Question

Balance each of the following oxidation-reduction reactions, which take place in acidic solution. a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{O}_{2}(g)\) b. \(\mathrm{BrO}_{3}^{-}(a q)+\mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Br}^{-}(a q)+\mathrm{Cu}^{2+}(a q)\) c. \(\mathrm{HNO}_{2}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{NO}(g)+\mathrm{I}_{2}(a q)\)

Short answer

The final balanced equations for the given redox reactions in acidic solutions are: a. \(4 \mathrm{MnO}_{4}^{-}(a q) + 10 \mathrm{H}_{2} \mathrm{O}_{2}(a q) + 12 \mathrm{H}^{+}(a q) \rightarrow 4 \mathrm{Mn}^{2+}(a q) + 5 \mathrm{O}_{2}(g) + 16 \mathrm{H}_{2}\mathrm{O}(l)\) b. \(2 \mathrm{BrO}_{3}^{-}(a q) + 6 \mathrm{Cu}^{+}(a q) + 12 \mathrm{H}^{+}(a q) \rightarrow 3 \mathrm{Cu}^{2+}(a q) + 6 \mathrm{Br}^{-}(a q) + 6 \mathrm{H}_{2} \mathrm{O}(l)\) c. \(2 \mathrm{HNO}_{2}(a q) + 2 \mathrm{I}^{-}(a q) \rightarrow \mathrm{NO}(g) + \mathrm{I}_{2}(a q) + \mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-step solution

Identify the oxidation and reduction half-reactions

For each reaction, we need to first determine the oxidation states of the elements and identify which ones are being oxidized and which are being reduced. a. In $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{O}_{2}(g)$: Mn is on +7 oxidation state in \(\mathrm{MnO}_{4}^{-}\) and is reduced to the +2 oxidation state in \(\mathrm{Mn}^{2+}\). O is on -1 oxidation state in \(\mathrm{H}_{2} \mathrm{O}_{2}\) and is oxidized to the 0 oxidation state in \(\mathrm{O}_{2}\). b. In $\mathrm{BrO}_{3}^{-}(a q)+\mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Br}^{-}(a q)+\mathrm{Cu}^{2+}(a q)$: Br is on +5 oxidation state in \(\mathrm{BrO}_{3}^{-}\) and is reduced to the -1 oxidation state in \(\mathrm{Br}^{-}\). Cu is on +1 oxidation state in \(\mathrm{Cu}^{+}\) and is oxidized to the +2 oxidation state in \(\mathrm{Cu}^{2+}\). c. In $\mathrm{HNO}_{2}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{NO}(g)+\mathrm{I}_{2}(a q)$: N is on +3 oxidation state in \(\mathrm{HNO}_{2}\) and is reduced to the +2 oxidation state in \(\mathrm{NO}\). I is on -1 oxidation state in \(\mathrm{I}^{-}\) and is oxidized to the 0 oxidation state in \(\mathrm{I}_{2}\).

Balance atoms, charge, and H2O in each half-reaction

We balance atoms, charges, and water molecules in each half-reaction, adjusting the coefficients as needed to ensure that the number of atoms and charges are the same on both sides of each equation. a. \(\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)\): Balance Mn atoms: Already balanced Balance O atoms: Add 4H2O to the right side Balance H atoms: Add 8H+ to the left side Balance charge: Add 5 electrons to the right side Final balanced half-reaction: \(5 \mathrm{e}^{-} + \mathrm{MnO}_{4}^{-}(a q) + 8 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q) + 4 \mathrm{H}_{2} \mathrm{O}(l)\) a. \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow \mathrm{O}_{2}(g)\): Balance O atoms: Adjust coefficients to 2H2O2 and O2 Balance H atoms: Add 4H+ to the right side Balance charge: Add 4 electrons to the left side Final balanced half-reaction: \(4 \mathrm{e}^{-} + 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 4 \mathrm{H}^{+}(a q) + \mathrm{O}_{2}(g)\) b. Reduce the reactions to their simplest forms and combine the half-reactions via the least common multiple of electrons: a. For Half-reactions: \(5 \mathrm{e}^{-} + \mathrm{MnO}_{4}^{-}(a q) + 8 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q) + 4 \mathrm{H}_{2} \mathrm{O}(l)\) \(4 \mathrm{e}^{-} + 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 4 \mathrm{H}^{+}(a q) + \mathrm{O}_{2}(g)\) LCM of 5 and 4 is 20, so multiply the first half-reaction by 4 and the second half-reaction by 5: \(20 \mathrm{e}^{-} + 4 \mathrm{MnO}_{4}^{-}(a q) + 32 \mathrm{H}^{+}(a q) \rightarrow 4 \mathrm{Mn}^{2+}(a q) + 16 \mathrm{H}_{2} \mathrm{O}(l)\) \(20 \mathrm{e}^{-} + 10 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 20 \mathrm{H}^{+}(a q) + 5 \mathrm{O}_{2}(g)\) Add the half-reactions together and simplify: \(4 \mathrm{MnO}_{4}^{-}(a q) + 10 \mathrm{H}_{2} \mathrm{O}_{2}(a q) + 32 \mathrm{H}^{+}(a q) \rightarrow 4 \mathrm{Mn}^{2+}(a q) + 16 \mathrm{H}_{2} \mathrm{O}(l) + 20 \mathrm{H}^{+}(a q) + 5 \mathrm{O}_{2}(g)\) Combine the H+ and cancel the electrons: \(4 \mathrm{MnO}_{4}^{-}(a q) + 10 \mathrm{H}_{2} \mathrm{O}_{2}(a q) + 12 \mathrm{H}^{+}(a q) \rightarrow 4 \mathrm{Mn}^{2+}(a q) + 16 \mathrm{H}_{2} \mathrm{O}(l) + 5 \mathrm{O}_{2}(g)\) Final balanced equation for Reaction A: \(4 \mathrm{MnO}_{4}^{-}(a q) + 10 \mathrm{H}_{2} \mathrm{O}_{2}(a q) + 12 \mathrm{H}^{+}(a q) \rightarrow 4 \mathrm{Mn}^{2+}(a q) + 5 \mathrm{O}_{2}(g) + 16 \mathrm{H}_{2}\mathrm{O}(l)\) Following the above steps, balance the remaining reactions (b, c). Final balanced equations: b. \(2 \mathrm{BrO}_{3}^{-}(a q) + 6 \mathrm{Cu}^{+}(a q) + 12 \mathrm{H}^{+}(a q) \rightarrow 3 \mathrm{Cu}^{2+}(a q) + 6 \mathrm{Br}^{-}(a q) + 6 \mathrm{H}_{2} \mathrm{O}(l)\) c. \(2 \mathrm{HNO}_{2}(a q) + 2 \mathrm{I}^{-}(a q) \rightarrow \mathrm{NO}(g) + \mathrm{I}_{2}(a q) + \mathrm{H}_{2} \mathrm{O}(l)\)

Key concepts

Oxidation States

Understanding oxidation states is crucial in balancing redox reactions. An oxidation state, also known as oxidation number, is a figure that represents the total number of electrons that an atom either loses or gains when forming a chemical bond. In simple terms, it's like a bookkeeping device that helps chemists track electrons during a chemical reaction.

In balancing redox equations, identifying the change in oxidation states between reactants and products reveals which elements are oxidized and which are reduced. Here are some general rules to determine oxidation states:

• For an atom in its elemental form, the oxidation state is zero.
• In ionic compounds, oxidation states correspond to the charge of the ion.
• In covalent compounds, more electronegative elements are assigned negative oxidation states, while less electronegative elements are assigned positive states, often equal to the charge it would have if the compound consisted of ions.

Linking to the Step-by-Step Solution

In the provided exercise, we can see how changes in oxidation states influence the balancing process. For instance, Mn changes from a +7 to a +2 oxidation state, indicating it has gained electrons and thus, been reduced. Conversely, O goes from -1 in \(\mathrm{H}_{2} \mathrm{O}_{2}\) to 0 in \(\mathrm{O}_{2}\), showing it has lost electrons and has been oxidized.

Half-Reaction Method

The half-reaction method is one of the standard techniques for balancing redox equations, especially complex ones. This approach breaks down the overall redox reaction into two simpler 'half-reactions,' one for oxidation and one for reduction. Each half-reaction is balanced individually for mass and charge.

Steps to Balance Using the Half-Reaction Method

• Write separate equations for the oxidation and reduction processes.
• Balance all elements except oxygen and hydrogen.
• Balance oxygen atoms by adding \(\mathrm{H}_{2}\mathrm{O}\) molecules.
• Balance hydrogen atoms by adding \(\mathrm{H}^{+}\) ions (in acidic solutions) or \(\mathrm{OH}^{-}\) ions (in basic solutions).
• Balance the charge by adding electrons—electrons gained should equal electrons lost.
• Multiply the half-reactions to equalize the number of electrons transferred in both the oxidation and reduction halves.
• Add the half-reactions together and cancel out anything that appears on both sides.

Applying this strategy to our textbook problems ensures the atomic and charge balance, as seen in the step-by-step solution for the given reactions.

Acidic Solution Chemistry

Acidic solution chemistry is paramount when balancing redox reactions that occur in an acidic medium. The presence of an acid implies an abundance of \(\mathrm{H}^{+}\) ions, which are used in the balancing process—particularly, when balancing the oxygen and hydrogen atoms within a half-reaction.

Key Points for Acidic Medium

• To balance oxygen atoms, add \(\mathrm{H}_{2}\mathrm{O}\) molecules to the side deficient in oxygen.
• Balance the hydrogen atoms by adding \(\mathrm{H}^{+}\) ions to the side deficient in hydrogen.
• If there is a need to balance the charges, \(\mathrm{H}^{+}\) ions may be used in conjunction with the addition of electrons.

When students apply these concepts to balance equations in an acidic medium, they can understand the role of \(\mathrm{H}^{+}\) ions and water molecules in maintaining equilibrium. The reversal of these rules can be applied if the reaction were to take place in a basic solution where \(\mathrm{OH}^{-}\) ions are abundant instead. The provided exercise serves as a practical application for these acidic solution principles, as it showcases the addition of \(\mathrm{H}^{+}\) ions and water for balance.