Problem 25 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) Balanced equation: \(\mathrm{Cr}_{2} \mathrm{O}_{7}{\underline{\phantom{xx}}}^{2-}(a q)+2\Gamma^{-}(a q)+14\mathrm{H}^+ \longrightarrow 2\mathrm{Cr}^{3}(a q)+\mathrm{IO}_{3}^{-}(a q)+7\mathrm{H}_2\mathrm{O}\); Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), Reducing agent: \(\Gamma^{-}\) (b) Balanced equation: \(\mathrm{MnO}_{4}^{-}(a q)+2\mathrm{CH}_{3} \mathrm{OH}(a q)+6\mathrm{H}^+ \longrightarrow \mathrm{Mn}^{2+}(a q)+ 2\mathrm{HCO}_{2} \mathrm{H}(a q)+ 4\mathrm{H}_2\mathrm{O}\); Oxidizing agent: \(\mathrm{MnO}_{4}^{-}\), Reducing agent: \(\mathrm{CH}_{3} \mathrm{OH}\)

Step 1: Identify the oxidation states

The oxidation states of each element are given as follows: Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is +6, I in \(\Gamma^{-}\) is -1, Cr in \(\mathrm{Cr}^{3+}\) is +3, and I in \(\mathrm{IO}_{3}^{-}\) is +5.

Step 2: Write half-reactions

We can write the following half-reactions: Oxidation: \(2\Gamma^{-} \longrightarrow \mathrm{IO}_{3}^{-} +6e^{-}\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6e^{-} \longrightarrow 2\mathrm{Cr}^{3+} \)

Step 3: Balance the half-reactions and combine

The two half-reactions are already balanced in terms of electrons. So, we can combine them, and add 14H+ ions to compensate the protons: $$\mathrm{Cr}_{2} \mathrm{O}_{7}{\underline{\phantom{xx}}}^{2-}(a q)+2\Gamma^{-}(a q)+14\mathrm{H}^+ \longrightarrow 2\mathrm{Cr}^{3}(a q)+\mathrm{IO}_{3}^{-}(a q)+7\mathrm{H}_2\mathrm{O}$$

Step 4: Identify the oxidizing and reducing agents

The oxidizing agent is \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), and the reducing agent is \(\Gamma^{-}\). (b)

Step 1: Identify the oxidation states

Mn in \(\mathrm{MnO}_{4}^{-}\) is +7, C in \(\mathrm{CH}_{3} \mathrm{OH}\) is -2, Mn in \(\mathrm{Mn}^{2+}\) is +2, C in \(\mathrm{HCO}_{2} \mathrm{H}\) is +2.

Step 2: Write half-reactions

We can write the following half-reactions: Oxidation: \(\mathrm{CH}_{3} \mathrm{OH} \longrightarrow \mathrm{HCO}_{2} \mathrm{H} +2e^{-}\) Reduction: \(\mathrm{MnO}_{4}^{-}+5e^{-} \longrightarrow \mathrm{Mn}^{2+}\)

Step 3: Balance the half-reactions and combine

Balance the reduction half-reaction with two moles of oxidation half-reaction and add 6H+ ions to compensate the protons: $$\mathrm{MnO}_{4}^{-}(a q)+2\mathrm{CH}_{3} \mathrm{OH}(a q)+6\mathrm{H}^+ \longrightarrow \mathrm{Mn}^{2+}(a q)+ 2\mathrm{HCO}_{2} \mathrm{H}(a q)+ 4\mathrm{H}_2\mathrm{O}$$

Step 4: Identify the oxidizing and reducing agents

The oxidizing agent is \(\mathrm{MnO}_{4}^{-}\), and the reducing agent is \(\mathrm{CH}_{3} \mathrm{OH}\). Similar approach can be taken for the remaining parts (c), (d), (e), and (f).

Key Concepts

Oxidation States

Understanding oxidation states is fundamental in identifying and balancing redox reactions. Oxidation states represent the hypothetical charge an atom would have if all its bonds to atoms of different elements were completely ionic. These values are assigned based on a set of rules, for examples, oxygen usually has an oxidation state of -2 (except in peroxides), and hydrogen is typically +1 when bonded to nonmetals.

An increase in oxidation state through a chemical reaction indicates oxidation, whereas a decrease signifies reduction. By identifying changes in oxidation states, we can pinpoint the substances being oxidized and reduced in a redox process. In the provided exercise, the challenge is to identify the oxidation states of the elements involved and use these to determine the flow of electrons, which is critical for balancing the reaction.

Half-Reactions

In a redox reaction, half-reactions provide a clearer picture of the oxidation and reduction processes taking place. By separating the overall reaction into two halves--one for oxidation and one for reduction--we can focus on the changes occurring to each reactant independently.

Each half-reaction shows the electrons lost or gained by the reactants. Balancing these half-reactions individually for mass and charge is key to ensuring the overall reaction is balanced. Moreover, when writing half-reactions, additional elements such as H₂O, H⁺, and OH^- are sometimes used to balance the equation in acidic or basic conditions. This is seen in the solution for the exercise where protons (H⁺) are added

Oxidizing and Reducing Agents

Oxidizing and reducing agents are essential components of redox reactions. An oxidizing agent, or oxidant, gains electrons and is reduced in the process, whereas a reducing agent, or reductant, loses electrons and is oxidized.

To determine the oxidizing and reducing agents in a reaction, one must examine the changes in oxidation states of the reactants. The species that undergoes an increase in its oxidation state is the reducing agent, and the one that undergoes a decrease is the oxidizing agent. In the textbook exercise, identifying these agents not only aids in understanding the reaction but also confirms whether the half-reactions have been correctly balanced in terms of electron transfer.

Stoichiometry of Redox Reactions

The stoichiometry of redox reactions involves the quantitative relationship between the number of moles of reactants and products. To achieve a balanced redox reaction, we must ensure that the number of atoms of each element is the same on both sides of the reaction equation and that the total charges are equal. This balancing act requires adjustments of coefficients in front of compounds in the reaction.

In redox reactions, particular attention is given to balancing the number of electrons lost in the oxidation half-reaction with those gained in the reduction half-reaction. This ensures the conservation of charge. If the electrons do not balance out, coefficients for the entire half-reactions are adjusted. Balancing reactions in acidic or basic solutions may require adding H⁺ or OH^- ions and water molecules to achieve the final balanced equation, as seen in the exercise solutions provided.