Question

Balance the following reactions in acid: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Ni}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Sn}^{4+}(a q)\)

Short answer

Question: Balance the following redox reactions in an acidic solution: (a) Ni²⁺(aq) + H₂O₂(aq) → Ni³⁺(aq) + H₂O(l) (b) Cr₂O₇²⁻(aq) + Sn²⁺(aq) → Cr³⁺(aq) + Sn⁴⁺(aq) Answer: (a) The balanced redox reaction is: 2Ni²⁺(aq) + H₂O₂(aq) → 2Ni³⁺(aq) + 2H₂O(l) (b) The balanced redox reaction is: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 3Sn²⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l) + 3Sn⁴⁺(aq)

Step-by-step solution

Reaction (a)

Step 1: Identify the half-reactions Oxidation: \(Ni^{2+} \to Ni^{3+}\) Reduction: \(H_2O_2 \to H_2O\) Step 2: Balance the atoms in both half reactions Oxidation: \(Ni^{2+} \to Ni^{3+}\) Reduction: \(H_2O_2 \to 2H_2O\) Step 3: Balance the charges by adding electrons Oxidation: \(Ni^{2+} + e^- \to Ni^{3+}\) Reduction: \(H_2O_2 + 2e^- \to 2H_2O\) Step 4: Make sure the number of electrons gained and lost is the same for both half-reactions Oxidation: 1 electron is lost Reduction: 2 electrons are gained Multiply the oxidation half-reaction by 2: 2(\(Ni^{2+} + e^- \to Ni^{3+}\)) -> \(2Ni^{2+} + 2e^- \to 2Ni^{3+}\) Step 5: Add the half-reactions together and balance hydrogen ions (\(H^+\)) for the acidic condition \(2Ni^{2+} + 2e^- + H_2O_2 + 2e^- \to 2Ni^{3+} + 2H_2O\) Since there are no hydrogen ions present, no further balancing is needed. The balanced reaction is: \(2Ni^{2+}(aq) + H_2O_2(aq) \to 2Ni^{3+}(aq) + 2H_2O\)

Reaction (b)

Step 1: Identify the half-reactions Oxidation: \(Sn^{2+} \to Sn^{4+}\) Reduction: \(Cr_2O_7^{2-} \to 2Cr^{3+}\) Step 2: Balance the atoms in both half reactions Oxidation: \(Sn^{2+} \to Sn^{4+}\) Reduction: \(Cr_2O_7^{2-} + 14H_2O \to 2Cr^{3+} + 7O_2\) Step 3: Balance the charges by adding electrons Oxidation: \(Sn^{2+} + 2e^- \to Sn^{4+}\) Reduction: \(Cr_2O_7^{2-} + 6e^- + 14H^+ \to 2Cr^{3+} + 7H_2O\) Step 4: Make sure the number of electrons gained and lost is the same for both half-reactions Oxidation: 2 electrons are lost Reduction: 6 electrons are gained Multiply the oxidation half-reaction by 3: 3(\(Sn^{2+} + 2e^- \to Sn^{4+}\)) -> \(3Sn^{2+} + 6e^- \to 3Sn^{4+}\) Step 5: Add the half-reactions together and balance any remaining hydrogen ions (\(H^+\)) for the acidic condition \(Cr_2O_7^{2-} + 6e^- + 14H^+ + 3Sn^{2+} + 6e^- \to 2Cr^{3+} + 7H_2O + 3Sn^{4+}\) The balanced reaction is: \(Cr_2O_7^{2-}(aq) + 14H^+(aq) + 3Sn^{2+}(aq) \to 2Cr^{3+}(aq) + 7H_2O + 3Sn^{4+}(aq)\)

Key concepts

Oxidation and Reduction

Oxidation and reduction are two key concepts in the study of redox reactions, which involve the transfer of electrons between substances.

In an oxidation process, an element loses electrons, which increases its oxidation state. This electron loss can be easily remembered with the acronym OIL (Oxidation Is Loss). Conversely, reduction involves the gain of electrons by an element, leading to a decrease in its oxidation state, similar to the mnemonic RIG (Reduction Is Gain).

For example, in the reaction where \(Ni^{2+}\) is converted to \(Ni^{3+}\), the nickel ion is oxidized because it loses an electron. Likewise, when \(H_2O_2\) is converted to \(H_2O\), it is reduced because it gains electrons. Understanding the differences between oxidation and reduction is crucial for effectively balancing redox reactions.

Half-Reactions

Half-reactions are a way to break down the overall redox reaction into its two separate components: one that represents oxidation and the other reduction.

Each half-reaction is balanced independently, which includes balancing the number of atoms involved and their charges with electrons. In the example problems, the half-reactions for the oxidation processes and the reduction processes are identified and balanced separately before recombining them to form the full balanced equation.

By focusing on one half-reaction at a time, students can more easily ensure that the number of atoms and the charge are conserved. This step-by-step approach is particularly helpful for complex equations and is a critical strategy for accurate and complete balancing of redox reactions.

Balancing Chemical Equations

Balancing chemical equations involves making sure that the number of each type of atom and the total charge is the same on both sides of the equation.

In a redox reaction, it is important to balance the elements and charges separately for each half-reaction first. Once both half-reactions are balanced, they must be combined in such a way that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. This may involve multiplying one or both half-reactions by an appropriate factor.

Furthermore, when balancing redox reactions in acidic solutions, as seen in our examples, hydrogen ions (\(H^+\)) are added to balance excess oxygen atoms, and water molecules are included to balance excess hydrogen atoms. This ensures that the final equation reflects the conservation of mass and charge.

Electron Transfer

Electron transfer is the heart of a redox reaction. It involves the movement of electrons from one species to another, which constitutes the oxidation and reduction processes.

In our examples, the nickel ion \(Ni^{2+}\) loses an electron becoming \(Ni^{3+}\), and \(H_2O_2\) gains electrons to become \(H_2O\), with the half-reactions illustrating this electron transfer explicitly. The balancing of electrons ensures that the same number of electrons are transferred in both the oxidation and the reduction processes, which is crucial for the conservation of charge.

It's important to remember that no electrons are actually lost or gained in the reaction as a whole; they are simply shifted from one reactant to another. This is why the electron transfer must be equal and opposite when combining the half-reactions to achieve a balanced overall reaction.