Question

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 \mathrm{M}), \mathrm{Cr}^{3+}(2.0 \mathrm{M})\right|\left|\mathrm{Co}^{2+}(0.20 \mathrm{M})\right| \mathrm{Co} $$ 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.

Short answer

The cell potential (\(\mathscr{E}\)) for the given galvanic cell is approximately 0.848V, and the Gibbs free energy change (\(\Delta G\)) is approximately -1.63 × 10^5 J/mol.

Step-by-step solution

Identify the half-cell reactions

Before we proceed, let's identify the half-cell reactions occurring at the anode and the cathode in the given galvanic cell. Observe that this cell has the following structure: $$\mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 \mathrm{M}), \mathrm{Cr}^{3+}(2.0 \mathrm{M})\right|\left|\mathrm{Co}^{2+}(0.20 \mathrm{M})\right| \mathrm{Co}$$ At the anode (left-side), we have Cr2+ being oxidized to Cr3+: \[ \mathrm{Cr}^{2+}(aq) \longrightarrow \mathrm{Cr}^{3+}(aq) + e^- \] At the cathode (right-side), we have Co2+ being reduced to Co: \[ \mathrm{Co}^{2+}(aq) + 2e^- \longrightarrow \mathrm{Co}(s) \] Now that we have our half-cell reactions, let us proceed.

Using the Nernst Equation

The Nernst Equation is given by: \[ E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q \] To use this equation, we need to find the values for \(E^\circ_{cell}\), n, and Q.

Calculate the standard cell potential, \(E^\circ_{cell}\)

We can find the standard cell potential by using the relationship between the equilibrium constant (K) and standard cell potential: \(E^\circ_{cell} =\frac{0.0592}{n} \log K\) We are given the equilibrium constant, K = 2.79 × 10^7, and the stoichiometry of the electrons exchanged in the overall reaction is 2, so n = 2. Now we can calculate the standard cell potential: \(E^\circ_{cell} =\frac{0.0592 \times \log (2.79 \times 10^7)}{2} = 0.902V\)

Calculate the reaction quotient (Q)

The reaction quotient, Q, is given by: \(Q = \frac{[\mathrm{Cr}^{3+}]^{2}}{[\mathrm{Cr}^{2+}][\mathrm{Co}^{2+}]}\) Using the given concentrations for Cr2+, Cr3+, and Co2+, we can calculate Q: \(Q = \frac{(2.0)^{2}}{(0.3)(0.2)} =\frac {4}{0.06} = 66.67\)

Calculate the cell potential, E_{cell}

Now we have all the values required to use the Nernst Equation. Let's substitute these values into the equation: \(E_{cell} = 0.902 - \frac{0.0592}{2} \log 66.67 = 0.902 - 0.0296 \times 1.823 = 0.902 - 0.054 = 0.848V\)

Calculate the Gibbs free energy change, ∆G

Now we will use the relationship between the cell potential and Gibbs free energy change: \(\Delta G = -nF E_{cell}\) Where n is the stoichiometry of the electrons exchanged (n = 2), F is the Faraday constant (F = 96,485 C/mol), and \(E_{cell}\) is the cell potential (0.848 V): \(\Delta G = -2 \times 96485 \times 0.848 = -163140.104 J/mol ≈ -1.63 \times 10^5 J/mol\) Thus, we have found the cell potential (\(\mathscr{E}\)) and Gibbs free energy change (\(\Delta G\)) for the given galvanic cell to be approximately 0.848V and -1.63 × 10^5 J/mol, respectively.

Key concepts

Nernst Equation

The Nernst Equation is crucial for understanding how the potential of a galvanic cell changes under different conditions. This equation helps calculate the cell's potential, which varies with the concentration of the ions involved. It's represented as follows: \(E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q\). Here, \(E^\circ_{cell}\) is the standard cell potential, \(n\) is the number of moles of electrons exchanged, and \(Q\) is the reaction quotient.

• \(E_{cell}\) gives the non-standard potential of the cell.
• \(\log Q\) adjusts this potential based on the concentrations of ions in solutions.

Using the Nernst Equation, we can see how changes in concentration can increase or decrease the cell's potential relative to the standard state. This concept is fundamental when analyzing cell behavior in real-world applications.

Half-cell Reactions

Half-cell reactions are what occur at each electrode of a galvanic cell and are essential for understanding the processes that produce electrical energy. In our example, we have two reactions: oxidation at the anode and reduction at the cathode.

• Anode: \(\text{Cr}^{2+}(aq) \rightarrow \text{Cr}^{3+}(aq) + e^-\) showing oxidation, where \(\text{Cr}^{2+}\) loses an electron.
• Cathode: \(\text{Co}^{2+}(aq) + 2e^- \rightarrow \text{Co}(s)\) indicating reduction, as \(\text{Co}^{2+}\) gains electrons to form solid cobalt.

These half-reactions need to be balanced, especially in terms of electrons, as electrons lost in the oxidation half-reaction must equal electrons gained in the reduction half-reaction. This ensures electron flow, driving the cell's energy production.

Standard Cell Potential

The standard cell potential, \(E^\circ_{cell}\), is a critical parameter that describes the voltage of a cell under standard conditions (1 M concentration, 1 atm pressure, and 25°C). It tells us the inherent ability of a cell to drive an electric current. The formula used is: \(E^\circ_{cell} = \frac{0.0592}{n} \log K\), where \(K\) is the equilibrium constant of the overall reaction. In our example, \(K = 2.79 \times 10^7\), and \(n = 2\), so the calculation gives us \(E^\circ_{cell} = 0.902 \text{ V}\). This potential indicates how much work the cell can perform when moving electrons from the anode to cathode. It's also an index for spontaneity of the reaction, where a more positive \(E^\circ_{cell}\) typically signifies a more spontaneous reaction.

Gibbs Free Energy

Gibbs free energy, \(\Delta G\), is a thermodynamic function that relates to the amount of energy available to do work. In the context of a galvanic cell, it's related to the cell potential by the equation: \(\Delta G = -nFE_{cell}\). Here, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant (96,485 C/mol), and \(E_{cell}\) is the cell potential. For our galvanic cell, this calculation yields \(\Delta G = -1.63 \times 10^5 \text{ J/mol}\). This negative \(\Delta G\) confirms that the reaction is spontaneous under the given conditions, meaning the cell can generate energy without external input. A more negative value of \(\Delta G\) generally indicates a higher capacity of the cell to perform electrical work.