Question

Calculate \(E^{\circ}\) for the following cells: (a) \(\mathrm{Pb}\left|\mathrm{PbSO}_{4} \| \mathrm{Pb}^{2+}\right| \mathrm{Pb}\) (b) \(\mathrm{Pt}\left|\mathrm{Cl}_{2}\right| \mathrm{ClO}_{3}^{-} \| \mathrm{O}_{2}\left|\mathrm{H}_{2} \mathrm{O}\right| \mathrm{Pt}\) (c) \(\mathrm{Pt}\left|\mathrm{OH}^{-}\right| \mathrm{O}_{2} \| \mathrm{ClO}_{3}^{-}, \mathrm{Cl}^{-} \mid \mathrm{Pt} \quad\) (basic medium)

Short answer

Question: Calculate the standard electrode potential (E°) for three electrochemical cells: (a) Pb|PbSO4|Pb2+|Pb (b) Pt|Cl2|ClO3-|O2|H2O|Pt (c) Pt|OH-|O2|ClO3-, Cl-|Pt (basic medium) Answer: (a) E° = 0 (b) E° = E°Cl2/Cl- - E°ClO3-/O2 (insufficient information to find exact values) (c) E° = E°OH-/O2 - E°ClO3-/Cl- (insufficient information to find exact values)

Step-by-step solution

Identify half reactions at the cathode and anode

For each cell, we need to identify which reactions are taking place at the cathode (reduction) and the anode (oxidation). (a) Pb|PbSO4|Pb2+|Pb: Here, the cathode and anode are both made of lead, so the half-reactions are: Cathode (reduction): Pb2+ + 2e- → Pb Anode (oxidation): Pb → Pb2+ + 2e- (b) Pt|Cl2|ClO3-|O2|H2O|Pt: Here, we have three distinct half-reactions taking place. Cathode (reduction): Cl2 + 2e → 2Cl- Anode (oxidation): 2ClO3- → Cl2 + 2O2 + 2e- (c) Pt|OH-|O2|ClO3-, Cl-|Pt (basic medium): The half reactions taking place in basic medium include combining OH- ions. Cathode (reduction): O2 + 2H2O + 4e- → 4OH- Anode (oxidation): 2ClO3- + 12OH- → 6H2O + 2Cl- + 10e-

Apply the Nernst Equation

Using the Nernst equation, we can calculate the E° potential. The equation is given by: E° = E°cathode - E°anode (a) For cell (a), the half-reactions have the same E°, which cancels out, resulting in: E° = E°Pb2+/Pb - E°Pb2+/Pb = 0 (b) For cell (b), use the E° values for Cl2/Cl- and ClO3-/O2 half reactions: E° = E°Cl2/Cl- - E°ClO3-/O2 Unfortunately, we do not have enough information provided in this exercise to directly find the values for these half-reactions. They would usually be found in a table of standard electrode potentials. (c) For cell (c), use the E° values for OH-/O2 and ClO3-/Cl- half reactions: E° = E°OH-/O2 - E°ClO3-/Cl- Similar to cell (b), we don't have enough information to directly find the value for these half-reactions without a table of standard electrode potentials.

Key concepts

Half Reactions

In electrochemical cells, half reactions are fundamental components that dictate the direction of electron flow. Each cell consists of two half-reactions: one occurring at the cathode and the other at the anode. These half-reactions are separated by a double vertical line notation, indicating them working together to complete a redox reaction.
For example, in the cell represented by **Pb | PbSO4 || Pb²+ | Pb**, the lead undergoes both oxidation and reduction. At the anode (oxidation), lead metal ( ext{Pb}) is converted into lead ions ( ext{Pb}^{2+}) by losing two electrons. Meanwhile, at the cathode (reduction), ext{Pb}^{2+} ions are converted back to lead metal by gaining electrons.
This process ensures the conservation of charge while facilitating the flow of electrons from anode to cathode, thus generating electric current.
Understanding half reactions helps in determining the overall cell reaction by adding the oxidation and reduction reactions. It also allows for the calculation of variables like the cell potential, represented by the next concept.

Standard Electrode Potential

The standard electrode potential, often denoted as \(E^{\circ}\), is a critical measurement indicating how likely a substance is to be reduced. It's measured in volts and recorded under standard conditions of 298 K, 1 atm pressure, and 1 M concentration for all reactants and products.
Each half-reaction has its own standard electrode potential, which can be found in reference tables. This potential helps predict the direction of electron flow in an electrochemical cell. The cathode's standard potential is compared with that of the anode to determine the overall cell potential. A higher \(E^{\circ}\) value at the cathode means it's more inclined to draw electrons toward itself, thus promoting reduction.
The overall cell potential, as derived from the half-reactions, is calculated by subtracting the anode potential from the cathode potential, represented by the formula:

• \(E^{\circ} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)

This calculation tells us how much voltage an electrochemical cell can potentially generate under standard conditions. Understanding and calculating these potentials allows one to assess cell viability and energy efficiency.

Nernst Equation

The Nernst Equation provides a mathematical model for understanding how the cell potential changes with varying conditions such as concentration, temperature, and pressure. It is an extension of the standard electrode potential, taking real-world conditions into account.
The equation is typically expressed as:
\[ E = E^{\circ} - \frac{RT}{nF} \ln Q \]
where:

• \(E\) is the cell potential under non-standard conditions.
• \(E^{\circ}\) is the standard cell potential.
• \(R\) is the universal gas constant \(8.314 \, J/mol \, K\).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons exchanged in the reaction.
• \(F\) is the Faraday constant \(96,485 \, C/mol\).
• \(Q\) is the reaction quotient, representing the ratio of product and reactant concentrations.

As concentrations shift from standard conditions, \(Q\) changes, leading to a different cell potential \(E\) than the standard potential \(E^{\circ}\). This equation is vital for calculations in real-world applications where conditions are rarely ideal. It enables the prediction of cell behavior under specific environments, ensuring optimal functionality of electrochemical cells.