Calculate ∂G and E_cell for the cell

Al/Al³⁺(0.01M) // Fe²⁺(0.02M)/Fe

E^oAl³⁺/Al = -1.60V

E^o_Fe²⁺_/Fe =0.44V.

To calculate the Gibbs free energy change (∂G) and the cell potential (E_cell) for the electrochemical cell involving aluminum and iron ions, we can follow a systematic approach. Let's break this down step by step.

Step 1: Determine the Standard Cell Potential (E°_cell)

The standard cell potential can be calculated using the standard reduction potentials of the half-reactions involved. The given standard reduction potentials are:

• E°_Al3+/Al = -1.60 V
• E°_Fe2+/Fe = 0.44 V

For the cell reaction, we need to reverse the aluminum half-reaction because aluminum is being oxidized. The oxidation potential for aluminum will be the negative of its reduction potential:

E°_cell = E°_cathode - E°_anode

Here, the cathode is where reduction occurs (Fe2+/Fe), and the anode is where oxidation occurs (Al/Al3+):

E°_cell = E°_Fe2+/Fe - E°_Al3+/Al

Substituting the values:

E°_cell = 0.44 V - (-1.60 V) = 0.44 V + 1.60 V = 2.04 V

Step 2: Calculate the Cell Potential (E_cell) Using the Nernst Equation

The Nernst equation allows us to calculate the cell potential under non-standard conditions:

E_cell = E°_cell - (RT/nF) * ln(Q)

Where:

• R = 8.314 J/(mol·K) (universal gas constant)
• T = temperature in Kelvin (assume 298 K for standard conditions)
• n = number of moles of electrons transferred (for Al to Al3+, n = 3; for Fe2+ to Fe, n = 2)
• F = 96485 C/mol (Faraday's constant)
• Q = reaction quotient

For the reaction:

2 Al + 3 Fe2+ → 2 Al3+ + 3 Fe

The reaction quotient Q can be calculated as:

Q = [Al3+]^2 / [Fe2+]^3

Given the concentrations:

[Al3+] = 0.01 M and [Fe2+] = 0.02 M, we can substitute these values:

Q = (0.01)^2 / (0.02)^3 = 0.0001 / 0.000008 = 12.5

Now, substituting the values into the Nernst equation:

E_cell = 2.04 V - (8.314 J/(mol·K) * 298 K / (6) * 96485 C/mol) * ln(12.5)

Calculating the term (RT/nF):

(8.314 * 298) / (6 * 96485) = 0.0005 V

Now, calculate ln(12.5):

ln(12.5) ≈ 2.5

Substituting this back into the equation:

E_cell = 2.04 V - (0.0005 * 2.5) = 2.04 V - 0.00125 V ≈ 2.03875 V

Step 3: Calculate the Gibbs Free Energy Change (∂G)

The Gibbs free energy change can be calculated using the formula:

∂G = -nFE_cell

Using n = 6 (the total number of electrons transferred in the balanced reaction), we can substitute the values:

∂G = -6 * 96485 C/mol * 2.03875 V

Calculating this gives:

∂G ≈ -6 * 96485 * 2.03875 ≈ -1177.5 kJ/mol

Summary of Results

To summarize, we have:

• Standard Cell Potential (E°_cell): 2.04 V
• Cell Potential (E_cell): 2.03875 V
• Gibbs Free Energy Change (∂G): -1177.5 kJ/mol

This indicates that the reaction is spontaneous under the given conditions, as evidenced by the negative Gibbs free energy change. If you have any further questions or need clarification on any part of this process, feel free to ask!