Question

Calculate the standard potential of the cell Pt(s) | cysteine(aq), cystine(aq) | | H+(aq | O2(g) | Pt(s) and the Gibbs standard energy and the enthalpy of reaction of the cell at 25°C

- estimate the ΔrGØ value at 35°C, use EØ= -0.34V for the cystine/cysteine ​​pair

Answer 1

The balanced cell reaction for the given cell schematic is H₂(g) + 2 Br¯(aq) → 2 H+ (aq) + 2 Br₂(aq). The standard cell potential (E°) is -1.07 V, indicating that the reaction is nonspontaneous under standard state conditions.

Step-by-step explanation:

The given cell schematic can be balanced and written as:

Pt(s) | H₂(g) | H+ (aq) || Br₂(aq), Br¯(aq) | Pt(s)

The balanced cell reaction is:

H₂(g) + 2 Br¯(aq) → 2 H+ (aq) + 2 Br₂(aq)

To calculate the standard cell potential (E°), we need the individual standard reduction potentials (E°) for each half-cell:

Standard Reduction Potential (E°) for H₂/H+ half-cell: 0 V

Standard Reduction Potential (E°) for Br₂/Br¯ half-cell: +1.07 V

Now, we can calculate the standard cell potential (E°) using the difference in the reduction potentials:

E°cell = E°cathode - E°anode = 0 V - (+1.07 V) = -1.07 V

Since the standard cell potential (E°) is negative, the reaction is nonspontaneous under standard state conditions.

Answer 2

The estimated standard Gibbs energy change at 35°C is approximately -6.16 × 10^5 J/mol.

To calculate the standard cell potential (E°), standard Gibbs energy change (ΔrG°), and standard enthalpy change (ΔrH°) of the given cell reaction, you can use the Nernst equation and thermodynamic relationships.

The given cell reaction is:

Pt(s) | cysteine(aq), cystine(aq) | H+(aq) | O2(g) | Pt(s)

First, write the balanced half-reactions and their standard reduction potentials (E°) at 25°C:

1. Cathode half-reaction (reduction at the cathode):

O2(g) + 4H+(aq) + 4e^- → 2H2O(l) (E° = 1.23 V)

2. Anode half-reaction (oxidation at the anode):

Cystine (Cys-Cys) → 2Cysteine (2Cys) + 2e^- (E° = -0.34 V)

Now, apply the Nernst equation to calculate the standard cell potential (E° cell):

E° cell = E° cathode - E° anode

E° cell = (1.23 V) - (-0.34 V)

E° cell = 1.57 V

Next, calculate the standard Gibbs energy change (ΔrG°) using the relationship:

ΔrG° = -nFE° cell

Where:

- n is the number of moles of electrons transferred (equal to 4 in this case since 4 electrons are transferred in the balanced cell reaction).

- F is the Faraday constant (96,485 C/mol).

ΔrG° = -(4 mol e^-) × (96,485 C/mol) × (1.57 V)

ΔrG° ≈ -6.16 × 10^5 J/mol (rounded to three significant figures)

Now, you want to estimate the ΔrG° value at 35°C. You can use the Gibbs-Helmholtz equation for this:

ΔrG°(T2) = ΔrG°(T1) - ΔrH°(T1) × (T2 - T1) / T1

Where:

- ΔrG°(T2) is the standard Gibbs energy change at the new temperature (35°C).

- ΔrG°(T1) is the standard Gibbs energy change at the initial temperature (25°C).

- ΔrH°(T1) is the standard enthalpy change at the initial temperature (25°C).

- T2 and T1 are the new and initial temperatures in Kelvin, respectively.

Given that ΔrG°(25°C) is -6.16 × 10^5 J/mol, ΔrH°(25°C) can be estimated from the enthalpy change of the half-reaction:

ΔrH°(25°C) = -nFE°(25°C) = -(4 mol e^-) × (96,485 C/mol) × (1.57 V)

Now, plug the values into the equation:

ΔrG°(35°C) = -6.16 × 10^5 J/mol - [-(4 mol e^-) × (96,485 C/mol) × (1.57 V)] × (35°C - 25°C) / (25°C + 273.15 K)

ΔrG°(35°C) ≈ -6.16 × 10^5 J/mol - (-0.00731 J/mol) × 10°C / (25°C + 273.15 K)

ΔrG°(35°C) ≈ -6.16 × 10^5 J/mol + (-0.00731 J/mol) × 10°C / (298.15 K)

ΔrG°(35°C) ≈ -6.16 × 10^5 J/mol + (-0.00731 J/mol) × 0.0336

ΔrG°(35°C) ≈ -6.16 × 10^5 J/mol - 2.46 J/mol

ΔrG°(35°C) ≈ -6.16 × 10^5 J/mol (rounded to two significant figures)

So, the estimated standard Gibbs energy change at 35°C is approximately -6.16 × 10^5 J/mol. Note that the estimation of ΔrH°(25°C) was used in this calculation, and the result was rounded to two significant figures for simplicity.