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 $$2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow{2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s)} \quad K=2.79 \times 10^{7}$$ 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 this galvanic cell is found using the Nernst Equation: \(\mathscr{E} = \mathscr{E}^{\circ} - \cfrac{RT}{nF} \ln K\). Balancing the redox reaction, the number of electrons transferred, \(n\), is 4. Plugging in the given values and constants, we find \(\mathscr{E} \approx 0.721\, \mathrm{V}\). To calculate the change in Gibbs free energy, \(\Delta G\), use the relationship \(\Delta G = -nF\mathscr{E}\), resulting in \(\Delta G \approx -278\, \mathrm{kJ\, mol^{-1}}\).

Step-by-step solution

Determine the standard potential for the cell reaction

Use the Nernst Equation: \(\mathscr{E} = \mathscr{E}^{\circ} - \cfrac{RT}{nF} \ln K\), where \(\mathscr{E}\) is the cell potential, \(\mathscr{E}^{\circ}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant, and \(K\) is the equilibrium constant. We are given \(K\), and we can find \(n\) by balancing the redox reaction. Then, we can solve for \(\mathscr{E}^{\circ}\). Step 2: Balance the redox reaction and find the number of electrons transferred (n)

Balance the redox reaction

The overall cell reaction is given as \(2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow{2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s)}\). Here, two electrons are transferred from each \(Cr^{2+}\) ion (total of 4 electrons) to the \(Co^{2+}\) ion. So, \(n = 4\). Step 3: Calculate the cell potential, \(\mathscr{E}\)

Calculate the cell potential using the Nernst Equation

Use the Nernst Equation: \(\mathscr{E} = \mathscr{E}^{\circ} - \cfrac{RT}{nF} \ln K\). We now have the values for all variables except \(\mathscr{E}^{\circ}\): \(T = 25^{\circ} \mathrm{C} + 273.15 = 298.15\, \mathrm{K}\) (convert Celsius to Kelvin), \(n = 4\), \(K = 2.79 \times 10^{7}\), \(R = 8.314\, \mathrm{J\, mol^{-1}K^{-1}}\), and \(F = 96485\, \mathrm{C\, mol^{-1}}\). Plug these values into the Nernst Equation and solve for \(\mathscr{E}\). Step 4: Calculate \(\Delta G\)

Calculate \(\Delta G\) using the relationship between \(\mathscr{E}\) and \(\Delta G\)

Use the relationship: \(\Delta G = -nF\mathscr{E}\), where \(\Delta G\) is the change in Gibbs free energy, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant, and \(\mathscr{E}\) is the cell potential. Plug in the values for \(n\), \(F\), and \(\mathscr{E}\) calculated in previous steps, and solve for \(\Delta G\).

Key concepts

Galvanic Cell

A galvanic cell, also known as a voltaic cell, is an electrochemical cell in which a spontaneous chemical reaction generates electrical energy. This process involves two half-reactions that occur at separate electrodes. These electrodes, usually metal or conductive materials, are immersed in solutions containing ions that participate in the reactions.

In galvanic cells, oxidation and reduction reactions occur in different compartments called half-cells. Oxidation (loss of electrons) happens at the anode, whereas reduction (gain of electrons) occurs at the cathode. An external wire connects these electrodes, allowing electrons released at the anode to travel to the cathode, generating electric current. Moreover, a salt bridge or porous membrane often links the solutions of the two half-cells, maintaining electrical neutrality by allowing ions to move between solutions.

• Anode: Negative electrode where oxidation occurs.
• Cathode: Positive electrode where reduction happens.
• Salt Bridge: Allows the flow of ions to maintain charge balance.

Understanding the basic setup of a galvanic cell helps in analyzing the cell's overall potential and reactivity, essential topics in electrochemistry.

Nernst Equation

The Nernst Equation is a crucial tool in electrochemistry that relates the cell potential to the concentrations of the reacting species. It helps us determine the cell potential (\(\mathscr{E}\)) under non-standard conditions.

The equation is expressed as follows: \[\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln Q\]Where:

• \(\mathscr{E}^{\circ}\) is the standard cell potential.
• \(R\) is the gas constant (8.314 J mol^-1K^-1).
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of electrons transferred in the reaction.
• \(F\) is the Faraday constant (96485 C mol^-1).
• \(Q\) is the reaction quotient reflecting the concentration ratio of products to reactants.

By adjusting the values of these variables, you can determine how different conditions affect the potential of a cell. This predictability makes the Nernst Equation very valuable for predicting the direction and feasibility of electrochemical reactions.

Cell Potential Calculation

Cell potential calculation is a vital skill in electrochemistry as it helps determine how much voltage an electrochemical cell can produce. The cell potential (\(\mathscr{E}\)) is influenced by the standard cell potential (\(\mathscr{E}^{\circ}\)), number of electrons transferred, temperature, and the concentrations of ionic species involved in the reactions.

To calculate the cell potential, you typically use the standard reduction potentials of the half-reactions. For the given reaction:

• Balance the redox reactions to find the number of electrons transferred (\(n\)).
• Use the Nernst Equation to find \(\mathscr{E}\) by inputting the known values, including the equilibrium constant \(K\), resulting in:\[\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln K\]

The careful application of these steps will enable the precise calculation of the cell potential, informing us about the spontaneity and efficiency of the electrochemical reactions taking place.

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) is a thermodynamic property that helps determine the spontaneity of a reaction. It provides insight into whether a chemical process can occur without outside intervention.

In electrochemistry, the relationship between Gibbs Free Energy and the cell potential is given by the formula:\[\Delta G = -nF\mathscr{E}\]Where:

• \(\Delta G\) is Gibbs Free Energy in joules.
• \(n\) is the number of electrons transferred.
• \(F\) is the Faraday constant.
• \(\mathscr{E}\) is the cell potential.

A negative \(\Delta G\) indicates that the reaction is spontaneous, while a positive value suggests it is non-spontaneous under the given conditions. Understanding this concept is crucial, as it bridges the connection between thermodynamics and electrochemistry, showing how energy transformations govern chemical reactions and processes.