Problem 79 - Chemistry Matter and Change

Question

Apply During electrolysis, an electrolytic cell releases bromine vapor and hydrogen gas. After electrolysis, the cell is found to contain a concentrated solution of potassium hydroxide. What was the composition of the cell before electrolysis began?

Step-by-Step Solution

Answer

The initial composition of the electrolytic cell before electrolysis began contained potassium ions (K^+), water (H_2O), and potassium bromide (KBr). This is inferred from the products of electrolysis (hydrogen gas and bromine vapor), the half-cell reactions, and the presence of a concentrated solution of potassium hydroxide (KOH) after electrolysis.

Step 1: 1. Identify the products of electrolysis

From the information given, we know that the products of the electrolysis are bromine vapor and hydrogen gas.

Step 2: 2. Determine the half-cell reactions for each product

Half-cell reactions are the two parts of a redox reaction. One half-cell involves oxidation, and the other half-cell involves reduction. We will now identify the half-cell reactions for the production of bromine vapor and hydrogen gas. The half-cell reaction for the production of hydrogen gas is: \(2H^+ (aq) + 2e^- → H_2(g)\) (Reduction) The half-cell reaction for the production of bromine vapor is: \(2Br^- (aq) → Br_2(g) + 2e^-\) (Oxidation)

Step 3: 3. Determine the overall redox reaction

By combining the oxidation half-cell reaction and the reduction half-cell reaction, we can find the overall redox reaction. This will help us identify the reactants present in the electrolytic cell. \(2H^+ (aq) + 2Br^- (aq) → H_2(g) + Br_2(g)\)

Step 4: 4. Find the source of H^+ and Br^- ions in the cell

We know that after electrolysis, the cell contains a concentrated solution of potassium hydroxide (KOH). This implies that KOH was present in the solution during the electrolysis. We can now infer the source of H^+ and Br^- ions in the cell: Since the cell contains a concentrated solution of KOH after electrolysis, the source of H^+ ions is likely due to the presence of K^+ ions and water (H_2O) in the solution: \(K^+ + H_2O → KOH + H^+\) The bromine ions (Br^-) must have come from a potassium bromide (KBr) salt in the initial cell solution: \(KBr → K^+ + Br^-\)

Step 5: 5. Determine the initial cell composition

Based on our analysis, the composition of the cell before electrolysis began would have contained potassium ions (K^+), water (H_2O), and potassium bromide (KBr).

Key Concepts

Redox Reactions

Redox reactions involve the transfer of electrons between two substances. They consist of two parts: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. These reactions are fundamental to various chemical processes, including electrolysis.
In the context of electrolysis in an electrolytic cell, redox reactions are artificially induced by an external voltage. For instance, when a potassium bromide (KBr) solution is subjected to electrolysis, we observe that bromine (Br₂) and hydrogen (H₂) gases are produced, indicating that redox reactions have taken place.

Understanding the Oxidation State

To identify the redox reactions, we look at the oxidation states of the reactants and products. Bromide ions (Br^-) lose electrons (oxidation) to form bromine gas, while hydrogen ions (H⁺) gain electrons (reduction) to form hydrogen gas. The electrons are transferred through the external circuit, satisfying the electrical neutrality of the overall reaction.

Half-Cell Reactions

Half-cell reactions refer to the individual oxidation and reduction processes that occur at the respective electrodes during electrolysis. Each half-cell reaction occurs in a separate compartment known as a half-cell, connected by a salt bridge or a porous barrier in a complete electrochemical cell.

Oxidation Half-Cell

In the oxidation half-cell of our problem, bromide ions (Br^-) give up their electrons to become bromine molecules (Br₂), releasing electrons to the anode. The oxidation half-cell reaction is represented as:
\(2Br^- (aq) \rightarrow Br_2(g) + 2e^-\).

Reduction Half-Cell

Simultaneously, in the reduction half-cell, hydrogen ions (H⁺) from the aqueous solution accept electrons from the cathode to form hydrogen gas (H₂). The reduction half-cell reaction is:
\(2H^+ (aq) + 2e^- \rightarrow H_2(g)\).
By recognizing these half-cell reactions, students can better understand how electrolysis separates the components of a solution through the flow of electrons.

Potassium Hydroxide

Potassium hydroxide (KOH) is a strong alkaline chemical compound that plays a crucial role in electrolysis involving potassium salts. It forms as a result of the reaction between potassium ions (K⁺) and hydroxide ions (OH^-) in water.

Electrolysis and Formation of KOH

When an aqueous solution of potassium bromide (KBr) undergoes electrolysis, KOH is produced as a byproduct in a concentrated form. This indicates that water molecules (H₂O) are also involved in the electrolytic reactions, leading to the production of hydroxide ions. The hydroxide ions combine with the potassium ions remaining in solution after the removal of bromide ions through oxidation.
The presence of potassium hydroxide in the cell after electrolysis helps us understand the initial components and the chemical changes that occurred during the process, enabling us to infer that both potassium bromide and water were present before electrolysis began.