Calculating Cell Potential Changes In A Lead-Nickel Voltaic Cell

Calculate the cell potential in volts (V) for a voltaic cell with $Pb^{2+}/Pb$ and $Ni^{2+}/Ni$ half-cells under standard conditions, where each compartment has a volume of 355 mL, after delivering 0.150 A for 26.0 hours at 25.0 ${ }^{\circ} C$.

Understanding the behavior of voltaic cells, also known as galvanic cells, is fundamental in electrochemistry. These cells harness spontaneous redox reactions to generate electrical energy. A typical voltaic cell comprises two half-cells, each containing an electrode immersed in an electrolyte solution. The electrodes are connected externally, allowing electrons to flow and produce an electric current. This article delves into the calculation of cell potential changes in a voltaic cell composed of $Pb^{2+}/Pb$ and $Ni^{2+}/Ni$ half-cells under specific conditions. We will explore how factors such as current, time, and temperature influence the cell potential, providing a comprehensive understanding of the electrochemical processes involved.

Understanding Voltaic Cells and Cell Potential

Voltaic cells, at their core, are electrochemical devices that convert chemical energy into electrical energy through spontaneous redox reactions. This conversion is the backbone of many everyday technologies, from batteries powering our smartphones to the energy systems in hybrid vehicles. The cell potential, often measured in volts, is a crucial parameter that indicates the potential difference between the two electrodes (half-cells) and reflects the cell's ability to drive electric current through an external circuit. Understanding cell potential is vital because it determines the voltage a cell can provide, which is critical for the cell's intended application.

Key Components of a Voltaic Cell

A voltaic cell consists primarily of two half-cells, each characterized by an electrode immersed in an electrolyte solution. The electrodes are typically made of conductive materials, often metals, which participate in the redox reactions. The electrolyte solutions contain ions of the same metal as the electrode, ensuring a medium for ion transport. These half-cells are connected via two pathways: an external circuit that allows for the flow of electrons (the electric current) and a salt bridge or porous membrane that facilitates the movement of ions, maintaining electrical neutrality within the cell. This setup enables the oxidation reaction to occur at one electrode (the anode) and the reduction reaction at the other (the cathode), thereby generating an electric current. The standard conditions, such as 298 K (25°C) and 1 atm pressure, are often used as reference points for measuring and comparing cell potentials, making it easier to evaluate the performance of different voltaic cells.

Calculating Standard Cell Potential

The standard cell potential ($E°_{cell}$) is a theoretical value calculated under standard conditions and is essential for predicting the feasibility and voltage output of a voltaic cell. To calculate $E°_{cell}$, one must first identify the half-reactions occurring at the anode (oxidation) and the cathode (reduction). The standard reduction potentials ($E°$) for these half-reactions can be found in standard electrochemical tables. The standard cell potential is then calculated using the formula:

$E°_{cell} = E°_{cathode} - E°_{anode}$

Where $E°_{cathode}$ is the standard reduction potential of the half-cell where reduction occurs, and $E°_{anode}$ is the standard reduction potential of the half-cell where oxidation occurs. This calculation provides a baseline understanding of the cell's potential, which can then be adjusted using the Nernst equation to account for non-standard conditions, such as variations in temperature and concentration. Understanding how to calculate the standard cell potential is a fundamental skill in electrochemistry, as it allows for the rational design and optimization of electrochemical devices.

Problem Statement: A Lead-Nickel Voltaic Cell

Consider a voltaic cell constructed from $Pb^{2+}/Pb$ and $Ni^{2+}/Ni$ half-cells operating under standard conditions, where each compartment contains a volume of 355 mL. The challenge is to determine the cell potential after the cell has been delivering a constant current of 0.150 A for 26.0 hours at a temperature of 25.0 °C. To solve this problem, we need to integrate several electrochemical concepts, including standard reduction potentials, the Nernst equation, and the relationship between current, time, and the amount of substance involved in the redox reactions. This problem not only tests our understanding of these concepts but also their application in a practical scenario involving the operation of a voltaic cell over an extended period. By systematically working through the problem, we can gain a deeper appreciation of the factors that affect cell potential and the performance of electrochemical cells in real-world applications.

Identifying Half-Reactions

The first step in solving this problem is to identify the half-reactions occurring at each electrode. In the $Pb^{2+}/Pb$ half-cell, the reaction is the reduction of lead(II) ions to lead metal:

$Pb^{2+}(aq) + 2e^- → Pb(s)$

In the $Ni^{2+}/Ni$ half-cell, the reaction is the reduction of nickel(II) ions to nickel metal:

$Ni^{2+}(aq) + 2e^- → Ni(s)$

To determine which reaction occurs at the anode (oxidation) and which at the cathode (reduction), we need to compare their standard reduction potentials. The standard reduction potential for $Pb^{2+}/Pb$ is -0.13 V, and for $Ni^{2+}/Ni$ is -0.25 V. Since nickel has a more negative reduction potential, it is more easily oxidized. Therefore, nickel will be oxidized at the anode, and lead(II) ions will be reduced at the cathode. The half-reactions are:

Anode (oxidation): $Ni(s) → Ni^{2+}(aq) + 2e^-$

Cathode (reduction): $Pb^{2+}(aq) + 2e^- → Pb(s)$

Identifying these half-reactions correctly is crucial because it sets the stage for all subsequent calculations, including the determination of the overall cell reaction and the application of the Nernst equation.

Calculating the Standard Cell Potential

Once the half-reactions have been identified, the next step is to calculate the standard cell potential ($E°_{cell}$) using the formula:

$E°_{cell} = E°_{cathode} - E°_{anode}$

Here, $E°_{cathode}$ is the standard reduction potential for the reduction of $Pb^{2+}$ (-0.13 V), and $E°_{anode}$ is the standard reduction potential for the reduction of $Ni^{2+}$ (-0.25 V). Note that even though nickel is being oxidized at the anode, we use its standard reduction potential in the calculation. Plugging in the values:

$E°_{cell} = -0.13 V - (-0.25 V) = 0.12 V$

This calculation reveals that the standard cell potential for the $Pb^{2+}/Pb$ and $Ni^{2+}/Ni$ voltaic cell under standard conditions is 0.12 V. This value serves as a reference point for understanding the cell's potential under non-standard conditions, which will be addressed using the Nernst equation. Calculating the standard cell potential is a fundamental step in electrochemistry as it provides a baseline understanding of the cell's potential under ideal conditions.

Applying the Nernst Equation

Understanding the Nernst Equation

The Nernst equation is a cornerstone in electrochemistry, providing a means to calculate the cell potential ($E_{cell}$) under non-standard conditions. These conditions deviate from the standard state (298 K, 1 atm, 1 M concentrations) and can include variations in temperature and ion concentrations. The equation directly relates the cell potential to the standard cell potential ($E°_{cell}$), temperature (T), the number of moles of electrons transferred in the balanced redox reaction (n), and the reaction quotient (Q). The reaction quotient is a measure of the relative amounts of products and reactants present in a reaction at a given time and is crucial for understanding the direction a reversible reaction will proceed to reach equilibrium. By incorporating these factors, the Nernst equation allows for a more accurate prediction of cell potential in real-world scenarios, making it an indispensable tool in electrochemical analysis and applications. Its general form is:

$E_{cell} = E°_{cell} - (RT/nF) * ln(Q)$

Where:

• $E_{cell}$ is the cell potential under non-standard conditions,
• $E°_{cell}$ is the standard cell potential,
• R is the ideal gas constant (8.314 J/(mol·K)),
• T is the temperature in Kelvin,
• n is the number of moles of electrons transferred in the balanced redox reaction,
• F is the Faraday constant (96485 C/mol),
• Q is the reaction quotient.

Calculating the Reaction Quotient (Q)

To utilize the Nernst equation effectively, the reaction quotient (Q) must be calculated. The reaction quotient provides a measure of the relative amounts of products and reactants at a given time and is essential for determining how the cell potential changes as the reaction progresses. For the voltaic cell involving $Pb^{2+}$ and $Ni^{2+}$, the overall balanced cell reaction is:

$Ni(s) + Pb^{2+}(aq) → Ni^{2+}(aq) + Pb(s)$

The reaction quotient (Q) is then expressed as the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. In this case, the solids (Ni and Pb) do not appear in the expression because their activities are considered to be 1. Therefore, the reaction quotient is:

$Q = [Ni^{2+}] / [Pb^{2+}]$

The initial concentrations of $Ni^{2+}$ and $Pb^{2+}$ are 1 M under standard conditions. However, as the cell operates, these concentrations change, which affects the cell potential. Accurately calculating Q requires understanding how the concentrations of the ions change over time, which is linked to the amount of charge transferred during the cell's operation.

Determining Changes in Ion Concentrations

The key to finding the cell potential after a specific time lies in determining how the ion concentrations change due to the flow of current. The total charge (q) passed through the cell is the product of the current (I) and the time (t):

$q = I * t$

In this case, the current is 0.150 A, and the time is 26.0 hours. First, we need to convert the time to seconds:

$t = 26.0 hours * (3600 seconds/hour) = 93600 seconds$

Now, we can calculate the total charge:

$q = 0.150 A * 93600 seconds = 14040 Coulombs$

Next, we use the Faraday constant (F = 96485 C/mol) to find the number of moles of electrons transferred (n):

$n_{electrons} = q / F = 14040 C / 96485 C/mol ≈ 0.1455 mol$

From the balanced half-reactions, we know that 2 moles of electrons are transferred per mole of reaction. Therefore, the change in the number of moles of $Ni^{2+}$ and $Pb^{2+}$ is:

$Δn = n_{electrons} / 2 = 0.1455 mol / 2 ≈ 0.07275 mol$

This change in the number of moles will affect the concentrations of $Ni^{2+}$ and $Pb^{2+}$ in the solution. Calculating these changes is critical for accurately determining the reaction quotient (Q) and, subsequently, the cell potential using the Nernst equation.

Calculating New Concentrations

After determining the change in the number of moles of ions, the next step is to calculate the new concentrations of $Ni^{2+}$ and $Pb^{2+}$ in the solution. Recall that each compartment has a volume of 355 mL, which is equivalent to 0.355 L. The change in concentration ($Δ[Ni^{2+}]$) can be calculated by dividing the change in the number of moles by the volume of the solution:

$Δ[Ni^{2+}] = Δn / Volume = 0.07275 mol / 0.355 L ≈ 0.205 M$

Since nickel is oxidized at the anode, the concentration of $Ni^{2+}$ increases. Thus, the new concentration of $Ni^{2+}$ is:

$[Ni^{2+}]_{new} = [Ni^{2+}]_{initial} + Δ[Ni^{2+}] = 1 M + 0.205 M = 1.205 M$

Similarly, for $Pb^{2+}$, the concentration decreases as it is reduced at the cathode:

$Δ[Pb^{2+}] = -Δn / Volume = -0.07275 mol / 0.355 L ≈ -0.205 M$

The new concentration of $Pb^{2+}$ is:

$[Pb^{2+}]_{new} = [Pb^{2+}]_{initial} + Δ[Pb^{2+}] = 1 M - 0.205 M = 0.795 M$

With these new concentrations, we can now calculate the reaction quotient (Q) more accurately. These new concentrations are crucial for using the Nernst equation to find the cell potential under the non-standard conditions after the cell has been operating for 26.0 hours.

Applying the Nernst Equation with New Concentrations

Now that we have calculated the new concentrations of $Ni^{2+}$ and $Pb^{2+}$, we can determine the reaction quotient (Q) using the expression:

$Q = [Ni^{2+}] / [Pb^{2+}] = 1.205 M / 0.795 M ≈ 1.516$

With the value of Q, we can now apply the Nernst equation to calculate the cell potential ($E_{cell}$) under the given non-standard conditions:

$E_{cell} = E°_{cell} - (RT/nF) * ln(Q)$

We have $E°_{cell} = 0.12 V$, $R = 8.314 J/(mol·K)$, $T = 25.0 °C = 298 K$, $n = 2$, and $F = 96485 C/mol$. Plugging these values into the Nernst equation:

$E_{cell} = 0.12 V - (8.314 J/(mol·K) * 298 K) / (2 * 96485 C/mol) * ln(1.516)$

$E_{cell} = 0.12 V - (2477.572 J/mol) / (192970 C/mol) * ln(1.516)$

$E_{cell} = 0.12 V - 0.01284 J/C * 0.416$

$E_{cell} = 0.12 V - 0.00534 V$

$E_{cell} ≈ 0.1147 V$

Therefore, the cell potential after delivering 0.150 A for 26.0 hours at 25.0 °C is approximately 0.1147 V. Applying the Nernst Equation in this context provides a precise understanding of how the cell potential deviates from its standard value due to changes in ion concentrations, demonstrating the practical utility of the Nernst equation in electrochemistry.

Conclusion

In summary, the cell potential of the $Pb^{2+}/Pb$ and $Ni^{2+}/Ni$ voltaic cell after delivering 0.150 A for 26.0 hours at 25.0 °C is approximately 0.1147 V. This calculation involved several key steps, including identifying the half-reactions, calculating the standard cell potential, determining the changes in ion concentrations due to the current flow, and applying the Nernst equation to find the cell potential under non-standard conditions. This comprehensive approach demonstrates the interconnectedness of electrochemical concepts and their practical application in understanding the behavior of voltaic cells. This detailed calculation not only answers the specific problem but also provides a template for analyzing other electrochemical scenarios, reinforcing the importance of electrochemistry in various scientific and technological fields.