Examples of Redox Reactions and Cell Potential Calculations

Introduction to Redox Reactions and Cell Potential Calculations

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two species. These reactions are fundamental in various chemical processes, including batteries, corrosion, and metabolic pathways. Understanding the cell potential calculations is crucial for predicting the feasibility of these reactions and determining the voltage produced in electrochemical cells. Here, we present three diverse examples of redox reactions along with their respective cell potential calculations.

Example 1: Zinc-Copper Galvanic Cell

In this example, we will consider a galvanic cell consisting of zinc and copper electrodes, commonly used in batteries. This setup helps illustrate how redox reactions generate electrical energy.

In a galvanic cell, zinc (Zn) is oxidized, and copper (Cu) is reduced. The half-reactions can be represented as follows:

• Oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻
• Reduction: Cu²⁺(aq) + 2e⁻ → Cu(s)

To calculate the standard cell potential (E°) at 25°C, we can use the standard reduction potentials from a reference table:

• E°(Zn²⁺/Zn) = -0.76 V
• E°(Cu²⁺/Cu) = +0.34 V

The overall cell potential can be calculated as follows:

E°cell = E°(reduction) - E°(oxidation) = (+0.34 V) - (-0.76 V) = +1.10 V

This positive cell potential indicates that the reaction is spontaneous, meaning the galvanic cell will produce electrical energy.

Notes

• The cell potential can vary with concentration and temperature, affecting the overall voltage.
• This reaction is commonly observed in everyday batteries.

Example 2: Electrolysis of Water

The electrolysis of water is a crucial process in hydrogen production. In this example, we will explore the redox reactions occurring during the electrolysis of water to generate hydrogen and oxygen gases.

During electrolysis, an electric current is passed through water, leading to the following half-reactions:

• Oxidation at the anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
• Reduction at the cathode: 4H⁺(aq) + 4e⁻ → 2H₂(g)

To calculate the standard cell potential for the water electrolysis reaction:

• E°(H⁺/H₂) = 0.00 V
• E°(O₂/H₂O) = +1.23 V

Using these values, we compute the overall cell potential:

E°cell = E°(reduction) - E°(oxidation) = (0.00 V) - (+1.23 V) = -1.23 V

This negative cell potential indicates that the reaction is non-spontaneous and requires an external voltage to proceed.

Notes

• The efficiency of the electrolysis process can be influenced by factors such as temperature, pressure, and electrode materials.
• This method is essential for renewable energy applications like hydrogen fuel cells.

Example 3: Corrosion of Iron

Corrosion is an electrochemical process that leads to the degradation of metals, with iron rusting being a common example. Understanding the redox reactions involved can help in corrosion prevention strategies.

The corrosion of iron can be expressed through the following half-reactions:

• Oxidation: Fe(s) → Fe²⁺(aq) + 2e⁻
• Reduction: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)

To calculate the standard cell potential for the corrosion process:

• E°(Fe²⁺/Fe) = -0.44 V
• E°(O₂/H₂O) = +1.23 V

Thus, the overall cell potential is:

E°cell = E°(reduction) - E°(oxidation) = (+1.23 V) - (-0.44 V) = +1.67 V

This positive cell potential indicates that the corrosion reaction is spontaneous, which is why iron reacts with oxygen and moisture in the environment.

Notes

• Preventive measures against corrosion include galvanization and applying protective coatings.
• Understanding the electrochemical basis of corrosion can lead to more effective strategies for material preservation.