Problem 66 - General Chemistry: Principles and Modern Applications

Question

An aqueous solution of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) is electrolyzed by means of Pt electrodes. (a) Which of the following gases should form at the anode: \(\mathrm{O}_{2}, \mathrm{H}_{2}, \mathrm{SO}_{2}, \mathrm{SO}_{3} ?\) Explain. (b) What product should form at the cathode? Explain. (c) What is the minimum voltage required? Why is the actual voltage needed likely to be higher than this value?

Step-by-Step Solution

Answer

a) Oxygen (\(\mathrm{O}_{2}\)) would form at the anode because the oxidation of water (\(\mathrm{H}_{2}\mathrm{O}\)) is more likely than the oxidation of sulfate (\(\mathrm{SO}_{4}^{2-}\)). b) Hydrogen (\(\mathrm{H}_{2}\)) would form at the cathode because the reduction of water (\(\mathrm{H}_{2}\mathrm{O}\)) is more thermodynamically favorable than the reduction of potassium ions (\(\mathrm{K}^{+}\)). c) The minimum voltage required for electrolysis to take place is 2.06 V. However, the actual voltage will likely be higher due to overpotential, which accounts for energy losses in the system.

Step 1: Identifying the Anodic Reaction

In electrolysis, the anode is where oxidation occurs. Here, either water (\(\mathrm{H}_{2}\mathrm{O}\)) or sulfate (\(\mathrm{SO}_{4}^{2-}\)) can be oxidized. The oxidation of water produces oxygen gas (\(\mathrm{O}_{2}\)) and hydrogen ions (\(\mathrm{H}^{+}\)) while the oxidation of sulfate would produce sulfur dioxide (\(\mathrm{SO}_{2}\)) or sulfur trioxide (\(\mathrm{SO}_{3}\)). However, the oxidation potential of \(\mathrm{SO}_{4}^{2-}\) is much higher than \(\mathrm{H}_{2}\mathrm{O}\). Therefore, it's more likely to have \(\mathrm{O}_{2}\) gas evolving at the anode.

Step 2: Identifying the Cathodic Reaction

At the cathode, reduction takes place. Given that the electrolyte is in an aqueous solution, the possible ions that can be reduced are the potassium ions (\(\mathrm{K}^{+}\)) and the water molecules (\(\mathrm{H}_{2}\mathrm{O}\)). The reduction of \(\mathrm{K}^{+}\) is less likely due to its position on the electrochemical series which makes the reduction of water to form \(\mathrm{H}_{2}\) gas and \(\mathrm{OH}^{-}\) ions more thermodynamically favorable.

Step 3: Calculating the Minimum Voltage Required

The minimum voltage required is determined by the difference in reduction potentials of the anode and cathode reactions, also known as the cell potential. The oxidation potential of \(\mathrm{H}_{2}\mathrm{O}\) to \(\mathrm{O}_{2}\) is 1.23 V and the reduction potential of \(\mathrm{H}_{2}\mathrm{O}\) to \(\mathrm{H}_{2}\) is -0.83 V. Hence, the minimum voltage required for the electrolysis to take place is 1.23 - (-0.83) = 2.06 V.

Step 4: Discussing the Higher Voltage Required in Practice

The actual voltage will likely be higher due to what is known as overpotential. Overpotential denotes energy losses in an electrolytic system due to various factors such as ohmic resistance, mass transfer and kinetics of electrode reactions. Overpotential makes it necessary to use a higher voltage so that electrolysis can proceed at a practical rate.

Key Concepts

Anodic Reaction

During electrolysis, reactions at the anode involve oxidation processes. In the scenario given, potential substances for oxidation include water (\(\mathrm{H}_{2}\mathrm{O}\)) and sulfate ion (\(\mathrm{SO}_{4}^{2-}\)). Water oxidation has a lower oxidation potential compared to sulfate, making it more favorable.
In this context, oxidation of \(\mathrm{H}_{2}\mathrm{O}\) produces oxygen gas \(\mathrm{O}_{2}\) and hydrogen ions \(\mathrm{H}^{+}\).
On the other hand, oxidizing \(\mathrm{SO}_{4}^{2-}\) would yield sulfur dioxide ( \(\mathrm{SO}_{2}\) or sulfur trioxide \(\mathrm{SO}_{3}\)), which requires significantly more energy.
Hence, during the electrolysis of \(\mathrm{K}_{2} \mathrm{SO}_{4}\), oxygen is the gas formed at the anode.

• Water is the preferred substance for oxidation at the anode.
• Thus, oxygen gas evolves as a byproduct of water oxidation.

Understanding which reactions occur at the anode is crucial for predicting products and identifying energy needs for electrolytic processes.

Cathodic Reaction

The cathode is the site for reduction reactions in electrolysis. In an aqueous solution, the electrolyte components might include ions such as potassium (\(\mathrm{K}^{+}\)) and water molecules (\(\mathrm{H}_{2}\mathrm{O}\)). The reduction at the cathode typically depends on which species is easier to reduce.
Here, the reduction of water is more feasible than reducing potassium ions because of their position on the electrochemical series.
Reducing water yields hydrogen gas (\(\mathrm{H}_{2}\)) and hydroxide ions (\(\mathrm{OH}^{-}\)).

• The cathodic reaction mainly leads to hydrogen gas evolution.
• The reduction of potassium ions is less favorable.

This understanding helps predict that hydrogen gas is formed at the cathode during electrolysis of \(\mathrm{K}_{2} \mathrm{SO}_{4}\).

Minimum Voltage Required

The minimum voltage required for a particular electrolytic process, also known as cell potential, is determined by the potential difference between anodic and cathodic reactions. Specifically, it is calculated from their respective standard reduction potentials.
In this electrolysis example, the oxidation potential for turning water into oxygen (\(\mathrm{O}_{2}\)) is 1.23 V,
while the reduction potential for converting water to hydrogen gas (\(\mathrm{H}_{2}\)) is -0.83 V.
The essential voltage needed is simply the sum of these values:
1.23 V - (-0.83 V) = 2.06 V.

• This 2.06 V represents the theoretical minimum voltage.
• It gives a benchmark for starting electrolysis.

Understanding the minimum voltage helps in determining if an electrolytic reaction will occur as intended.

Overpotential

In practical applications of electrolysis, the actual voltage required is often higher than the theoretically calculated minimum. This phenomenon is known as overpotential. Overpotential arises from several factors that introduce energy inefficiencies:

• Ohmic Resistance: Resistance within the electrolyte and electrodes due to material properties.
• Mass Transfer: The movement of ions in the solution may not be immediate, causing delays.
• Reaction Kinetics: Slower rate of reaction at the electrode surfaces.

As a result, the applied voltage must be increased beyond the theoretical minimum to achieve desired reaction rates at a practical level. Understanding overpotential is crucial for optimizing electrolytic processes and ensuring efficiency in energy consumption.