Question

Suppose a galvanic cell was constructed at \(25^{\circ} \mathrm{C}\) using a \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell (in which the molar concentration of \(\mathrm{Cu}^{2+}\) was \(1.00 \mathrm{M}\) ) and a hydrogen electrode having a partial pressure of \(\mathrm{H}_{2}\) equal to 1 atm. The hydrogen electrode dips into a solution of unknown hydrogen ionconcentration, and the two half- cells are connected by a salt bridge. The precise value of \(E_{\mathrm{cell}}^{\circ}\) is \(+0.3419 \mathrm{~V}\). (a) Derive an equation for the \(\mathrm{pH}\) of the solution with the unknown hydrogen ion concentration, expressed in terms of \(E_{\text {cell }}\) and \(E_{\text {cell }}^{\circ}\) (b) If the \(\mathrm{pH}\) of the solution were \(5.15,\) what would be the observed potential of the cell? (c) If the potential of the cell were \(0.645 \mathrm{~V}\), what would be the \(\mathrm{pH}\) of the solution?

Short answer

a) pH = -\(\frac{2}{0.0592}\)(E_{cell} - E_{cell}^{\text{o}}), b) E_{cell} when pH = 5.15 is E_{cell} = E_{cell}^{\text{o}} + 0.0592 \times 5.15, c) pH when E_{cell} = 0.645 V is pH = \(\frac{2(0.645 - 0.3419)}{0.0592}\)

Step-by-step solution

Write down the Nernst Equation

Begin by writing the Nernst equation, which relates the cell potential to its standard cell potential and the reaction quotient (Q): \[\begin{equation}E_{cell} = E_{cell}^{\text{o}} - \frac{0.0592}{n} \times \text{log(Q)}\end{equation}\]where - \(E_{cell}\) is the cell potential,- \(E_{cell}^{\text{o}}\) is the standard cell potential, - \(n\) is the number of moles of electrons transferred in the half-reactions of the cell, and - \(Q\) is the reaction quotient.

Determine the Half-Reactions and Calculate Q

Identify the half-reactions for the galvanic cell:Anode (oxidation): \(\mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2\ e^-\)Cathode (reduction): \(\mathrm{Cu}^{2+} + 2\ e^- \rightarrow \mathrm{Cu}\)The reaction quotient (Q) for the overall reaction is given by:\[Q = \frac{[\mathrm{H}^+]^2}{[\mathrm{Cu}^{2+}]}\]Since the concentration of copper ions \(\mathrm{Cu}^{2+}\) is 1.00 M, the expression for Q simplifies to:\[Q = [\mathrm{H}^+]^2\]

Express pH in Terms of Cell Potential

With \(n = 2\) electrons transferred and given that \(\text{pH} = -\log[\mathrm{H}^+]\), rearrange the Nernst equation to solve for \(\text{pH}\) in terms of \(E_{cell}\):\[\begin{equation}E_{cell} = E_{cell}^{\text{o}} - \frac{0.0592}{2} \times \text{log}([\mathrm{H}^+]^2)\end{equation}\]\[\begin{equation}-\text{log}([\mathrm{H}^+]) = -\frac{2}{0.0592}(E_{cell} - E_{cell}^{\text{o}})\end{equation}\]\[\begin{equation}\text{pH} = -\frac{2}{0.0592}(E_{cell} - E_{cell}^{\text{o}})\end{equation}\]This equation relates the pH of the solution to the cell potential.

Derive the Cell Potential from pH

For part (b), where the pH is given as 5.15, use the derived equation in reverse to find the cell potential.First, calculate the hydrogen ion concentration from the pH:\[[\mathrm{H}^+] = 10^{-\text{pH}} = 10^{-5.15}\]Then, insert this value into the Nernst equation and solve for \(E_{cell}\):\[\begin{equation}E_{cell} = E_{cell}^{\text{o}} - \frac{0.0592}{2} \times \text{log}(10^{-5.15 \times 2})\end{equation}\]\[\begin{equation}E_{cell} = E_{cell}^{\text{o}} - \frac{0.0592}{2} \times (-5.15 \times 2)\end{equation}\]Insert the value for \(E_{cell}^{\text{o}}\), and calculate for \(E_{cell}\).

Calculate pH from Cell Potential

For part (c), where the cell potential is given as 0.645 V, use the derived equation to find the pH.By inserting the given \(E_{cell}\) and the standard cell potential, solve for pH:\[\begin{equation}0.645 = 0.3419 - \frac{0.0592}{2} \times \text{log}(\mathrm{[H^+]}^2)\end{equation}\]Rearrange and solve for \([\mathrm{H}^+]\), then convert to pH:\[\begin{equation}-\log([\mathrm{H}^+]) = \frac{2(0.645 - 0.3419)}{0.0592}\end{equation}\]\[\begin{equation}\text{pH} = \frac{2(0.645 - 0.3419)}{0.0592}\end{equation}\]

Key concepts

Nernst Equation

The Nernst equation is a fundamental relation in electrochemistry that provides a link between the electrochemical cell potential and the concentrations of the reactants and products involved. It involves the reaction quotient, temperature, number of electrons transferred in the cell reaction, and the standard cell potential. The general form of the Nernst equation can be expressed as:
\[E_{cell} = E_{cell}^{\text{o}} - \frac{0.0592}{n} \times \log(Q)\]
where:

• \(E_{cell}\) is the current cell potential,
• \(E_{cell}^{\text{o}}\) is the standard cell potential,
• \(n\) represents the number of moles of electrons transferred in the half-reactions, and
• \(Q\) is the reaction quotient.

When it comes to pH calculations, the Nernst equation allows us to relate the hydrogen ion concentration (and hence pH) to the cell potential by rearranging the formula to isolate the term containing the concentration of hydrogen ions.

Cell Potential

Cell potential, often represented by \(E_{cell}\), is the measure of the voltage difference between two half-cells in an electrochemical cell. It's a critical factor that determines the direction and extent to which a chemical reaction can generate electrical energy.
In a standard condition, where all reactants and products are at their standard states (1M concentration for solutes, 1 bar pressure for gases, and pure solids or liquids), the cell potential is termed as \(E_{cell}^{\text{o}}\). The standard cell potential is a constant value that provides the driving force for the electrochemical reaction and can be calculated by subtracting the standard potential of the anode from that of the cathode:
\[E_{cell}^{\text{o}} = E_{cathode}^{\text{o}} - E_{anode}^{\text{o}}\]
The cell potential is also temperature-dependent, and the Nernst equation allows us to adjust for changes in temperature and reactant/product concentrations.

Hydrogen Electrode

The hydrogen electrode is a reference electrode based on the redox reaction involving hydrogen gas and is often used to measure the pH of a solution. It is set up by dipping a platinum electrode into an acidic solution and bubbling hydrogen gas through it at a pressure of 1 atm.
The half-cell reaction at the hydrogen electrode is:
\[\mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2e^-\]
The standard potential of the hydrogen electrode is defined as 0 V, making it a convenient reference point. When we integrate the hydrogen electrode into cell potential measurements, we effectively measure the tendency of another half-cell to gain or lose electrons relative to the hydrogen half-cell. This provides a crucial connection between electrochemical reactions and pH, as seen in the exercise, allowing us to use the Nernst equation to find the pH of the solution with an unknown hydrogen ion concentration.

Reaction Quotient

The reaction quotient (Q) is a measure of the relative amounts of products and reactants present during a reaction at a particular point in time. It is defined as the ratio of the concentrations of the products raised to the power of their stoichiometric coefficients to the concentrations of the reactants raised to the power of their stoichiometric coefficients.
For the cell reaction involving hydrogen gas and copper ions, the reaction quotient would be expressed as:
\[Q = \frac{[\mathrm{H}^+]^2}{[\mathrm{Cu}^{2+}]}\]
In the context of the Nernst equation, the reaction quotient is vital as it allows us to calculate the actual cell potential at any moment of the reaction, differentiating it from the standard cell potential where all reactants and products are in their standard states. By substituting Q into the Nernst equation, we can relate it to the pH of the solution, since the concentration of hydrogen ions \([\mathrm{H}^+]\) directly impacts the reaction quotient.