Question

The equilibrium constant for the reaction: $$\operatorname{Sr}(s)+\mathrm{Mg}^{2+}(a q) \rightleftharpoons \mathrm{Sr}^{2+}(a q)+\mathrm{Mg}(s)$$ is \(2.69 \times 10^{12}\) at \(25^{\circ} \mathrm{C}\). Calculate \(E^{\circ}\) for a cell made up of \(\mathrm{Sr} / \mathrm{Sr}^{2+}\) and \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) half-cells.

Short answer

The standard cell potential \( E^{\circ} \) is approximately 0.762 V.

Step-by-step solution

Identify the Reaction and Equilibrium Constant

Firstly, note the given reaction: \( \operatorname{Sr}(s)+\mathrm{Mg}^{2+}(a q) \rightleftharpoons \mathrm{Sr}^{2+}(a q)+\mathrm{Mg}(s) \), and the equilibrium constant \( K = 2.69 \times 10^{12} \). This constant reflects the balance between reactants and products in the reaction.

Use the Relationship Between E° and K

The relationship between the standard cell potential \( E^{\circ} \) and the equilibrium constant \( K \) is given by the Nernst equation at equilibrium: \[ E^{\circ} = \frac{RT}{nF} \ln K \] where \( R \) is the universal gas constant \( 8.314 \, \text{J/mol}\, \text{K} \), \( T \) is the temperature in Kelvin, \( F \) is the Faraday's constant \( 96485 \, \text{C/mol} \), and \( n \) is the number of moles of electrons transferred in the reaction. Since the reaction \( \operatorname{Sr}(s) + \mathrm{Mg}^{2+} \rightleftharpoons \mathrm{Sr}^{2+} + \mathrm{Mg}(s) \) involves a two-electron transfer (\( n = 2 \)), we can use these constants to find \( E^{\circ} \).

Convert Temperature to Kelvin

Convert the given temperature from degrees Celsius to Kelvin: \( T = 25^{\circ} \text{C} + 273.15 = 298.15 \, \text{K} \). This is necessary for using the Nernst equation.

Substitute Values into the Nernst Equation

Now, substitute the known values into the Nernst equation: \[ E^{\circ} = \frac{(8.314)(298.15)}{(2)(96485)} \ln (2.69 \times 10^{12}) \].Calculate this value to determine \( E^{\circ} \).

Calculate the Natural Logarithm

Calculate the natural logarithm of the equilibrium constant: \( \ln (2.69 \times 10^{12}) \approx 29.335 \).

Complete the Calculation for E°

Now, compute the expression:\[ E^{\circ} = \frac{(8.314)(298.15)}{(2)(96485)} \cdot 29.335 \].Carry out the multiplication and division: \[ E^{\circ} \approx 0.762 \, \text{V} \]. This is the standard cell potential for the reaction at standard conditions.

Key concepts

Equilibrium Constant

The equilibrium constant, symbolized as \( K \), is a vital concept in chemistry that defines the balance between the concentrations of the products and reactants in a reaction. It is expressed in terms of the concentration (in case of solutions) or pressure (in case of gases) of the products and reactants of the balanced chemical reaction.
For the reaction \( \operatorname{Sr}(s) + \mathrm{Mg}^{2+}(aq) \rightleftharpoons \mathrm{Sr}^{2+}(aq) + \mathrm{Mg}(s) \), the equilibrium constant \( K = 2.69 \times 10^{12} \) is remarkably high.
This suggests that, at equilibrium, the formation of products is heavily favored. In simpler terms, almost all of the reactants convert into products because the value is so large.
In electrochemistry, \( K \) is used in conjunction with other principles to determine various aspects such as the standard cell potential \( E^\circ \), providing insight into the likelihood and extent of electron transfers happening spontaneously.

Nernst Equation

The Nernst equation is a fundamental tool in electrochemistry used to relate the cell potential with the concentrations of the reactants and products involved. For a cell at equilibrium, the Nernst equation simplifies to relate the standard cell potential \( E^\circ \) to the equilibrium constant \( K \): \[ E^\circ = \frac{RT}{nF} \ln K \] Here, \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, \( n \) represents the number of electrons transferred in the reaction, and \( F \) is Faraday's constant.
This equation allows us to calculate \( E^\circ \) by substituting the given equilibrium constant and known constants, turning a seemingly abstract equilibrium constant into a tangible electrical measurement.
It is particularly useful because it aligns thermodynamic qualities (\( K \)) with electrochemical measurements (\( E^\circ \)).

Cell Potential

The cell potential, also known as electromotive force (emf), represents the capacity of an electrochemical cell to drive electric current through a circuit.
In simple terms, it measures how strongly electrons are pushed from one side of the cell to the other.
The standard cell potential \( E^\circ \) is specifically for conditions where all reactants and products are at a concentration of 1 M, and it represents the maximum potential difference in these conditions. In the process of solving the exercise, we found \( E^\circ \approx 0.762 \, \text{V} \), which indicates a high tendency for the current to flow.
The higher the cell potential, the greater the potential energy available to do work, meaning this reaction is thermodynamically favored. This is in agreement with the large \( K \) value, indicating that the chemical reaction is capable of generating electricity effectively.

Electron Transfer

Electron transfer plays a crucial role in electrochemical reactions. It involves the movement of electrons from one substance to another, typically from a reducing agent to an oxidizing agent.
In the cell composed of \( \mathrm{Sr}/\mathrm{Sr}^{2+} \) and \( \mathrm{Mg}/\mathrm{Mg}^{2+} \) half-cells, electrons are transferred from strontium metal \( \operatorname{Sr}(s) \) to magnesium ions \( \mathrm{Mg}^{2+}(aq) \).
This transfer is spontaneous given the calculated positive \( E^\circ \) and high equilibrium constant.
The balance of electron flow is central to determining the reaction's electromotive force, and in this instance, involves two moles of electrons being transferred (\( n = 2 \)), aligning the concept with calculations via the Nernst equation. Electron transfer is key, as it essentially translates into the cell's ability to generate electricity when part of a closed circuit.