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)} \\ {\text { (basic solution) }} $$

Short answer

The balanced reactions and the oxidizing and reducing agents are: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^+ + 6\mathrm{I}^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + \mathrm{I}_{2}\) Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) Reducing agent: \(\mathrm{I}^-\) (b) \(\mathrm{MnO}_{4}^{-}(aq) + 5\mathrm{CH}_{3}\mathrm{OH}(aq) + 2\mathrm{H}^+ \rightarrow \mathrm{Mn}^{2+}(aq) + 5\mathrm{HCOOH}(aq)\) Oxidizing agent: \(\mathrm{MnO}_{4}^-\) Reducing agent: \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\mathrm{I}_{2}(s) + 2\mathrm{OCl}^{-}(aq) + 6\mathrm{H}^{+}(aq) \rightarrow 2\mathrm{IO}_{3}^{-}(aq) + 4\mathrm{Cl}^{-}(aq) + 6\mathrm{H}^{+}(aq)\) Oxidizing agent: \(\mathrm{OCl}^-\) Reducing agent: \(\mathrm{I}_{2}\) (d) \(2\mathrm{As}_{2}\mathrm{O}_{3}(s) + 3\mathrm{NO}_{3}^{-}(aq) + 10\mathrm{H}^+(aq) \rightarrow 4\mathrm{H}_{3}\mathrm{AsO}_{4}(aq) + 3\mathrm{N}_{2}\mathrm{O}_{3}(aq)\) Oxidizing agent: \(\mathrm{NO}_3^{-}\) Reducing agent: \(\mathrm{As}_{2}\mathrm{O}_{3}\) (e) \(\mathrm{MnO}_{4}^{-}(aq) + 2\mathrm{Br}^{-}(aq) + 4\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{MnO}_{2}(s) + 2\mathrm{BrO}_{3}^{-}(aq) + 4\mathrm{OH}^{-}(aq)\) Oxidizing agent: \(\mathrm{MnO}_{4}^-\) Reducing agent: \(\mathrm{Br}^-\) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}^{2-}(aq) + 2\mathrm{ClO}^{-}(aq) \rightarrow \mathrm{PbO}_{2}(s) + 2\mathrm{Cl}^{-}(aq)+ 4\mathrm{OH}^{-}(aq)\) Oxidizing agent: \(\mathrm{ClO}^-\) Reducing agent: \(\mathrm{Pb}(\mathrm{OH})_{4}^{2-}\)

Step-by-step solution

Identify the changes in oxidation state

: Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\): +6 Cr in \(\mathrm{Cr}^{3+}\): +3 I in \(\mathrm{I}^-(aq)\): -1 I in \(\mathrm{IO}_{3}^{-}\): +5

Write the unbalanced half-reactions

: Oxidation: \(\mathrm{I}^- \rightarrow \mathrm{IO}_{3}^-\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow \mathrm{Cr}^{3+}\)

Balance the half-reactions

: Oxidation: \[\mathrm{2I}^- \rightarrow \mathrm{I}_{2} + 6e^-\] Reduction (using 14H+): \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6e^- + 14\mathrm{H}^+ \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\]

Combine the balanced half-reactions

: \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^+ + 6\mathrm{I}^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} + \mathrm{I}_{2}\] Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) Reducing agent: \(\mathrm{I}^-\) #b:# For brevity, we will follow the same procedure for reactions (b), (c), (d), (e), and (f). The balanced reactions are given below along with the oxidizing and reducing agents: \\(\text{(b)}: \mathrm{MnO}_{4}^{-}(aq) + 5\mathrm{CH}_{3}\mathrm{OH}(aq) + 2\mathrm{H}^+ \rightarrow \mathrm{Mn}^{2+}(aq) + 5\mathrm{HCOOH}(aq)\\) Oxidizing agent: \(\mathrm{MnO}_{4}^-\) Reducing agent: \(\mathrm{CH}_{3} \mathrm{OH}\) \\(\text{(c)}: \mathrm{I}_{2}(s) + 2\mathrm{OCl}^{-}(aq) + 6\mathrm{H}^{+}(aq) \rightarrow 2\mathrm{IO}_{3}^{-}(aq) + 4\mathrm{Cl}^{-}(aq) + 6\mathrm{H}^{+}(aq)\\) Oxidizing agent: \(\mathrm{OCl}^-\) Reducing agent: \(\mathrm{I}_{2}\) \\(\text{(d)}: 2\mathrm{As}_{2}\mathrm{O}_{3}(s) + 3\mathrm{NO}_{3}^{-}(aq) + 10\mathrm{H}^+(aq) \rightarrow 4\mathrm{H}_{3}\mathrm{AsO}_{4}(aq) + 3\mathrm{N}_{2}\mathrm{O}_{3}(aq)\\) Oxidizing agent: \(\mathrm{NO}_3^{-}\) Reducing agent: \(\mathrm{As}_{2}\mathrm{O}_{3}\) \\(\text{(e)}: \mathrm{MnO}_{4}^{-}(aq) + 2\mathrm{Br}^{-}(aq) + 4\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{MnO}_{2}(s) + 2\mathrm{BrO}_{3}^{-}(aq) + 4\mathrm{OH}^{-}(aq)\\) Oxidizing agent: \(\mathrm{MnO}_{4}^-\) Reducing agent: \(\mathrm{Br}^-\) \\(\text{(f)}: \mathrm{Pb}(\mathrm{OH})_{4}^{2-}(aq) + 2\mathrm{ClO}^{-}(aq) \rightarrow \mathrm{PbO}_{2}(s) + 2\mathrm{Cl}^{-}(aq)+ 4\mathrm{OH}^{-}(aq)\\) Oxidizing agent: \(\mathrm{ClO}^-\) Reducing agent: \(\mathrm{Pb}(\mathrm{OH})_{4}^{2-}\)

Key concepts

Oxidation and Reduction Agents

Understanding oxidation and reduction agents is crucial in redox chemistry. These agents are substances that have the capacity to either gain electrons (reduction) or lose electrons (oxidation). An oxidizing agent accepts electrons from another substance, thereby being reduced itself. For example, in exercise (a), \(\mathrm{Cr}_2 \mathrm{O}_{7}^{2-}\) is an oxidizing agent because it gains electrons and reduces from a +6 to a +3 oxidation state. On the other hand, a reducing agent donates electrons, undergoing oxidation itself. In the same reaction, iodide ion \(\mathrm{I}^-\) acts as the reducing agent because it loses electrons and its oxidation state increases from -1 to +5. Identifying these agents in a redox reaction helps in comprehending the flow of electrons and the overall process of the reaction.

Remember, the oxidizing agent gets reduced and the reducing agent gets oxidized. This may seem counterintuitive at first, but the terms refer to the agents' actions on other substances, not on themselves!

Half-reaction Method

Balancing redox reactions can be simplified by using the half-reaction method. This technique involves splitting the overall redox reaction into two separate half-reactions—one for oxidation and one for reduction. Each half-reaction is balanced independently for mass and charge. In the problem provided, the oxidation half-reaction \(\mathrm{I}^- \rightarrow \mathrm{IO}_{3}^-\) is balanced by adding electrons to compensate for the change in oxidation number. Similarly, the reduction half-reaction \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow \mathrm{Cr}^{3+}\) is balanced using \(\mathrm{H}^+\) ions and water molecules. After both half-reactions are balanced, they can be combined to give the balanced overall equation. This method is particularly helpful as it allows you to focus on each part of the reaction one step at a time, making it easier to ensure that all atoms and charges are accounted for.

Oxidation States

The concept of oxidation states or oxidation numbers is foundational in identifying how electrons are redistributed in redox processes. An oxidation state is a hypothetical charge that an atom would have if all bonds were completely ionic. When balancing redox reactions, it's critical to determine the change in oxidation states of the elements involved to figure out which species are oxidized and which are reduced. For instance, chromium's oxidation state changes from +6 in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to +3 in \(\mathrm{Cr}^{3+}\). The change in oxidation state also helps in determining how many electrons are gained or lost in the half-reactions. In exercises, identifying oxidation states is often the first step to understanding the electron transfer that constitutes the core of the reaction.

Acidic and Basic Solutions

The environment of a redox reaction, whether it is in an acidic or basic solution, has an impact on the reaction's balancing. For redox reactions in acidic solutions, \(\mathrm{H}^+\) ions are used to balance hydrogen atoms, and water molecules are used to balance oxygen atoms. In contrast, for reactions in basic solutions, \(\mathrm{OH}^-\) ions and water are used for balancing purposes. Like in exercise (e), where \(\mathrm{OH}^-\) ions appear on the product side of the balanced reaction for the basic medium. Being aware of the reaction environment is essential because it determines which additional species (\(\mathrm{H}^+\) or \(\mathrm{OH}^-\)) will be used in the balancing process. Balancing reactions in different environments requires careful attention to the specific substances involved, ensuring both mass and charge are balanced.