Question

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\\ \hline{M^{4+}+4 e^{-} \longrightarrow M} & {0.66} \\ {N^{3+}+3 e^{-} \longrightarrow N} & {0.39}\end{array}$$ What is the value of \(\Delta G^{\circ}\) and \(K\) for this cell?

Short answer

For the given galvanic cell with half-reactions \(M^{4+} + 4e^- \rightarrow M\) and \(N^{3+} + 3e^- \rightarrow N\), the overall balanced redox reaction is \(3M^{4+} + 12N^{3+} \rightarrow 3M + 4N\). With standard electrode potentials of 0.66V and 0.39V, the standard cell potential is \(0.27V\). This yields a standard Gibbs free energy change, \(\Delta G^{\circ}\), of \(-312000 J/mol\) and an equilibrium constant, K, of \(3.4 \times 10^{42}\).

Step-by-step solution

1. Determine the overall balanced redox reaction

To find the overall balanced redox reaction, we need to combine the given half-reactions while ensuring that the same number of electrons are gained and lost, hence we need to multiply the first half-reaction by 3 and the second half-reaction by 4 to balance the electrons: \(3(M^{4+} + 4e^- \rightarrow M) \\ 4(N^{3+} + 3e^- \rightarrow N) \) Now, we can sum up the half-reactions to obtain the full redox reaction: \(3M^{4+} + 12e^- + 12 N^{3+} \rightarrow 3M + 12e^- + 4N \) The electrons cancel out, and the balanced redox reaction is \(3M^{4+} + 12N^{3+} \rightarrow 3M + 4N \)

2. Calculate the standard cell potential

We will use the standard electrode potentials given in the problem (\(\color{blue}{\mathscr{E}_1^{\circ} = 0.66V }\) and \(\color{blue}{\mathscr{E}_2^{\circ} = 0.39V}\)) to calculate the standard cell potential: \[\mathscr{E}_{cell}^{\circ} = \mathscr{E}_{cathode}^{\circ} - \mathscr{E}_{anode}^{\circ}\] Since a spontaneous redox reaction has a positive cell potential (\(\mathscr{E}_{cell}^{\circ} > 0\)), we assign the half-cell with a higher standard electrode potential as the cathode and the other as the anode. \(\mathscr{E}_{cell}^{\circ} = 0.66V - 0.39V = \color{blue}{0.27V}\)

3. Calculate the Gibbs free energy change

We can calculate the standard Gibbs free energy change (\(\Delta G^{\circ}\)) using the standard cell potential (\(\mathscr{E}_{cell}^{\circ}\)) and the number of electrons transferred (n) in the following equation: \[\Delta G^{\circ} = -n F \mathscr{E}_{cell}^{\circ}\] where F is the Faraday constant, which is approximately 96485 C/mol of electrons. In this case, the number of electrons transferred is 12 (from the balanced redox reaction). Therefore, \(\Delta G^{\circ} = -(12)(96485C/mol)(0.27V)\) \(\Delta G^{\circ} = \color{blue}{-312000 J/mol}\)

4. Calculate the equilibrium constant (K)

We can find the equilibrium constant (K) using the standard Gibbs free energy change (\(\Delta G^{\circ}\)) and the equation: \[K = e^{-\frac{\Delta G^{\circ}}{RT}}\] where R is the universal gas constant, approximately 8.314 J/(mol K), and T is the temperature in Kelvin. Assuming a temperature of 298 K, \(K = e^{-\frac{-312000 J/mol}{(8.314 J/(mol K))(298 K)}}\) \(K = \color{blue}{3.4 \times 10^{42}}\) The equilibrium constant, K, for the cell is approximately 3.4 x 10^42, and the standard Gibbs free energy change is -312000 J/mol.

Key concepts

Redox Reaction

In a galvanic cell, the redox reaction is central to its function. Redox reactions involve the transfer of electrons between chemical species. Each redox reaction is composed of two half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained.
In our example, we have two half-reactions:

• Oxidation: \( M^{4+} + 4 e^{-} \rightarrow M \)
• Reduction: \( N^{3+} + 3 e^{-} \rightarrow N \)

To make sure electrons are balanced in the entire reaction, the half-reactions must be multiplied by integer factors so that equal numbers of electrons are transferred. In this exercise, the first reaction is multiplied by 3 and the second by 4, leading to a complete equation:\[ 3M^{4+} + 12N^{3+} \rightarrow 3M + 4N \]Understanding how electrons move in these reactions is crucial for analyzing galvanic cells.

Standard Cell Potential

The standard cell potential \( \mathscr{E}_{cell}^{\circ} \) is a measure of the voltage available from a galvanic cell under standard conditions. It is derived from the electrode potentials of the cathode and anode.
In this exercise, the standard electrode potentials given are:

• Cathode (\( \mathscr{E}_{cathode}^{\circ} = 0.66 \text{V} \))
• Anode (\( \mathscr{E}_{anode}^{\circ} = 0.39 \text{V}\))

To find the standard cell potential, subtract the anode potential from the cathode potential:\[ \mathscr{E}_{cell}^{\circ} = \mathscr{E}_{cathode}^{\circ} - \mathscr{E}_{anode}^{\circ} = 0.66 \text{V} - 0.39 \text{V} = 0.27 \text{V} \]
A positive standard cell potential indicates that the redox reaction is spontaneous, meaning it can proceed without external input.

Gibbs Free Energy

Gibbs free energy \( (\Delta G^{\circ}) \) is a critical concept in understanding the spontaneity of redox reactions. It is the energy difference that indicates whether a reaction will occur spontaneously. The relation between \( \Delta G^{\circ} \), standard cell potential \( \mathscr{E}_{cell}^{\circ} \), and the number of electrons transferred \( (n) \) is given by:\[ \Delta G^{\circ} = -nF\mathscr{E}_{cell}^{\circ} \]Here, \( F \) is the Faraday constant \( (96485 \, \text{C/mol}) \). In our example, 12 electrons are transferred, resulting in:\[ \Delta G^{\circ} = -(12)(96485 \text{C/mol})(0.27 \text{V}) = -312000 \text{J/mol} \]
A negative \( \Delta G^{\circ} \) indicates that the reaction is spontaneous, aligning with a positive standard cell potential.

Equilibrium Constant

The equilibrium constant \( (K) \) expresses the balance between reactants and products in a chemical equilibrium. For electrochemical reactions, it's related to \( \Delta G^{\circ} \) through:\[ K = e^{-\Delta G^{\circ} / RT} \]Where \( R \) is the universal gas constant \( (8.314 \, \text{J/(mol K)}) \) and \( T \) is the temperature in Kelvin. Assuming standard conditions at \( 298 \, \text{K} \), plug in the values from the exercise:\[ K = e^{-(-312000 \, \text{J/mol})/(8.314 \, \text{J/(mol K)} \times 298 \, \text{K})} = 3.4 \times 10^{42} \]
A large \( K \) value means the reaction heavily favors product formation, indicating an equilibrium position shifted towards the products, which is consistent with a spontaneous process.