Question

Write the equation for the reaction, if any, that occurs when each of the following experiments is performed under standard conditions. (a) Sulfur is added to mercury. (b) Manganese dioxide in acidic solution is added to liquid mercury. (c) Aluminum metal is added to a solution of potassium ions.

Short answer

Question: Predict and write balanced chemical equations for the possible reactions that could occur under standard conditions when given the following sets of reactants: (a) Sulfur and mercury (b) Manganese dioxide, an acidic solution, and mercury (c) Aluminum and potassium ions Answer: (a) Hg(l) + S(s) -> HgS(s) (b) 2 Hg(l) + MnO₂(s) + 4H⁺(aq) -> 2 Hg²⁺(aq) + Mn²⁺(aq) + 2H₂O(l) (c) 2 Al(s) + 6K⁺(aq) -> 2 Al³⁺(aq) + 6 K(s)

Step-by-step solution

(a) Sulfur and mercury reaction

Since sulfur is a non-metal and mercury is a transition metal, there is a possibility that they can form an ionic compound. The cation is Hg²⁺ and the anion is S²⁻. So, the formula for the compound (mercury(II) sulfide) will be HgS. The balanced chemical equation for this reaction is: Hg(l) + S(s) -> HgS(s)

(b) Manganese dioxide, acidic solution, and mercury reaction

In this reaction, manganese dioxide (MnO₂) is in an acidic solution and reacts with liquid mercury (Hg). MnO₂ can act as an oxidizing agent, oxidizing Hg to Hg²⁺. In the process, Mn from MnO₂ gets reduced to Mn²⁺. The balanced redox reaction for an acidic solution is: 2 Hg(l) + MnO₂(s) + 4H⁺(aq) -> 2 Hg²⁺(aq) + Mn²⁺(aq) + 2H₂O(l)

(c) Aluminum and potassium ion reaction

Aluminum (Al) is a more reactive metal than potassium (K), so it can displace potassium ions in a solution, where aluminum reacts with potassium ions to form aluminum ions and potassium metal. The balanced chemical equation for this reaction is: 2 Al(s) + 6K⁺(aq) -> 2 Al³⁺(aq) + 6 K(s)

Key concepts

Redox Reactions

Redox reactions are fundamental processes in chemistry involving the transfer of electrons between two substances. These reactions are composed of two half-reactions: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. In the context of the provided exercise, manganese dioxide reacts with mercury in an acidic solution. Here, mercury is oxidized from Hg to Hg²⁺ and manganese is reduced from MnO₂ to Mn²⁺. It's important to remember that in any redox reaction, the number of electrons lost in the oxidation process must be equal to the number of electrons gained in the reduction process to maintain charge balance.

For example, the exercise shows the redox reaction:
2 Hg(l) + MnO₂(s) + 4H⁺(aq) -> 2 Hg²⁺(aq) + Mn²⁺(aq) + 2H₂O(l).
Here, Hg is oxidized and MnO₂ is the oxidizing agent that gets reduced. Students should learn to identify what is being oxidized and what is being reduced in a redox reaction by looking at changes in oxidation states.

Ionic Compounds

Ionic compounds are formed from the electrostatic attraction between ions, which are atoms or groups of atoms that have gained or lost electrons and therefore have a net charge. Cations are positively charged ions, usually metals, and anions are negatively charged ions, usually non-metals or polyatomic ions. In the reaction between sulfur and mercury, an ionic compound, mercury(II) sulfide (HgS), is formed. This compound is composed of mercury ions (Hg²⁺) and sulfide ions (S²⁻).

When creating formulas for ionic compounds, it's crucial to ensure that the total charge is balanced. That means the total positive charge must equal the total negative charge. In the exercise, the balanced formula for mercury(II) sulfide accurately reflects this balance: Hg²⁺ + S²⁻ -> HgS.

Chemical Equation Balancing

Balancing chemical equations is essential in accurately representing chemical reactions. The Law of Conservation of Mass dictates that the mass of the reactants must be equal to the mass of the products. To balance a chemical equation, one must ensure that the same number of each type of atom appears on both sides of the equation. In the provided solution, the balancing of equations is demonstrated.

For example, aluminum reacting with potassium ions:
2 Al(s) + 6K⁺(aq) -> 2 Al³⁺(aq) + 6 K(s).
This equation shows a balanced exchange of atoms and charges. Balancing equations sometimes requires placing coefficients before the chemical formulas to achieve this atom-for-atom balance without altering the actual formulas of the compounds or elements involved.

Metal Reactivity Series

The metal reactivity series is a list of metals ranked by their reactivity with other elements. This series is crucial for predicting the outcomes of displacement reactions, where a more reactive metal can displace a less reactive metal from its compound. In the exercise, aluminum, which is higher in the reactivity series than potassium, displaces potassium ions from a solution to form aluminum ions and free potassium metal.

Understanding the reactivity series allows students to predict whether a reaction will occur based on the metals involved. For instance, because aluminum is above potassium in the reactivity series, the reaction proceeds to form Al³⁺ ions and solid potassium:
2 Al(s) + 6K⁺(aq) -> 2 Al³⁺(aq) + 6 K(s).
Learning the reactivity series can be helpful not only for solving chemical reaction problems but also for understanding the general behavior of metals in different chemical processes.