Question

Write the Nernst equation for the following processes at some temperature \(T\) : (a) \(\mathrm{Mg}(s)+\mathrm{Sn}^{2+}(a q) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\operatorname{Sn}(s)\) (b) \(2 \mathrm{Cr}(s)+3 \mathrm{~Pb}^{2+}(a q) \rightleftarrows 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{~Pb}(s)\)

Short answer

(a) \( E = E^0 - \frac{RT}{2F} \ln \left( \frac{[\text{Mg}^{2+}]}{[\text{Sn}^{2+}]} \right) \); (b) \( E = E^0 - \frac{RT}{6F} \ln \left( \frac{[\text{Cr}^{3+}]^2}{[\text{Pb}^{2+}]^3} \right) \)."

Step-by-step solution

Identify the Half-Reactions

For part (a), identify the oxidation and reduction half-reactions:- Oxidation: \( \text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^- \)- Reduction: \( \text{Sn}^{2+}(aq) + 2e^- \rightarrow \text{Sn}(s) \)For part (b), do the same:- Oxidation: \( 2\text{Cr}(s) \rightarrow 2\text{Cr}^{3+}(aq) + 6e^- \)- Reduction: \( 3\text{Pb}^{2+}(aq) + 6e^- \rightarrow 3\text{Pb}(s) \).

Write the Nernst Equation for Part (a)

Use the Nernst equation:\[ E = E^0 - \frac{RT}{nF} \ln Q \]For part (a), substitute the values:- \(n = 2\) (number of electrons transferred).- \( Q = \frac{[\text{Mg}^{2+}]}{[\text{Sn}^{2+}]} \) (reaction quotient).Thus, the Nernst equation is:\[ E = E^0 - \frac{RT}{2F} \ln \left( \frac{[\text{Mg}^{2+}]}{[\text{Sn}^{2+}]} \right) \].

Write the Nernst Equation for Part (b)

Use the Nernst equation for part (b):\[ E = E^0 - \frac{RT}{nF} \ln Q \]For part (b), substitute the values:- \(n = 6\) (number of electrons transferred).- \( Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Pb}^{2+}]^3} \).Thus, the Nernst equation is:\[ E = E^0 - \frac{RT}{6F} \ln \left( \frac{[\text{Cr}^{3+}]^2}{[\text{Pb}^{2+}]^3} \right) \].

Key concepts

Half-Reactions

When a redox reaction occurs, it involves the transfer of electrons between two species. This process can be broken down into two half-reactions: oxidation and reduction. In an oxidation half-reaction, a species loses electrons. In a reduction half-reaction, a species gains electrons.
For example, in part (a) of the original exercise:

• Oxidation: \( \text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^- \)
• Reduction: \( \text{Sn}^{2+}(aq) + 2e^- \rightarrow \text{Sn}(s) \)

These half-reactions showcase the electron flow from magnesium to tin, where magnesium is oxidized and tin is reduced. Similarly, part (b) follows the same principle with chromium and lead.
Recognizing and separating these half-reactions is crucial for understanding the overall redox process and calculating the cell potential using the Nernst equation.

Reaction Quotient

The reaction quotient, represented as \( Q \), is a measure of the relative quantities of products and reactants present during a reaction at any given point in time. For electrochemical cells, \( Q \) can be used to determine the direction in which the reaction will proceed.
In chemical terms, for a generic reaction, the quotient is written as:

• \( Q = \frac{[\text{products}]}{[\text{reactants}]} \)

For instance, in the original exercise part (a), the reaction quotient for the process involving magnesium and tin can be written as:

• \( Q = \frac{[\text{Mg}^{2+}]}{[\text{Sn}^{2+}]} \)

In part (b), it becomes:

• \( Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Pb}^{2+}]^3} \)

This expression shows how the concentrations of ions change as the cell reacts over time.
Understanding \( Q \) is key to applying the Nernst equation as it influences the cell potential, especially when the system is not at equilibrium.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the study of chemical processes that cause electrons to move. This electron transfer is the basis for many types of reactions, notably oxidation-reduction reactions. It merges chemical kinetics and thermodynamics with electron flow and energy generation.

• Electrochemical reactions are divided into two types: galvanic (voltaic) cells, which convert chemical energy into electrical energy, and electrolytic cells, which use electrical energy to drive chemical reactions.
• Key to electrochemistry is the understanding and application of the Nernst equation, which relates the reduction potential of an electrochemical cell to the standard electrode potential, temperature, and reaction quotient.

Essentially, electrochemistry is a vital science highly applicable in batteries, electroplating, corrosion prevention, and many other fields.
Grasping its fundamental principles helps students in handling both theoretical concepts and practical applications.

Oxidation-Reduction

Oxidation-reduction, or redox, reactions involve a change in oxidation states of the involved species. In simple terms:

• Oxidation is the loss of electrons and results in an increase in oxidation state.
• Reduction is the gain of electrons and results in a decrease in oxidation state.

Redox reactions commonly occur in electrochemical cells where electrons flow through an external circuit.
The combination of the oxidation and reduction half-reactions provides the full picture of how electrons are redistributed between molecules or ions. In electrochemical contexts, these processes are crucial for energy transformations and reactions efficiency.
Understanding redox reactions is fundamental not only to chemistry students but also for real-world applications such as energy storage and industrial synthesis.

Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa. They are composed of two half-cells, each containing a specific electrode and an electrolyte solution.

• In a galvanic cell, chemical reactions occur spontaneously to produce electrical energy. Batteries are common examples of galvanic cells.
• In an electrolytic cell, an external voltage is applied to bring about a chemical change. An example of this is the electrolysis of water.

Connecting the two half-cells is a salt bridge, allowing ions to move and maintain neutrality as the reaction proceeds.
These cells harness the energy from redox reactions that occur at the electrodes.
Electrochemical cells are used in many applications, from powering small devices to large-scale industrial processes, playing an essential role in modern technological applications.