Question

Consider the following half-reactions and \(E^{\circ}\) values: $$ \begin{array}{ll} \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & E^{\circ}=0.34 \mathrm{~V} \\ \mathrm{~Pb}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s) & E^{\circ}=-0.13 \mathrm{~V} \end{array} $$ (a) Which of these metals or ions is the strongest oxidizing agent? Which is the strongest reducing agent? (b) The half-reactions can be used to construct three different galvanic cells. Tell which cell delivers the highest voltage, identify the anode and cathode, and tell the direction of electron and ion flow. (c) Write the cell reaction for part (b), and calculate the values of \(E^{\circ}, \Delta G^{\circ}\) (in kilojoules), and \(K\) for this reaction at \(25^{\circ} \mathrm{C}\). (d) Calculate the voltage for the cell in part (b) if both ion concentrations are \(0.010 \mathrm{M}\).

Short answer

(a) Strongest oxidizing agent: \(\mathrm{Ag}^+\); Reducing agent: \(\mathrm{Pb}\). (b) Highest voltage cell: \(\mathrm{Ag}/\mathrm{Ag}^+\) - cathode, \(\mathrm{Pb}^{2+}/\mathrm{Pb}\) - anode; electrons flow \(\mathrm{Pb} \to \mathrm{Ag}\). (c) Reaction: \(2\mathrm{Ag}^{+} + \mathrm{Pb} \to 2\mathrm{Ag} + \mathrm{Pb}^{2+}\); \(E^\circ=0.93\,\text{V}, \Delta G^\circ=-179.5\,\text{kJ/mol}, K\approx 7.64\times 10^{31}\). (d) \(E\approx 0.89\,\text{V}\).

Step-by-step solution

Identify Oxidizing and Reducing Agents

The strongest oxidizing agent is the species that gets reduced, which corresponds to the highest standard reduction potential. Conversely, the strongest reducing agent is the species that gets oxidized, typically having the lowest (or most negative) standard reduction potential in its forward (oxidation) reaction.

Determine the Strongest Oxidizing Agent

Compare the given standard reduction potentials: \(\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}\) with \(E^\circ = 0.80\,\mathrm{V}\), \(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\) with \(E^\circ = 0.34\,\mathrm{V}\), and \(\mathrm{Pb}^{2+} + 2e^- \rightarrow \mathrm{Pb}\) with \(E^\circ = -0.13\,\mathrm{V}\). The strongest oxidizing agent is \(\mathrm{Ag}^+\) due to its highest potential of \(0.80\,\mathrm{V}\).

Determine the Strongest Reducing Agent

Since being a reducing agent means donating electrons, the lower the (or more negative) reduction potential, the stronger the reducing power. \(\mathrm{Pb}\) is the strongest reducing agent because it has the lowest potential of \(-0.13\,\mathrm{V}\).

Identify the Highest Voltage Cell

Calculate the cell potential for each possible pair: \(\text{Ag}^+/ \text{Ag} - \text{Pb}^{2+}/ \text{Pb}\), \(\text{Ag}^+/ \text{Ag} - \text{Cu}^{2+}/ \text{Cu}\), and \(\text{Cu}^{2+}/ \text{Cu} - \text{Pb}^{2+}/ \text{Pb}\). The cell potential is the difference between the cathode and anode potentials. The highest voltage will come from pairing \(\text{Ag}^+/ \text{Ag}\) as the cathode and \(\text{Pb}^{2+}/ \text{Pb}\) as the anode.

Calculate E° for the Highest Voltage Cell

The cell potential \(E^\circ_{\text{cell}}\) is given by the difference \(E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\). For \(\text{Ag}^+/ \text{Ag}\) as cathode and \(\text{Pb}^{2+}/ \text{Pb}\) as anode, \(E^\circ = 0.80 - (-0.13) = 0.93\,\mathrm{V}\).

Direction of Electron and Ion Flow

Electrons flow from anode to cathode externally, or \(\text{Pb}\rightarrow\text{Ag}\), while \(\text{Pb}^{2+}\) ions will migrate into the solution and \(\text{Ag}^+\) ions will deposit as \(\text{Ag}\) on the cathode.

Write the Cell Reaction

The balanced cell reaction is obtained by combining the two half-reactions: \(\mathrm{Ag}^{+}(aq) + \mathrm{e}^- \rightarrow \mathrm{Ag}(s)\) and \(\mathrm{Pb}(s) \rightarrow \mathrm{Pb}^{2+}(aq) + 2\mathrm{e}^-\). The overall reaction is: 2\(\mathrm{Ag}^+ + \mathrm{Pb} \rightarrow 2\mathrm{Ag} + \mathrm{Pb}^{2+}\).

Calculate \(\Delta G^\circ\)

Use the equation \(\Delta G^\circ = -nFE^\circ\), where \(n=2\) moles of electrons, \(F = 96485\,\text{C/mol}\), and \(E^\circ = 0.93\,\text{V}\). Thus, \(\Delta G^\circ = -2 \times 96485 \times 0.93 = -179.5\,\text{kJ/mol}\).

Calculate the Equilibrium Constant \(K\)

The relationship \(\ln K = \frac{nFE^\circ}{RT}\) can be rearranged to find \(K\): \(K = e^{\frac{nFE^\circ}{RT}}\). Using \(R = 8.314\,\text{J/K}\cdot \text{mol}\), \(T=298\,\text{K}\), and prior values, \(K \approx 7.64 \times 10^{31}\).

Adjust Ion Concentration to Find \(E\) at 0.01 M

Apply the Nernst equation: \(E = E^\circ - \frac{RT}{nF}\ln Q\). The reaction quotient \(Q = \frac{[\text{Pb}^{2+}]}{[\text{Ag}^+]^2}\) and substituting values gives \((Q = 10000)\), solving gives \(E \approx 0.89\,\text{V}\).

Key concepts

Standard Reduction Potential

Standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is measured under standard conditions: 1 M concentration for solutions and 1 atm pressure for gases in electrochemical cells at 25°C.
These values are crucial in determining which substances will oxidize or reduce in a redox reaction. Thus, they tell us which substances act as oxidizing or reducing agents.

• Oxidizing Agents: These species have a higher standard reduction potential. They tend to gain electrons during a reaction.
• Reducing Agents: These have a lower or more negative standard reduction potential. They tend to lose electrons during a reaction.

The standard reduction potentials themselves are reported as voltage, symbolized by \(E^{\circ}\). In our exercise, silver ion \( \text{Ag}^+ \) had the highest \(E^{\circ}\) of 0.80 V, making it the strongest oxidizing agent. Similarly, lead \( \text{Pb} \), with the lowest \(E^{\circ}\) of -0.13 V, functioned as the strongest reducing agent.

Galvanic Cells

Galvanic cells, also known as voltaic cells, are devices that convert chemical energy into electrical energy through spontaneous redox reactions. These cells have two half-cells connected by a salt bridge, with each half-cell hosting one of the redox reactions.
The cell can be characterized by its electrodes:

• Anode: The electrode where oxidation occurs, releasing electrons.
• Cathode: The electrode where reduction takes place, accepting electrons.

In our exercise, the greatest cell voltage is achieved using the combination where silver ions \( \text{Ag}^+ \) serve as the cathode and lead \( \text{Pb} \) as the anode. The voltage can be found by subtracting the anode's standard reduction potential from that of the cathode's, giving us \(0.93\,\text{V}\).
Electrons flow from the anode to the cathode through an external circuit, while ions move within the internal circuit through the salt bridge to maintain charge balance. This steady flow is what fuels devices powered by galvanic cells.

Nernst Equation

The Nernst Equation provides the means to calculate the cell potential at non-standard conditions. It adjusts the standard cell potential to account for differing ionic concentrations. The equation is:
\[ E = E^{\circ} - \frac{RT}{nF} \ln Q \] Where:

• \(E\) is the cell potential at non-standard conditions
• \(E^{\circ}\) is the standard cell potential
• \(R\) is the gas constant (8.314 J/mol K)
• \(T\) is the temperature in Kelvin
• \(n\) is the number of moles of electrons transferred
• \(F\) is Faraday’s constant (96485 C/mol)
• \(Q\) is the reaction quotient

In the context of our problem, the Nernst Equation enabled us to determine the cell potential under a situation where both ion concentrations were \(0.01 \text{M}\). Doing the calculations, we found the adjusted cell potential to be roughly \(0.89\,\text{V}\). By incorporating various initial concentrations, the Nernst Equation serves as a vital tool for predicting the behavior of galvanic cells in real-world applications.