Question

Consider the cell described below: $$\text { Al }\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb}$$ Calculate the cell potential after the reaction has operated long enough for the [Al \(^{3+} ]\) to have changed by 0.60 \(\mathrm{mol} / \mathrm{L}\) . (Assume \(T=25^{\circ} \mathrm{C} .\)

Short answer

The initial cell potential is \(E_{cell}^0 = 1.53 \ V\). After the reaction has operated long enough, the new concentrations are \([Al^{3+}]' = 0.40 \ M\) and \([Pb^{2+}]' = 1.40 \ M\). Using the Nernst equation, we can calculate the new cell potential: \(E_{cell} = 1.53 V - \frac{(8.314 J/(mol·K))(298 K)}{6(96485 C/mol)} \ln\frac{(1.40)^2(0.40)^3}{(1.00)^2(1.00)^3}\), which yields \(E_{cell} \approx 1.36 \ V\).

Step-by-step solution

Calculate the initial cell potential

First, we need to determine the initial cell potential of the galvanic cell. To do this, we'll need to use the standard reduction potentials for the half-cell reactions. The standard reduction potentials are: $$E^0_{Al^{3+}/Al} = -1.66 \ V$$ $$E^0_{Pb^{2+}/Pb} = -0.13 \ V$$ Now we can calculate the initial cell potential using the equation: $$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$$ $$E_{cell}^{0} = (-0.13V) - (-1.66V)$$

Calculate the new concentrations

We are given that the change in the concentration of Al^{3+} is 0.60 mol/L, and we need to find the new concentrations after the reaction has operated long enough. First, calculate the moles of electrons, \(n_e\), that would be required by the change in the concentration of \(Al^{3+}\): $$n_e = 3 \times \Delta [Al^{3+}] = 3 \times 0.60 \ mol$$ Now we can calculate the new concentration of \(Pb^{2+}\) by considering the stoichiometry of the reaction: $$\Delta[Pb^{2+}] = \frac{2}{3} \times \Delta[Al^{3+}] = \frac{2}{3} \times 0.60 \ mol$$ Lastly, we can calculate the new concentrations of \(Al^{3+}\) and \(Pb^{2+}\): $$[Al^{3+}'] = 1.00 \ M - 0.60 \ M$$ $$[Pb^{2+}'] = 1.00 \ M + \frac{0.60 \ M}{3} \times 2$$

Use the Nernst Equation to find the cell potential

Now we'll apply the Nernst equation to calculate the cell potential after the reaction has operated long enough: $$E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q$$ where R is the ideal gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred per reaction, F is Faraday's constant, and Q is the reaction quotient given by: $$Q = \frac{[Pb^{2+}'][Al^{3+}]^3}{[Al^{3+}'][Pb^{2+}]^2}$$ Plug in the values of the constants and the new concentrations of \(Al^{3+}\) and \(Pb^{2+}\), and calculate the new cell potential, \(E_{cell}\).

Key concepts

Nernst equation

The Nernst Equation is a critical tool in electrochemistry, as it relates the cell potential of an electrochemical cell to its concentrations and temperature. It allows us to calculate the cell potential under non-standard conditions.
Its formula is stated as follows:

• \[E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q\]

Let's break down the formula:

• \(E_{cell}^0\) is the standard cell potential, which is calculated using the standard reduction potentials of the half-cells.
• \(R\) is the ideal gas constant, \(8.314 \, \text{J/mol} \, \text{K}\).
• \(T\) is the temperature in Kelvin, typically 298 K for room temperature.
• \(n\) is the number of moles of electrons exchanged in the reaction.
• \(F\) is Faraday's constant, \(96485 \, \text{C/mol} \, \text{e}^-\).
• \(Q\) is the reaction quotient. It signifies the ratio of the concentrations of products to reactants at any point in time.

This equation reveals how the cell potential decreases as the reaction progresses and concentrations deviate from standard conditions. By evaluating this change, chemists and students can predict the performance of galvanic cells in realistic scenarios.

cell potential

The concept of cell potential, also known as electromotive force (EMF), is fundamental for understanding how galvanic cells work. Cell potential measures the driving force behind the movement of electrons from the anode to the cathode within a cell.
Electrochemical cells generate voltage by separating oxidation and reduction reactions. The difference in potential between these two reactions results in a measurable voltage, which can be used to do electrical work.

• The standard cell potential, \(E_{cell}^0\), is calculated from the standard electrode potentials of the two half-cells, defined as \(E_{cathode}^0 - E_{anode}^0\).
• A positive cell potential indicates a spontaneous reaction, whereas a negative cell potential suggests a non-spontaneous one.
• This potential is influenced by the concentrations of ions and the temperature of the cell, as dictated by the Nernst equation.

By understanding cell potential, students can learn how energy is harnessed in electrochemical processes, such as in batteries and corrosion prevention systems. It equips them with the knowledge needed to predict and manipulate electrical outputs effectively.

galvanic cell

A galvanic cell, also known as a voltaic cell, converts chemical energy into electrical energy through spontaneous redox reactions. These cells are the human-made design behind common batteries.
The basic structure of a galvanic cell includes:

• Two electrodes, called the anode and the cathode, immersed in electrolyte solutions capable of conducting ions.
• A salt bridge that maintains electrical neutrality by permitting the exchange of ions between two half-cells.
• External circuitry, through which electrons migrate from the anode, where oxidation occurs, to the cathode, where reduction happens.

Here's how it works:

• The anode releases electrons when a metal undergoes oxidation, forming positive ions.
• The electrons travel through the conductive wire to the cathode.
• Meanwhile, in the cathode, a reduction reaction takes place as electrons integrate into ions turning into metals or other reduced products.

Galvanic cells are a practical application of electrochemical principles and offer insight into a variety of technologies, from portable power sources to industrial electrolysis. Understanding how they work helps students grasp the real-world implications of chemical reactions and energy conversion.