Calculating Standard Gibbs Energy Change

The question asks us to calculate the standard Gibbs energy change (ΔG°) for a given electrochemical reaction, using the provided standard cell potential (E°_cell) and temperature.

The reaction is:
\(2Au^{3+}_{(aq)} + 6Br^-_{(aq)} \rightarrow 2Au_{(s)} + 3Br_{2(l)}\)

The given standard cell potential is \(E^\circ_{cell} = 2.49 \text{ V}\) at 298 K.

The relationship between standard Gibbs energy change (ΔG°) and standard cell potential (E°_cell) is given by the equation:

\(\Delta G^\circ = -nFE^\circ_{cell}\)

Where:

• \(\Delta G^\circ\) is the standard Gibbs energy change.
• \(n\) is the number of moles of electrons transferred in the balanced reaction.
• \(F\) is the Faraday constant, approximately \(96485 \text{ C/mol}\).
• \(E^\circ_{cell}\) is the standard cell potential.

Determining the Number of Electrons Transferred (n)

To find the number of electrons transferred, we need to look at the oxidation and reduction half-reactions involved in the overall cell reaction:

Overall reaction: \(2Au^{3+}_{(aq)} + 6Br^-_{(aq)} \rightarrow 2Au_{(s)} + 3Br_{2(l)}\)

Let's identify the species being reduced and oxidized:

• Gold (Au) is reduced from an oxidation state of +3 in \(Au^{3+}\) to 0 in \(Au_{(s)}\).
• Bromine (Br) is oxidized from an oxidation state of -1 in \(Br^-\) to 0 in \(Br_2\).

Now, let's write the balanced half-reactions:

• Reduction half-reaction: \(Au^{3+}_{(aq)} + 3e^- \rightarrow Au_{(s)}\)
• Oxidation half-reaction: \(2Br^-_{(aq)} \rightarrow Br_{2(l)} + 2e^-\)

To get the overall balanced reaction, the number of electrons lost in oxidation must equal the number of electrons gained in reduction. We multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3:

• \(2 \times (Au^{3+}_{(aq)} + 3e^- \rightarrow Au_{(s)}) \implies 2Au^{3+}_{(aq)} + 6e^- \rightarrow 2Au_{(s)}\)
• \(3 \times (2Br^-_{(aq)} \rightarrow Br_{2(l)} + 2e^-) \implies 6Br^-_{(aq)} \rightarrow 3Br_{2(l)} + 6e^-\)

Adding the balanced half-reactions gives:

\(2Au^{3+}_{(aq)} + 6e^- + 6Br^-_{(aq)} \rightarrow 2Au_{(s)} + 3Br_{2(l)} + 6e^-\)

The electrons cancel out, resulting in the given overall reaction:

\(2Au^{3+}_{(aq)} + 6Br^-_{(aq)} \rightarrow 2Au_{(s)} + 3Br_{2(l)}\)

From the balanced half-reactions, we can see that the number of electrons transferred (\(n\)) is 6.

Calculating ΔG°

Now we can plug the values into the formula \(\Delta G^\circ = -nFE^\circ_{cell}\):

• \(n = 6 \text{ mol}\)
• \(F = 96485 \text{ C/mol}\)
• \(E^\circ_{cell} = 2.49 \text{ V}\)

\(\Delta G^\circ = -(6 \text{ mol}) \times (96485 \text{ C/mol}) \times (2.49 \text{ V})\)

Since \(1 \text{ V} = 1 \text{ J/C}\), the units will be Joules (J):

\(\Delta G^\circ = -(6 \times 96485 \times 2.49) \text{ J}\)

\(\Delta G^\circ = -(578910 \times 2.49) \text{ J}\)

\(\Delta G^\circ = -1441980.9 \text{ J}\)

Converting ΔG° to Kilojoules

The options are given in kilojoules (kJ). We need to convert Joules to Kilojoules using the conversion factor \(1 \text{ kJ} = 1000 \text{ J}\).

\(\Delta G^\circ = \frac{-1441980.9 \text{ J}}{1000 \text{ J/kJ}}\)

\(\Delta G^\circ = -1441.9809 \text{ kJ}\)

Rounding this value to two decimal places gives -1441.98 kJ, which is approximately -1442 kJ.

Comparing this result with the given options, we find that it is closest to -1442 kJ.

The standard Gibbs energy change for the reaction is approximately -1442 kJ.

Summary of Gibbs Energy Calculation Steps

Here is a summary of the steps taken to calculate the standard Gibbs energy change from standard cell potential:

1. Identify the balanced electrochemical reaction.
2. Break down the overall reaction into balanced oxidation and reduction half-reactions.
3. Determine the number of electrons (\(n\)) transferred in the balanced reaction. This is the number of electrons that cancel out when adding the half-reactions.
4. Use the formula \(\Delta G^\circ = -nFE^\circ_{cell}\) to calculate the standard Gibbs energy change.
5. Ensure units are consistent; typically, the result will be in Joules if E°_cell is in Volts and F is in C/mol.
6. Convert the result to the desired units, such as kilojoules, if necessary.

Revision Table: Key Concepts in Electrochemistry and Thermodynamics

Additional Information on Gibbs Energy and Cell Potential

The relationship \(\Delta G^\circ = -nFE^\circ_{cell}\) is a fundamental equation connecting thermodynamics (\(\Delta G^\circ\)) and electrochemistry (\(E^\circ_{cell}\)). It shows that the standard Gibbs energy change is directly proportional to the standard cell potential. The negative sign indicates that a spontaneous reaction (which has a negative \(\Delta G^\circ\)) will have a positive standard cell potential \(E^\circ_{cell}\), capable of doing electrical work.

The quantity \(nF\) represents the total charge transferred during the reaction per mole of reaction as written. The units of \(n\) are moles of electrons, and \(F\) is in Coulombs per mole of electrons, so \(nF\) is in Coulombs per mole of reaction.

The unit of potential, the Volt (V), is defined as 1 Joule per Coulomb (1 V = 1 J/C). Therefore, when multiplying \(nF\) (in Coulombs) by \(E^\circ_{cell}\) (in Volts or J/C), the result is in Joules (C × J/C = J), which is the unit of energy.

This equation holds true under standard conditions. For non-standard conditions, the Gibbs energy change (\(\Delta G\)) is related to the cell potential (\(E_{cell}\)) by \(\Delta G = -nFE_{cell}\). The non-standard cell potential \(E_{cell}\) can be calculated using the Nernst equation: \(E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q\), where R is the ideal gas constant, T is the temperature, and Q is the reaction quotient.