Problem 97 - Chemistry: Principles and Reactions

Question

Consider the cell $$ \mathrm{Pt}\left|\mathrm{H}_{2}\right| \mathrm{H}^{+} \| \mathrm{H}^{+}\left|\mathrm{H}_{2}\right| \mathrm{Pt} $$ In the anode half-cell, hydrogen gas at \(1.0 \mathrm{~atm}\) is bubbled over a platinum electrode dipping into a solution that has a \(\mathrm{pH}\) of \(7.0 .\) The other half-cell is identical to the first except that the solution around the platinum electrode has a pH of \(0.0\). What is the cell voltage?

Step-by-Step Solution

Answer

The cell voltage is approximately 0.414 V.

Step 1: Identify the half reactions for anode and cathode

The given cell can be written as: $$ \mathrm{Pt}\left|\mathrm{H}_{2}\right| \mathrm{H}^{+} \| \mathrm{H}^{+}\left|\mathrm{H}_{2}\right| \mathrm{Pt} $$ And the half reactions are: Anode: \(\mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2e^-\) Cathode: \(\mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2e^-\)

Step 2: Calculate the cell potential difference using Nernst equation

Since the half-cell reactions are the same, we can focus on the difference in hydrogen ion concentrations. The Nernst equation relates the cell potential difference to the concentrations of the ions involved in the redox reactions: $$E_{cell}=E^\circ_{cell}-\dfrac{RT}{nF}\ln Q $$ Here, \(E^\circ_{cell}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred in the redox reaction, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. As both half-cell reactions are standard pHydrogen electrodes, \(E^\circ_{cell} =0V\). Given the pH values, the concentrations of \(\mathrm{H}^+\) ions in both half-cells are as follows: Anode: \([\mathrm{H}^{+}] = 10^{-7}\, \mathrm{M}\), as \(\mathrm{pH}=7\) Cathode: \([\mathrm{H}^{+}] = 10^{0} = 1.0\, \mathrm{M}\), as \(\mathrm{pH}=0\) Using these values, we can find the reaction quotient \(Q\): $$ Q = \dfrac{[\mathrm{H}^{+}]_\text{cathode}}{[\mathrm{H}^{+}]_\text{anode}} = \dfrac{1.0\mathrm{M}}{10^{-7}\,\mathrm{M}} = 1\times 10^7$$ With \(Q\), now we can find the cell potential difference: $$E_{cell} = 0 - \dfrac{(8.314\,\mathrm{J\, mol^{-1} K^{-1}}) (298\,\mathrm{K})}{(2)(96485\,\mathrm{C\, mol^{-1}})} \ln 1\times 10^7$$

Step 3: Calculate the cell voltage

Now we can plug values of the constants into the equation to calculate the cell voltage: $$E_{cell} = \dfrac{(8.314\,\mathrm{J\, mol^{-1} K^{-1}}) (298\,\mathrm{K})}{(2)(96485\,\mathrm{C\, mol^{-1}})} \ln 1\times 10^7$$ $$E_{cell} \approx 0.05916 \,\mathrm{V} \times \log(1\times 10^7)$$ $$E_{cell} \approx 0.05916 \,\mathrm{V} \times 7$$ $$E_{cell} \approx 0.4141\, \mathrm{V}$$ Therefore, the cell voltage is approximately \(0.414 \mathrm{V}\).

Key Concepts

Nernst Equation

The Nernst Equation is a fundamental formula in electrochemistry used to calculate the electromotive force (EMF) of a cell under non-standard conditions. It helps us determine the cell potential by considering the concentrations of the reacting ions. The equation is given by:\[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \]Where:

• \(E^0_{cell}\) is the standard cell potential when concentrations are at standard state (1 M for solutions).
• \(R\) is the universal gas constant, \(8.314 \text{ J mol}^{-1}\, \text{K}^{-1}\).
• \(T\) represents the temperature in Kelvin.
• \(n\) is the number of electrons transferred in the redox reaction.
• \(F\) is Faraday's constant, \(96,485 \text{ C mol}^{-1}\).
• \(Q\) is the reaction quotient, expressed as the ratio of product concentrations to reactant concentrations at any given point in time.

The Nernst Equation allows chemists to predict the cell potential by simply adjusting for differing ion concentrations. It is a powerful tool when dealing with half-cells like hydrogen electrodes which we've discussed in this exercise.
When applying it as seen in the given problem, use the hydrogen ion concentrations from the two half-cells to determine the voltage potential across the electrochemical cell.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical reactions where the oxidation state of molecules, atoms, or ions change by gaining or losing electrons. The foundational principles of redox reactions are essential in electrochemistry, playing a significant role in the flow of electrons in cells which generate electric current.
In our exercise, we have a hydrogen electrode cell, where the redox reactions are represented as follows:- **Anode Reaction (Oxidation):** \(\mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2e^-\) Here, hydrogen molecules lose electrons, which is the oxidation process.- **Cathode Reaction (Reduction):** \(\mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2e^-\) Similarly, hydrogen molecules receiving electrons are reduced.Redox reactions are always coupled, meaning that for every oxidation, there must be a corresponding reduction. This coupling leads to the flow of electrons which can be harnessed to do work, in the form of electricity, like in galvanic cells or batteries.
Understanding redox reactions helps students to predict the behavior of electrochemical cells, including factors like voltage and current outcomes.

pH and Cell Potential

The pH of a solution is intimately associated with the concentration of hydrogen ions \( [\mathrm{H}^+] \), and thus directly affects the cell potential of an electrochemical cell.

pH and Hydrogen Ion Concentration

The pH scale helps determine the acidity or basicity of a solution. Being a logarithmic scale, each unit change in the pH value corresponds to ten-fold change in \([\mathrm{H}^+]\). This connection links directly to the Nernst Equation when calculating cell potential in hydrogen-based electrochemical cells.

• A pH of 7 implies a neutral solution, with \([\mathrm{H}^+] = 10^{-7} \text{ M}\).
• A pH of 0 indicates a strongly acidic solution, with \([\mathrm{H}^+] = 1 \text{ M}\).

Effect on Cell Potential

In this exercise, the difference in pH between the anode and cathode affects the reaction quotient \( Q \), as it depends on \([\mathrm{H}^+]\) in both half-cells. By maximizing this difference, a higher cell potential can be achieved due to more significant ion concentration gradient:

• In the anode: \([\mathrm{H}^+] = 10^{-7} \text{ M}\)
• In the cathode: \([\mathrm{H}^+] = 1 \text{ M}\)

The potential difference calculated using the Nernst Equation estimates how changes in pH and hydrogen ion concentrations result in the electromotive force driving the electrons through the circuit, further illustrating how interconnected these concepts are in electrochemical applications.