Question

Calculate the pressure of \(\mathrm{H}_{2}\) (in atm) required to maintain equilibrium with respect to the following reaction at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Pb}(s)+2 \mathrm{H}^{+}(a q) \rightleftarrows \mathrm{Pb}^{2+}(a q)+\mathrm{H}_{2}(g)$$ given that \(\left[\mathrm{Pb}^{2+}\right]=0.035 M\) and the solution is buffered at \(\mathrm{pH} 1.60\).

Short answer

The required pressure of H₂ is approximately 5,926 atm.

Step-by-step solution

Identify Relevant Equation

The given equilibrium reaction is:\[ \mathrm{Pb}(s) + 2 \mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{Pb}^{2+}(a q) + \mathrm{H}_{2}(g) \]This is a redox reaction and the Nernst equation will be used to calculate the equilibrium pressure of \( \mathrm{H}_{2} \).

Write the Nernst Equation

The Nernst equation at standard temperature and pressure is given by:\[ E = E^0 - \frac{RT}{nF} \ln Q \]where \(E\) is the cell potential, \(E^0\) is the standard cell potential, \(R\) is the universal gas constant \(8.314 \, \text{J/mol·K}\), \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant \(96485 \, \text{C/mol}\), and \(Q\) is the reaction quotient.

Calculate Standard Cell Potential (E⁰)

The standard reduction potentials are:- \( \mathrm{Pb}^{2+} + 2 e^- \rightarrow \mathrm{Pb}(s) \), \(E^0_1 = -0.13 \, \text{V} \)- \( 2 \mathrm{H}^+ + 2 e^- \rightarrow \mathrm{H}_2(g) \), \(E^0_2 = 0.00 \, \text{V} \)The standard oxidation potential for the reverse reaction is \(E^0_2 = -0.00 \, \text{V} \) (since it is the reverse of reduction).The cell potential for the reaction is given by the difference:\[ E^0 = E^0_2 - E^0_1 = 0.00 \, \text{V} - (-0.13 \, \text{V}) = 0.13 \, \text{V} \]

Calculate the Reaction Quotient (Q)

The reaction quotient \(Q\) is defined as:\[ Q = \frac{[\mathrm{Pb}^{2+}][\mathrm{H}_{2}]}{[\mathrm{H}^{+}]^2} \]From the problem, we have \([\mathrm{Pb}^{2+}] = 0.035 \, \text{M}\), let's call the pressure of \(\mathrm{H}_2\) as \(P_{\mathrm{H}_2}\), and the pH is 1.60, so \([\mathrm{H}^+] = 10^{-1.60} \, \text{M}\).Therefore, \[ Q = \frac{0.035 \times P_{\mathrm{H}_2}}{(10^{-1.60})^2} \]

Apply the Nernst Equation for Equilibrium

At equilibrium, \(E = 0\). Thus, substitute the values into the Nernst equation:\[ 0 = 0.13 - \frac{8.314 \times 298}{2 \times 96485} \ln \left( \frac{0.035 \times P_{\mathrm{H}_2}}{(10^{-1.60})^2} \right) \]This can be simplified to:\[ \ln \left( \frac{0.035 \times P_{\mathrm{H}_2}}{10^{-3.2}} \right) = \frac{2 \times 96485 \times 0.13}{8.314 \times 298} \]

Solve for Pressure of Hydrogen (P_H₂)

Calculate the constant from the simplified equation:\[ \frac{2 \times 96485 \times 0.13}{8.314 \times 298} = 10.65 \]Thus, the equation becomes:\[ 10.65 = \ln(\frac{0.035 \times P_{\mathrm{H}_2}}{10^{-3.2}}) \]Exponentiate both sides:\[ e^{10.65} = \frac{0.035 \times P_{\mathrm{H}_2}}{10^{-3.2}} \]Solve for \(P_{\mathrm{H}_2}\):\[ P_{\mathrm{H}_2} = e^{10.65} \times 10^{-3.2} / 0.035 \approx 207.4 / 0.035 \]This gives \(P_{\mathrm{H}_2} \approx 5,926 \, \text{atm}\).

Key concepts

Redox Reaction

A redox reaction is a chemical process where electrons are transferred between two substances. In the given exercise, we are dealing with the reaction \( \text{Pb}(s) + 2 \text{H}^+(aq) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{H}_2(g) \). This involves the transfer of electrons from lead to hydrogen ions. Lead (\( \text{Pb}(s) \)), acting as a reducing agent, donates electrons to hydrogen ions (\( \text{H}^+ \)), which are reduced to form hydrogen gas (\( \text{H}_2 \)), while the lead is oxidized to lead ions (\( \text{Pb}^{2+} \)).

Some key points about redox reactions include:

• They always occur simultaneously with oxidation and reduction.
• Oxidation involves the loss of electrons, while reduction involves the gain of electrons.
• In the above reaction, hydrogen ions gain electrons (reduced), and lead loses electrons (oxidized).

Understanding redox reactions is essential in calculating cell potentials and using the Nernst equation effectively.

Equilibrium Pressure

Equilibrium pressure refers to the pressure at which the rate of the forward reaction equals the rate of the backward reaction. In the context of gas reactions, it represents the pressure at which reactants and products are in balance with each other.

For the provided redox reaction, equilibrium needs be maintained with respect to the hydrogen gas pressure. This means we calculate the pressure at which the concentration of \( \text{H}_2(g) \) does not change over time. This is done using the Nernst equation, where the cell potential equals zero at equilibrium.

At equilibrium, the chemical reaction has no net change in the concentration of the reacting species, meaning:

• The forward and reverse reactions occur at equal rates.
• The concentrations (or pressures, if gases) of the products and reactants remain constant.

Calculating equilibrium pressure helps determine the conditions necessary for a reaction to maintain its balance over time.

Standard Cell Potential

The standard cell potential \((E^0)\) is the voltage potential difference between two half-cells in a galvanic cell when all components are in their standard states, typically \(1 \, ext{M} \) for solutions and \(1 \, ext{atm} \) for gases, at a specified temperature (usually 25°C).

In our scenario, the standard cell potential is calculated using standard reduction potentials of the involved reactions. These are:

• \( \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}(s) \) with \(E^0_1 = -0.13 \, ext{V} \)
• \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g) \) with \(E^0_2 = 0.00 \, ext{V} \)

The cell potential is obtained by subtracting \(E^0_1\) from \(E^0_2 \), resulting in \(E^0 = 0.13 \, ext{V} \).

Standard cell potentials indicate the tendency of a redox reaction to occur. A positive \(E^0\) suggests that the reaction is spontaneous under standard conditions.

Reaction Quotient

The reaction quotient \((Q)\) is a ratio that compares the concentrations or pressures of the products to the reactants at any point in time during a reaction. It is similar to the equilibrium constant \((K)\), but while \(K\) pertains to equilibrium conditions, \(Q\) applies to any point before equilibrium is reached.

For our redox reaction, the reaction quotient is calculated as:
\[ Q = \frac{[\text{Pb}^{2+}][\text{H}_{2}]}{[\text{H}^{+}]^2} \]
Given the data:

• \([\text{Pb}^{2+}] = 0.035 \, ext{M}\)
• Let \(P_{\text{H}_2}\) represent the equilibrium pressure of \(\text{H}_2\)
• From pH = 1.60, \([\text{H}^+] = 10^{-1.60} \, ext{M}\)

Plug these values into the expression for \(Q\) to get:
\[ Q = \frac{0.035 \times P_{\text{H}_2}}{(10^{-1.60})^2} \]

Knowing \(Q\) helps in using the Nernst equation to find when the system reaches equilibrium, where simplifies solving for unknown variables such as the pressure of \(\text{H}_2\).