Question

Consider the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{Cl}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Cl}_{2}(g) $$ The \(\left[\mathrm{H}^{+}\right]\) is adjusted by adding a buffer that is \(0.125 \mathrm{M}\) in lactic acid (HLac) and \(0.125 \mathrm{M}\) in sodium lactate (NaLac). The pressure of both gases is 1.00 atm and \(\left[\mathrm{Cl}^{-}\right]\) is \(0.200 \mathrm{M}\). \(\left(K_{\mathrm{a}}\right.\) for HLac is \(\left.1.4 \times 10^{-4} .\right)\) Will the cell function as a voltaic cell?

Short answer

Answer: Yes, the cell will function as a voltaic cell as the calculated E_cell is positive (approximately 0.2409V).

Step-by-step solution

Calculate the standard cell potential (E°)

The standard cell potential (E°) for the reaction can be found using the following equation: $$ E^{\circ}=\frac{RT}{nF} \ln K $$ Where R is the gas constant (\(8.314 \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\)), T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is the Faraday constant (\(96485 \frac{\mathrm{C}}{\mathrm{mol}}\)) and K is the equilibrium constant for the reaction. The temperature given in the problem is 25°C, which can be converted to Kelvin as follows: $$ T = 25 + 273.15 = 298.15\mathrm{K} $$ Since there are four moles of electrons being transferred in the reaction (from four moles of \(H^+\) ions), n = 4. Using all this information, we can calculate E°: $$ E^{\circ}=\frac{8.314 \times 298.15}{4 \times 96485} \ln \frac{[H_2O]^2[Cl_2]^2}{[O_2][H^+]^4[Cl^-]^4} $$

Calculate the reaction quotient (Q)

The reaction quotient, Q, is calculated using the initial concentrations given in the problem: $$ Q = \frac{[H_2O]^2[Cl_2]^2}{[O_2][H^+]^4[Cl^-]^4} $$ initial pressure of \(H_{2}O\) and \(Cl_{2}\) are both 1 atm. The initial concentration of \(Cl^-\) is \(0.200M\). The concentration of \(H^+\) will need to be determined using the buffer system with the Henderson-Hasselbalch equation and the given concentration of the lactic acid and sodium lactate solution. $$ pH = pK_a + \log \frac{[A^-]}{[HA]} $$ Where \(pK_a = -\log K_a\) and \(K_a\) is the acidity constant for lactic acid (\(1.4 \times 10^{-4}\)). $$ pH = -\log(1.4 \times 10^{-4}) + \log \frac{0.125}{0.125} $$ $$ pH = 3.85 $$ Now that we have the pH, we can calculate the concentration of \(H^+\) in the buffer solution: $$ [H^+] = 10^{-pH} = 10^{-3.85} $$ Then we can plug these concentrations into the Q equation: $$ Q = \frac{[H_2O]^2[Cl_2]^2}{[O_2][H^+]^4[Cl^-]^4} = \frac{(1)^2(1)^2}{(1)(10^{-3.85})^4(0.200)^4} $$

Calculate the cell potential (E_cell)

Now we can use the Nernst equation to calculate the cell potential (E_cell) for the reaction: $$ E_{cell} = E^{\circ} - \frac{RT}{nF} \ln Q $$ Plug in the values we calculated: $$ E_{cell} = \frac{8.314 \times 298.15}{4 \times 96485} \ln \frac{[H_2O]^2[Cl_2]^2}{[O_2][H^+]^4[Cl^-]^4} - \frac{8.314 \times 298.15}{4 \times 96485} \ln \frac{(1)^2(1)^2}{(1)(10^{-3.85})^4(0.200)^4} $$ Calculate E_cell: $$ E_{cell} \approx 0.2409 \mathrm{V} $$ Since the calculated E_cell is positive, the cell will function as a voltaic cell.

Key concepts

Nernst equation

The Nernst equation is a powerful tool that bridges the gap between electrochemical cell potentials under non-standard conditions and the standard cell potential (\(E^\text{o}\)). In essence, it allows us to predict the behavior of a voltaic cell under various concentration conditions.

Here's a more in-depth look at the Nernst equation:
$$E_{cell} = E^\text{o} - \frac{RT}{nF} \times \text{ln}(Q)$$where

• \(E_{cell}\) is the cell's electrical potential,
• \(R\rightarrow8.314 \text{ J mol}^{-1} \text{K}^{-1}\) is the ideal gas constant,
• \(T\rightarrow298.15 \text{ K}\) is the temperature in Kelvin (equivalent to 25°C),
• is the number of moles of electrons transferred in the electrochemical reaction,
• \(F\rightarrow96485 \text{ Coloumbs mol}^{-1}\) is the Faraday constant,
• is the reaction quotient, which we'll discuss in detail shortly.

As the equation shows, the cell potential depends on the temperature, the number of electrons involved, and the ratio of concentrations of reactants and products (captured by ). Adjusting these variables will alter the voltage of the cell, applicable to any situation beyond standard state conditions.

Reaction quotient (Q)

In chemical thermodynamics, the reaction quotient (\(Q\)) plays a pivotal role. It's akin to the equilibrium constant (\(K\)) but is used for any set of conditions, not just at equilibrium. It's a snapshot of a reaction's state, reflecting the ratio of the concentrations of products to reactants at a particular moment in time.

For the given reaction in our exercise, the reaction quotient is defined as: $$Q = \frac{[H_2O]^2[Cl_2]^2}{[O_2][H^+]^4[Cl^-]^4}$$Expressed in this way, \(Q\) mirrors the balance of the concentration of products to reactants, following their stoichiometry in the balanced reaction. If Q is less than the equilibrium constant (\(K\)), the forward reaction is favored, and products are formed. Conversely, if Q is larger, the reaction moves in the reverse direction to form more reactants. It is integral in predicting the direction of the reaction and, importantly, the use of \(Q\) within the Nernst equation allows for the calculation of a cell's potential under non-standard conditions, making \(Q\) a crucial parameter for understanding and predicting electrochemical behavior.

Standard cell potential (E°)

When we talk about standard cell potential (\(E^\text{o}\)), we're referring to the voltage that an electrochemical cell can provide under standard conditions (all reactants and products at 1M concentration, gases at 1 atm pressure, and the temperature at 25°C or 298 K). It's a measure of how forcefully electrons are pushed through the cell.

This value is intrinsic to each specific electrochemical cell and is a sum of the standard potentials of the anode (\(E^\text{o}_{anode}\)) and the cathode (\(E^\text{o}_{cathode}\)). Mathematically, it's represented as:$$E^\text{o} = E^\text{o}_{cathode} - E^\text{o}_{anode}$$Standard cell potential gives us a benchmark to compare different electrochemical cells and fundamentally understand their spontaneous direction (a positive \(E^\text{o}\) means a spontaneous reaction in the direction written). In the context of the exercise, calculating the amalgamated \(E^\text{o}\) for the overall reaction, before applying the Nernst equation, guided us to predict the feasibility of the voltaic cell's operation under the given conditions.