Question

Complete and balance the following equations: (a) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{H}^{+}(a q) \longrightarrow\) (b) \(\mathrm{Fe}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow\) (c) \(\mathrm{I}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{H}^{+}(a q) \longrightarrow\) (d) \(\mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{H}^{+}(a q) \longrightarrow\) (e) \(\mathrm{I}^{-}(a q)+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow \mathrm{I}_{2}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{OH}^{-}(a q)\)

Short answer

\( (a) \ \mathrm{MnO}_4^-(aq)+5H_2\mathrm{O}_2(aq)+8H^+(aq) \rightarrow 2H_2\mathrm{O}(l)+\mathrm{Mn}^{2+}(aq) \) \( (b) \ \mathrm{Fe}^{2+}(aq)+2H_2\mathrm{O}_2(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + 4H^+(aq) + 4O^{2-}(aq) \) \( (c) \ 2\mathrm{I}^-(aq)+2H_2\mathrm{O}_2(aq)+4H^+(aq) \rightarrow \mathrm{I}_2(s) + 4H_2\mathrm{O}(l) \) \( (d) \ \mathrm{Cu(s)}+2H_2\mathrm{O}_2(aq)+4H^+(aq) \rightarrow \mathrm{Cu}^{2+}(aq)+4H_2\mathrm{O}(l) \) \( (e) \ 2\mathrm{I}^-(aq)+\mathrm{O}_3(g) \rightarrow \mathrm{I}_2(s)+\mathrm{O}_2(g)+2\mathrm{OH}^-(aq) \)

Step-by-step solution

(a)

First, let's identify the half-reactions in the given equation. The permanganate ion (\(\mathrm{MnO}_4^-\)) is the oxidizing agent capable of being reduced, and hydrogen peroxide (\(\mathrm{H}_2\mathrm{O}_{2}\)) is the reducing agent capable of being oxidized. 1. Write the half-reactions: i) \(\mathrm{MnO}_4^-(aq)\) → \(Mn^{2+}(aq)\) + \(4O^{2-}\) ii) \(\mathrm{H}_2\mathrm{O}_2(aq)\) → \(2H^+(aq)\) + \(2O^{2-}\) 2. Balance the half-reactions: i) \(\mathrm{MnO}_4^-(aq)\) + \(4e^-\) → \(Mn^{2+}(aq)\) + \(4O^{2-}\) ii) \(2H_2\mathrm{O}_2(aq)\) → \(4H^+(aq)\) + \(4O^{2-}+4e^-\) 3. Add half-reactions: \(\mathrm{MnO}_4^-(aq)+5H_2\mathrm{O}_2(aq)+8H^+(aq)\) → \(2H_2\mathrm{O}(l)+Mn^{2+}(aq)+4O^{2-}(aq)+4O^{2-}(aq)+4H^+(aq)\) 4. Simplify and leave the final equation: \(\mathrm{MnO}_4^-(aq)+5H_2\mathrm{O}_2(aq)+8H^+(aq)\) → \(2H_2\mathrm{O}(l)+Mn^{2+}(aq)+4O^{2-}(aq)+4H^+(aq)\)

(b)

1. Write the half-reactions: i) \(\mathrm{Fe}^{2+}(aq)\) → \(\mathrm{Fe}^{3+}(aq)\) ii) \(\mathrm{H}_2\mathrm{O}_2(aq)\) → \(2H^+(aq)\) + \(2O^{2-}\) 2. Balance the half-reactions: i) \(\mathrm{Fe}^{2+}(aq) + e^-\) → \(\mathrm{Fe}^{3+}(aq)\) ii) \(2H_2\mathrm{O}_2(aq)\) → \(4H^+(aq)\) + \(4O^{2-}(aq) + 2e^-\) 3. Add half-reactions and simplify: \(\mathrm{Fe}^{2+}(aq)+2H_2\mathrm{O}_2(aq)\) → \(\mathrm{Fe}^{3+}(aq) + 4H^+(aq) + 4O^{2-}(aq)\)

(c)

1. Write the half-reactions: i) \(\mathrm{I}^-(aq)\) → \(\mathrm{I}_2 (s)\) ii) \(\mathrm{H}_2\mathrm{O}_2(aq)\) → \(2H^+(aq)\) + \(2O^{2-}\) 2. Balance the half-reactions: i) \(2\mathrm{I}^-(aq) + 2e^-\) → \(\mathrm{I}_2 (s)\) ii) \(2H_2O_2(aq)\) → \(4H^+(aq)\) + \(4O^{2-}(aq) + 2e^-\) 3. Add half-reactions and simplify: \(2\mathrm{I}^-(aq)+2H_2\mathrm{O}_2(aq)+4H^+(aq)\) → \(\mathrm{I}_2 (s) + 4H_2\mathrm{O}(l)\)

(d)

** 1. Write the half-reactions: i) \(\mathrm{Cu(s)}\) → \(\mathrm{Cu}^{2+}(aq)\) ii) \(\mathrm{H}_2\mathrm{O}_2(aq)\) → \(2H^+(aq)\) + \(2O^{2-}\) 2. Balance the half-reactions: i) \(\mathrm{Cu(s)} + 2e^-\) → \(\mathrm{Cu}^{2+}(aq)\) ii) \(2H_2\mathrm{O}_2(aq)\) → \(4H^+(aq)\) + \(4O^{2-}(aq) + 2e^-\) 3. Add half-reactions and simplify: \(\mathrm{Cu(s)}+2H_2\mathrm{O}_2(aq)+4H^+(aq)\) → \(\mathrm{Cu}^{2+}(aq)+4H_2\mathrm{O}(l)\)

(e)

This final equation is a nonredox reaction between aqueous iodide ions, gaseous ozone, and the resulting solid iodine, gaseous oxygen, and hydroxide ions in solution. The only unbalanced specie is \(\mathrm{H}^+\), so we just add some 2 waters to balance hydrogen: \(2\mathrm{I}^-(aq)+\mathrm{O}_3(g) \rightarrow \mathrm{I}_2(s)+\mathrm{O}_2(g)+2\mathrm{OH}^-(aq)\)

Key concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, commonly known as redox reactions, are chemical processes where electrons are transferred between two substances. In redox reactions, one substance gets oxidized (loses electrons) while the other gets reduced (gains electrons). An easy way to remember this is through the acronym OIL RIG - Oxidation Is Loss, Reduction Is Gain.

For example, when manganese (VII) oxide reacts with hydrogen peroxide in an acidic solution, manganese (VII) oxide is reduced to manganese (II) ions, and hydrogen peroxide is oxidized to water.

• MnO₄^- (oxidized) + H₂O₂ (reduced) + H⁺ → Mn²⁺ + O₂ + H₂O

Half-Reaction Method

The half-reaction method is a systematic approach to balancing redox equations. It involves separating the overall redox reaction into two half-reactions - one for oxidation and one for reduction. Each half-reaction is balanced independently for mass and charge.

As seen in the textbook solution, hydrogen peroxide is oxidized and permanganate ion is reduced:

• Oxidation half-reaction: H₂O₂ → O₂ + 2H⁺ + 2e^-
• Reduction half-reaction: MnO₄^- + 5e^- → Mn²⁺

After balancing each half-reaction for the number of atoms and charge, they can be combined to give the balanced chemical equation.

Chemical Equation Balancing

The balanced chemical equation is essential for accurately describing a chemical reaction. Each element must have the same number of atoms on both sides of the equation. Additionally, charges should also balance to comply with the conservation of charge. In redox reactions, this balance is achieved using the number of electrons lost in the oxidation half-reaction to equal the electrons gained in the reduction half-reaction.

Considering the given reactions, each element and charge is meticulously balanced to maintain equivalence from reactants to products. Substances like H⁺ and electrons are essentially bookkeeping tools that help us keep track of these conservation laws.

Stoichiometry

Stoichiometry is the field of chemistry that relates to calculating the quantities of reactants and products in a chemical reaction. It's based on the law of conservation of mass and the balanced chemical equation. Stoichiometry allows chemists to predict how much of each reactant is necessary to completely react without any excess, and how much of each product will form.

For instance, in the reaction between iodide ions and ozone, calculating the molar ratios of reactants to products helps us understand the exact amounts at play:

• 2I^- + O₃ → I₂ + O₂ + 2OH^-

In this stoichiometric relationship, every mole of ozone reacts with two moles of iodide ions to produce one mole of iodine, one mole of oxygen, and two moles of hydroxide ions.