Problem 41 - Chemistry and Chemical Reactivity

Question

An aqueous solution of KBr is placed in a beaker with two inert platinum electrodes. When the cell is attached to an external source of electrical energy, electrolysis occurs. (a) Hydrogen gas and hydroxide ion form at the cathode. Write an equation for the half-reaction that occurs at this electrode. (b) Bromine is the primary product at the anode. Write an equation for its formation.

Step-by-Step Solution

Answer

(a) The half-reaction at the cathode is \(2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)\). (b) The half-reaction at the anode is \(2Br^-(aq) \rightarrow Br_2(g) + 2e^-\).

Step 1: Understanding Electrolysis

In electrolysis, a current is passed through an aqueous solution to drive a chemical reaction that wouldn’t normally occur without the added energy. In this scenario, the aqueous solution of KBr is subject to electrolysis.

Step 2: Identifying the Electrodes

During electrolysis, reduction occurs at the cathode while oxidation occurs at the anode. In the solution of KBr, the cathode will attract cations and the anode will attract anions.

Step 3: Electrolysis at the Cathode

At the cathode, the cations in the solution accept electrons. In aqueous solutions, water can also be reduced alongside K⁺ ions, but it is more favorable to reduce water. The reduction of water at the cathode yields hydrogen gas (H₂) and hydroxide ions (OH⁻). The half-reaction is written as follows: \[2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)\]

Step 4: Electrolysis at the Anode

At the anode, anions lose electrons. Bromide ions (Br⁻) from KBr solution are oxidized to form bromine gas. The anode half-reaction for this oxidation is: \[2Br^-(aq) \rightarrow Br_2(g) + 2e^-\]

Step 5: Balancing the Cell Reaction (Optional)

Combining the half-reactions for a complete balanced representation is often useful to understand the full scope of electrolytic reactions:Cathode: \(2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)\)Anode: \(2Br^-(aq) \rightarrow Br_2(g) + 2e^-\)Overall: \(2H_2O(l) + 2Br^-(aq) \rightarrow H_2(g) + 2OH^-(aq) + Br_2(g)\)

Key Concepts

Redox reactions

Redox reactions are a fundamental type of chemical reaction where oxidation and reduction occur simultaneously. A handy way to remember this process is through the acronym "OIL RIG." It stands for "Oxidation is Loss, Reduction is Gain" of electrons. In electrochemical cells, like those used in electrolysis, these reactions involve the transfer of electrons between two substances.

During electrolysis, there are two main sites of reactions: the cathode and the anode. **Oxidation** occurs at the anode where the substance loses electrons, while **reduction** happens at the cathode where a substance gains electrons. In the electrolysis of aqueous KBr, the K⁺ ions and water molecules migrate toward the cathode, while Br⁻ ions move to the anode to undergo oxidation. Understanding the direction of electron flow is crucial to mastering redox reactions.

Hydrogen gas formation

In electrolysis, hydrogen gas formation occurs at the cathode. The process involves the reduction of water molecules, where each molecule gains electrons to form hydrogen gas and hydroxide ions. This occurs because it's energetically more favorable to reduce water than to reduce the potassium ions present in the KBr solution.

The half-reaction for this process is written as: \[2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)\] This equation shows that two electrons are needed to produce one molecule of hydrogen gas (consisting of two hydrogen atoms) and two hydroxide ions.

• The cathode attracts cations and water molecules.
• Water molecules are reduced, leading to hydrogen gas formation.
• Hydroxide ions remain in solution, increasing the mixture's pH.

Seeing this in action helps to visualize the changes occurring at the cathode during the electrolysis of KBr.

Bromine gas formation

At the anode, bromine gas formation occurs as bromide ions undergo oxidation. This half-reaction involves the loss of electrons as bromide (\[Br^-\]) from the potassium bromide solution becomes bromine gas (\[Br_2\]).

The half-reaction for this process is represented as: \[2Br^-(aq) \rightarrow Br_2(g) + 2e^-\] Here, two electrons are released during the conversion of two bromide ions to one molecule of bromine gas.

• The anode attracts anions from the solution.
• Bromide ions lose electrons and form bromine gas.
• This gas may appear as a brown vapor or collect as a liquid at the anode.

These steps highlight the transformation at the anode, showcasing how anionic bromide becomes bromine gas in the electrolysis procedure. This conversion is vital to understanding the comprehensive reactions in this electrochemical cell.