Problem 87 - Chemistry An Atoms First Approach

Question

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$\begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K=2.79 \times 10^{7} \end{aligned}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

Step-by-Step Solution

Answer

The cell potential (E) for this galvanic cell is approximately 0.851 V, and the Gibbs free energy change (ΔG) for the given reaction at these conditions is approximately -164000 J/mol.

Step 1: Identify the half-reactions

Since we have the overall reaction, let's determine the half-reactions as reduction: (i) For Cr: \(Cr^{3+}(aq) + e^{-} \rightarrow Cr^{2+}(aq)\) (ii) For Co: \(Co^{2+}(aq) + 2e^{-}\rightarrow Co(s)\)

Step 2: Write and calculate the Nernst equation for the cell potential

The Nernst equation is: \(\mathscr{E} = \mathscr{E}^{\circ} - \frac{0.05916}{n} \log Q\) Here, we need to find the standard cell potential, E°, which could be calculated by combining the standard reduction potentials for each half-cell. But since we are given the equilibrium constant, we can use the relationship between E° and K instead: \(\mathscr{E}^{\circ} = \frac{0.05916}{n} \log K\) We know that n = the number of electrons transferred, which is 2 for this reaction. And K = 2.79 × 10^7, so we can now find E°: \(\mathscr{E}^{\circ} = \frac{0.05916}{2} \log (2.79 \times 10^{7}) \approx 1.026\, V\) Now, we need to find the reaction quotient, Q. The reaction formula for Q is: \(Q = \frac{[\mathrm{Cr}^{3+}]^{2}}{[\mathrm{Cr}^{2+}]^{2}[\mathrm{Co}^{2+}}\) Given the concentrations: [Cr³⁺] = 2.0 M, [Cr²⁺] = 0.30 M, [Co²⁺] = 0.20 M, we calculate Q: \(Q = \frac{(2.0)^2}{(0.30)^2(0.20)} = 444.44\) Now, we can calculate E using the Nernst equation: \(\mathscr{E} = 1.026 - \frac{0.05916}{2} \log 444.44 \approx 0.851\, V\)

Step 3: Calculate ΔG using the cell potential

We can find the Gibbs free energy change (ΔG) using the formula: \(\Delta G = -nFE\) Where n is the number of electrons transferred (n = 2), F is the Faraday constant (F ≈ 96485 C/mol), and E is the cell potential we calculated in the previous step (E ≈ 0.851 V). Now, we can calculate ΔG: \(\Delta G = - (2)(96485 C/mol)(0.851 V) \approx -164000\, J\, mol^{-1}\) So, the cell potential (E) for this galvanic cell is approximately 0.851 V, and the Gibbs free energy change (ΔG) for the given reaction at these conditions is approximately -164000 J/mol.

Key Concepts

Nernst Equation

The Nernst equation is a fundamental concept in electrochemistry that provides a way to calculate the electric potential of a galvanic cell under non-standard conditions. It accounts for the effect of ion concentration on the cell potential. This makes the Nernst equation essential for understanding how batteries work, how cells sustain their membrane potential, and other applications where the ion concentration is crucial.

The general form of the Nernst equation is as follows:
\[\mathscr{E} = \mathscr{E}^{\circ} - \left(\frac{0.05916}{n}\right) \log Q\]
where \(\mathscr{E}\) is the cell potential under non-standard conditions, \(\mathscr{E}^{\circ}\) is the standard cell potential, \(n\) is the number of moles of electrons transferred in the electrochemical reaction, and \(Q\) is the reaction quotient reflecting the ratio of product and reactant activities.

In practice, the Nernst equation allows us to assess how cell potential will change as the concentration of reactants and products shifts, which helps us understand the behavior of electrochemical cells in a dynamic environment.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationships between electrical energy and chemical reactions. This field of study is crucial for the development of technology such as batteries, fuel cells, corrosion prevention, and electroplating. In the exercise involving the galvanic cell, we delve into the heart of electrochemistry, exploring how chemical energy is converted into electrical energy and vice versa.

The galvanic cell is a classic electrochemical setup in which a spontaneous redox reaction generates an electrical current. This process entails the flow of electrons from the anode (where oxidation occurs) to the cathode (where reduction takes place) through an external circuit. Understanding galvanic cells reveals the practical application of redox reactions and provides a real-world context for the thermodynamic principles at play in electrochemistry.

Gibbs Free Energy

Gibbs free energy (\(\Delta G\)) is a thermodynamic quantity that measures the maximum amount of work that can be done by a thermodynamic process at constant temperature and pressure. In the context of electrochemistry, the change in Gibbs free energy is directly related to the electrical work that can be harnessed from a galvanic cell.

The relationship between the Gibbs free energy change and cell potential is given by the equation:
\[\Delta G = -nFE\]
where \(\Delta G\) is the change in Gibbs free energy, \(n\) is the amount of substance (measured in moles of electrons transferred), \(F\) is the Faraday constant, and \(E\) is the cell potential. A negative value for \(\Delta G\) suggests that the reaction is spontaneous, driving the electrons through the circuit, while a positive value indicates a non-spontaneous reaction requiring an external source of energy to proceed.

Equilibrium Constant

The equilibrium constant (\(K\)) expresses the extent to which a chemical reaction occurs until reaching a state of equilibrium. In the context of the electrochemical cell, the equilibrium constant is related to the standard cell potential (\(\mathscr{E}^{\circ}\)) by the equation:
\[\mathscr{E}^{\circ} = \left(\frac{0.05916}{n}\right) \log K\]
In the given galvanic cell, we deal with a high equilibrium constant value, which indicates that the products of the redox reaction are heavily favored at equilibrium. This is characteristic of a reaction that will proceed forward to a significant extent under standard conditions, producing a strong driving force for the flow of electrons, and thus a higher cell potential.

Understanding the equilibrium constant in relation to electrochemical cells helps students predict the direction and extent of reactions, the likelihood of a cell to perform work, and how changes in concentration affect cell voltage and free energy.