Question

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{~V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{~V} \end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, \(\mathrm{HA}\), at an initial concentration of \(1.00 M .\) If the observed cell potential is \(0.333 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Short answer

The Ka value for the weak acid (HA) in the given galvanic cell is approximately 5.73 x 10⁻⁶.

Step-by-step solution

Write down the overall cell reaction.

The overall cell reaction can be found by adding the given two half-reactions. Remember to reverse the reaction with the lower standard potential to maintain spontaneity: \[ \mathrm{Fe}^{2+}(aq) + 2 \mathrm{H}^{+}(aq) \longrightarrow \mathrm{Fe}(s) + \mathrm{H}_{2}(g) \]

Write the Nernst equation for the cell.

The Nernst equation relates the cell potential (E) to standard cell potential (E°), the reaction quotient (Q), temperature (T), and the number of moles of transferred electrons (n). It can be written as: \[ \mathrm{E}=\mathrm{E}^{\circ}-\frac{RT}{nF} \ln Q \] where R is the gas constant, 8.314 J/(mol·K), T is the temperature in Kelvin (25°C = 298 K), F is the Faraday constant, 9.6485 x 10^4 C/mol, n is the number of transferred electrons in the balanced overall cell reaction (n = 2 in our case).

Calculate the standard cell potential.

The standard cell potential (E°) for the cell can be calculated by subtracting the standard potential of the anode (Fe²⁺/Fe) from that of the cathode (H⁺/H₂). Remember to reverse the reaction with the lower standard potential before calculating E°: \[ \mathrm{E}^{\circ} = \mathscr{E}^{\circ}_{\mathrm{cath.}} - \mathscr{E}^{\circ}_{\mathrm{anode}} = 0.000 \ \mathrm{V} - (-0.440 \ \mathrm{V}) = 0.440 \ \mathrm{V} \]

Write the reaction quotient (Q).

Using the overall spontaneous cell reaction: \[ Q=\frac{P_{\mathrm{H}_{2}}}{[\mathrm{Fe}^{2+}][\mathrm{H}^{+}]^{2}}=\frac{1.00\mathrm{~ atm}}{[1.00\times10^{-3}\mathrm{~M}][\mathrm{H}^{+}]^{2}} \] Since the weak acid HA initially has 1.00 M concentration and dissociates into H⁺ and A⁻ ions, we can approximate [H⁺] by the amount of HA that dissociated, x. Thus, [H⁺] = x, and [A⁻] = x, and [HA] = 1.00 M – x. Then, Ka = x² / (1.00 – x), and Q = 1.00 / x². We can ignore the change in [H⁺] due to the dissociation of water as Ka is typically very small.

Substitute the known values into the Nernst equation.

Plug in the observed cell potential (0.333 V), the calculated standard cell potential (0.440 V), n (2), R (8.314 J/(mol·K)), T (298 K), F (9.6485 x 10^4 C/mol), and Q to the Nernst equation: \[ 0.333 \ \mathrm{V}= 0.440 \ \mathrm{V} -\frac{8.314 \, \text{J/(\(mol\cdot K\))}\times 298 \, \text{K}}{2\times9.6485\times10^{4} \, \text{C/mol}} \ln \frac{1.00}{x^{2}} \]

Solve for x.

Rearrange the equation and solve for the unknown x which represents the concentration of H⁺ ions: \[ \ln \frac{1.00}{x^{2}}=\frac{(0.440 - 0.333)\times2\times9.6485\times10^{4} \, \text{C/mol}}{8.314\, \text{J/(\(mol\cdot K\))}\times 298\, \text{K}}= 13.48 \] The next step is to find the anti-log of 13.48 to solve for x²: \[ \frac{1.00}{x^{2}}=e^{13.48}\implies x^{2}=\frac{1}{e^{13.48}} \implies x=\sqrt{\frac{1}{e^{13.48}}} \] Calculating x, we get: \[ x = 2.394 \times 10^{-3} \, \mathrm{M} \]

Calculate the Ka value.

Finally, we can calculate the Ka value for the weak acid HA using the formula for Ka and the value of x we found: \[ \mathrm{Ka} = \frac{x^{2}}{1.00 M - x} = \frac{\left(2.394 \times 10^{-3}\right)^2}{1.00 - 2.394 \times 10^{-3}} = 5.73\times10^{-6} \] Therefore, the Ka value for the weak acid (HA) is approximately 5.73 x 10⁻⁶.

Key concepts

Nernst Equation

The Nernst equation is of vital importance in electrochemistry as it allows us to calculate the cell potential under non-standard conditions. Understanding this equation helps you predict how voltage will change with concentration, pressure, and temperature. The Nernst equation is normally expressed as:
\[\begin{equation}E = E^{\text{o}} - \frac{RT}{nF} \ln Q\end{equation}\]
Here, E is the cell potential at non-standard conditions, E^{\text{o}} 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 transferred in the reaction, F is the Faraday constant (approximately 96485 C/mol), and Q is the reaction quotient. The reaction quotient Q represents the ratio of products to reactants, each raised to their respective stoichiometric coefficients.
In the context of a galvanic cell, using this equation allows us to calculate the cell potential by considering changes in concentration of the electrolyte solutions, partial pressures of gases involved, and environmental temperature - which in the real world, are hardly ever at 'standard' conditions.

Standard Cell Potential

The standard cell potential (E^{\text{o}}) is a key concept that reflects the inherent voltage produced by a cell when all reagents are at standard conditions, which typically means solutes at a concentration of 1 M, gases at a pressure of 1 atm, and a temperature of 298 K (25°C). This value is determined by the difference in potential energies between the anode and cathode under these conditions. It is calculated using the standard reduction potentials of the half reactions involved:
\[\begin{equation}E^{\text{o}} = E^{\text{o}}_{cathode} - E^{\text{o}}_{anode}\end{equation}\]
For the given galvanic cell, the standard cell potential helps us gauge how much voltage we can expect from the reaction of iron cations and hydrogen ions to produce iron metal and hydrogen gas. The more positive the E^{\text{o}} value, the greater the cell's ability to produce electrical energy spontaneously. In real-world scenarios, this translates to how well a battery would work under ideal conditions.

Reaction Quotient

The reaction quotient (Q) is crucial for understanding the direction in which a reaction will proceed at any given point in time. It is similar to the equilibrium constant but is used for non-equilibrium conditions. Its role in the Nernst equation is vital as it demonstrates the effect of changing concentrations on the cell potential. The reaction quotient is determined by the expression:
\[\begin{equation}Q = \frac{[products]}{[reactants]}\end{equation}\]
where the concentrations of the gaseous products and reactants are raised to the power of their stoichiometric coefficients. For the provided exercise, this quantifies the ratio of the pressure of hydrogen gas to the concentration of iron ions and the concentration of hydrogen ions squared. Any change in these concentrations will affect the resulting cell potential, which is integral when working with batteries or electrochemical cells in various states of discharge or under different conditions.

Acid Dissociation Constant (Ka)

The acid dissociation constant (Ka) tells us about the strength of an acid in solution — specifically, its tendency to donate a proton to the solution. It is a specific form of equilibrium constant that applies to acids. For a weak acid (HA) that dissociates into a proton (H^+) and its conjugate base (A^−), the expression is as follows:
\[\begin{equation}Ka = \frac{[H^+][A^−]}{[HA]}\end{equation}\]
Hence, a larger Ka value represents a stronger acid. In the case of our galvanic cell, knowing the Ka value allows us to estimate the proton concentration at equilibrium, which in turn, is used in computing the reaction quotient (Q) for the Nernst equation. When dealing with weak acids, we often assume that the concentration of the acid does not change significantly upon dissociation. However, one must be mindful that as ions accumulate, they can affect the reaction's dynamics, which might require adjustments to the equilibrium calculations.