Problem 60 - Chemistry: An Atoms-Focused Approach

Question

In a voltaic cell based on the \(\mathrm{Zn} / \mathrm{Cu}^{2+}\) cell reaction, $$ \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) $$ there is exactly one mole of each reactant and product. A second cell based on the following cell reaction: $$ \mathrm{Cd}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow \mathrm{Cu}(s)+\mathrm{Cd}^{2+}(a q) $$ also has exactly one mole of each reactant and product. Which of the following statements about these two cells is true? a. Their cell potentials are the same. b. The masses of their electrodes are the same. c. The quantities of electric charge that they can produce are the same. d. The quantities of electric energy that they can produce are the same.

Step-by-Step Solution

Answer

Answer: The quantities of electric charge that the two cells can produce are the same.

Step 1: Examining the cell potentials

To determine whether the two cells have the same cell potentials, we need to look at their standard reduction potentials. The standard reduction potentials for Zn and Cd are: \(\mathrm{Zn}^{2+} + 2e^{-} \rightarrow \mathrm{Zn}(s) \ \ E^{\circ} = -0.76\ \mathrm{V}\) (Zn electrode) \(\mathrm{Cd}^{2+} + 2e^{-} \rightarrow \mathrm{Cd}(s) \ \ E^{\circ} = -0.40\ \mathrm{V}\) (Cd electrode) Since the reduction potentials are different, the cell potentials of these two voltaic cells are also different. So, statement A is false.

Step 2: Comparing the electrode masses

In both cells, electrodes are made of \(\mathrm{Zn}\) and \(\mathrm{Cd}\). We are given that each cell has exactly one mole of each reactant and product. The molar masses of these metals are different: \(\mathrm{Zn}: 65.38\ \mathrm{g/mol}\) \(\mathrm{Cd}: 112.41\ \mathrm{g/mol}\) Since the molar masses of \(\mathrm{Zn}\) and \(\mathrm{Cd}\) are different and there is one mole of each electrode, their masses will also be different. So, statement B is false.

Step 3: Comparing the quantities of electric charge produced

According to Faraday's Law of Electrolysis, the amount of charge transferred during a redox reaction is directly proportional to the moles of electrons involved. For both Zn/Cu2+ and Cd/Cu2+ reactions, the reduction half-reaction involves the transfer of 2 moles of electrons: \(2 \ \mathrm{e^{-}} \rightarrow \mathrm{Zn}^{2+}\) for the Zn/Cu2+ cell \(2 \ \mathrm{e^{-}} \rightarrow \mathrm{Cd}^{2+}\) for the Cd/Cu2+ cell Since the electron transfer is the same for both cells and there is one mole of each reactant and product, the quantities of electric charge produced will also be the same. Therefore, statement C is true.

Step 4: Comparing the quantities of electric energy produced

The amount of electric energy produced by a cell is given by the equation: \(W = nFE\) Where \(W\) is the electrical work, \(n\) is the number of moles of electrons, \(F\) is Faraday's constant (96485 C/mol), and \(E\) is the cell potential. Since we demonstrated in step 1 that the cell potentials of these two voltaic cells are different, the quantities of electric energy that they can produce will also be different. So, statement D is false. So, the statement C is the correct answer: The quantities of electric charge that the two cells can produce are the same.

Key Concepts

Cell Potential

The cell potential, also known as electromotive force (EMF), is the driving force behind the flow of electrons in a voltaic cell. It is determined by the difference in standard reduction potentials of the two electrodes involved in the cell reaction. In our example, the Zn/Cu²⁺ cell and the Cd/Cu²⁺ cell have different reduction potentials, meaning their cell potentials will not be the same. The formula to calculate cell potential is:

• For the Zn/Cu²⁺ cell:
  \[ E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \]

This implies different production capabilities for work done or electricity generated. By comparing reduction potentials of Zn (-0.76 V) and Cd (-0.40 V), we can see a clear distinction. Thus, the cell potential varies due to differences in material properties, crucial for determining how much work can potentially be extracted from each cell.

Electrode Mass

Electrode mass is influenced by the molar mass of the metals used as electrodes in a voltaic cell. Each cell in our example contains one mole of electrode material. The molar mass for Zn is 65.38 g/mol, and for Cd, it is 112.41 g/mol. This distinction means that, despite having equal moles, the masses of the electrodes in the Zn/Cu²⁺ and Cd/Cu²⁺ cells are inherently different. This difference in mass does not affect the quantity of electric charge produced but is important when considering the practical construction and material requirements of a cell.

Electric Charge

Electric charge in a voltaic cell is generated through the movement of electrons during the redox reactions. Faraday's Law of Electrolysis tells us the quantity of charge is directly linked to the moles of electrons transferred. In both the Zn/Cu²⁺ and Cd/Cu²⁺ cells, the same number of electrons (2 moles) is involved in the reaction, resulting in an equal amount of electric charge. The constant used here is Faraday's constant, 96485 Coulombs per mole of electrons. This equality implies that both cells can produce the same quantity of electric charge despite differences in other properties like electrode material or potential.

Electric Energy

Electric energy produced by a voltaic cell depends on both the electric charge moved and the cell potential. This relationship is captured by the formula:

• \[ W = nFE \]

where \(W\) is the work done or energy produced, \(n\) is the number of electrons, \(F\) is Faraday’s constant, and \(E\) is the cell potential. Since the cell potentials for zn/Cu²⁺ and Cd/Cu²⁺ cells differ, as demonstrated earlier, the electric energy each can produce will also differ. Even with equal quantities of charge, the varying cell potentials lead to a difference in the total electric energy output, highlighting the importance of cell potential in energy considerations for voltaic cells.