Question

In basic solution, aluminum metal is a strong reducing agent and is oxidized to \(\mathrm{AlO}_{2}^{-}\). Give balanced equations for the reaction of \(\mathrm{Al}\) in basic solution with the following: (a) \(\mathrm{NaNO}_{3}\), to give ammonia; (b) water, to give hydrogen; (c) \(\mathrm{Na}_{2} \mathrm{SnO}_{3}\), to give metallic tin.

Short answer

(a) 8Al + 3NO3⁻ + 12H2O → 8AlO2⁻ + 3NH3 + 9OH⁻. (b) 2Al + 6H2O → 3H2 + 2AlO2⁻ + 2OH⁻. (c) 4Al + 3SnO32⁻ + 2H2O → 3Sn + 4AlO2⁻ + 4OH⁻.

Step-by-step solution

Balance the Reaction with NaNO3

In basic solution, aluminum (Al) reacts with sodium nitrate (NaNO3) to produce aluminum oxide ion \( \mathrm{AlO}_2^- \) and ammonia (NH₃). First, identify that aluminum is being oxidized and nitrogen is being reduced. The half-reactions for oxidation and reduction are \( \mathrm{Al} \rightarrow \mathrm{AlO}_2^- + 3e^- \) and \( \mathrm{NO_3^-} + 10e^- + 12\mathrm{H_2O} \rightarrow \mathrm{NH_3} + 9\mathrm{OH^-} \), respectively. Balance both reactions by adding appropriate OH⁻ and water molecules, then scale the reactions to balance the electron transfer and combine to form the overall balanced equation: \[ 8\mathrm{Al} + 3\mathrm{NO_3^-} + 12\mathrm{H_2O} \rightarrow 8\mathrm{AlO}_2^- + 3\mathrm{NH_3} + 9\mathrm{OH^-} \].

Balance the Reaction with Water

Aluminum reacts with water (H₂O) in a basic solution to produce hydrogen gas (H₂) and aluminum oxide ion \( \mathrm{AlO}_2^- \). The reaction starts with \( \mathrm{Al} + \mathrm{H_2O} \rightarrow \mathrm{H_2} + \mathrm{AlO}_2^- \), but it needs to be balanced. First write the oxidation half-reaction \( \mathrm{Al} \rightarrow \mathrm{AlO}_2^- + 3e^- \) and the reduction half-reaction \( \mathrm{H_2O} + e^- \rightarrow \mathrm{H_2} + \mathrm{OH^-} \). Scale these reactions to balance the electrons: multiply the reduction reaction by 3, then combine them. The balanced reaction is: \[ 2\mathrm{Al} + 6\mathrm{H_{2}O} \rightarrow 3\mathrm{H_{2}} + 2\mathrm{AlO_{2}^{-}} + 2\mathrm{OH^-} \].

Balance the Reaction with Na2SnO3

For the reaction of aluminum with sodium stannate (Na₂SnO₃) producing metallic tin (Sn), write the half-reactions: \( \mathrm{Al} \rightarrow \mathrm{AlO}_2^- + 3e^- \) for oxidation and \( \mathrm{SnO_3^{2-}} + 4\mathrm{H_2O} + 4e^- \rightarrow \mathrm{Sn} + 8\mathrm{OH^-} \) for reduction. Balance the electrons (multiply the oxidation half by 4 and the reduction half by 3). Combine the half-reactions: \[ 4\mathrm{Al} + 3\mathrm{SnO_3^{2-}} + 2\mathrm{H_2O} \rightarrow 3\mathrm{Sn} + 4\mathrm{AlO_2^-} + 4\mathrm{OH^-} \].

Key concepts

Oxidation

Oxidation is a fundamental aspect of redox reactions, where an element loses electrons. In the context of our exercise, aluminum (\( \mathrm{Al} \)) undergoes oxidation. It loses multiple electrons, transforming into the aluminum oxide ion (\( \mathrm{AlO}_2^- \)). This transformation is evident in the half-reaction: aluminum giving up its electrons. The electrons lost in this process play a crucial role in balancing equations, as we'll see later on. Oxidation not only changes the charge of the element but also its chemical properties. Keep in mind, oxidation does not always mean gaining oxygen, it simply indicates the loss of electrons. In reactions involving metals like aluminum, oxidation can lead to significant energy exchange.

Reduction

Reduction complements oxidation in redox reactions. It involves the gain of electrons by an element. For example, in our example, the nitrate ion (\( \mathrm{NO_3^-} \)) is reduced to ammonia (\( \mathrm{NH_3} \)), while the stannate ion (\( \mathrm{SnO_3^{2-}} \)) is reduced to metallic tin. Reduction is always paired with oxidation; as one substance loses electrons (oxidation), another gains them (reduction). In chemical terms, the reduction reaction also involves the balancing of charges by the addition of hydrogen atoms or producing hydroxide ions. Understanding reduction helps grasp the energy transfer and chemical changes occurring during the reaction.

Balanced Equations

A balanced equation ensures that the same number of each type of atom appears on both sides of the reaction. This maintains the principle of mass conservation. Balancing equations involves equating the electrons transferred during oxidation and reduction. Steps include: writing separate half-reactions for oxidation and reduction, scaling them to equalize electron transfer, and combining them. This helps accurately represent the chemical reaction occurring. In redox reactions, balancing often involves adding water and hydroxide ions in a basic solution. It's like a puzzle; you need to fit all parts correctly for the whole picture to make sense.

Half-Reactions

Half-reactions are a tool to simplify the process of analyzing redox reactions. By breaking down the overall reaction into oxidation and reduction components, each half-reaction focuses on just one element or ion gaining or losing electrons. This separation helps in easily balancing more complex reactions. Starting with half-reactions, we can understand how electrons are shuffled around. Electrons lost in the oxidation half-reaction equal those gained in the reduction, maintaining charge balance. After balancing electrons in both halves, they are recombined to form the overall balanced reaction. This systematic approach makes complex chemical equations easier to handle.

Basic Solution

A basic solution in redox reactions means the presence of excess hydroxide ions (\( \mathrm{OH^-} \)). Balancing redox reactions in basic solutions often requires adding water or hydroxide ions to both sides of the equation. This ensures that atoms, charge, and the medium's basicity are all considered. The steps involve writing separate half-reactions and adding hydroxide ions where necessary to maintain balance. This practice ensures that, not only elements but also the solution's nature are maintained correctly throughout the reaction. Understanding how a basic solution influences reactions can significantly aid in predicting product outcomes and reaction behavior.