Chemisry question

Question

Consider a galvanic cell based on the half reactions:

𝐹𝑒2 + (𝑎𝑞) + 2𝑒 ― →𝐹𝑒 (𝑠) 𝐸°

𝑟𝑒𝑑 = ―0.447 𝑉

2𝐻 + (𝑎𝑞) + 2𝑒 ― →𝐻2 (𝑔) 𝐸°

𝑟𝑒𝑑 = 0.000 𝑉

The iron compartment contains an iron electrode in a solution of Fe2+ with a concentration

of 1.00 × 10–2 M. The hydrogen compartment contains a platinum electrode where the

partial pressure of H2 is 1.00 atm and the concentration of H+ is supplied by a weak acid

that has an initial concentration of 1.00 M. The measured cell potential is 0.269 V at 25 °C.

What is the standard cell potential of this galvanic cell?

6. What is the concentration of H+ in solution in the hydrogen compartment?

7. What is the value for Ka for the weak acid in the hydrogen compartment?

8. What is the standard free energy in kJ for the acid dissociation reaction that is

occurring in the hydrogen compartment?

9. Iron solid adopts a body-centered cubic structure with a unit cell edge length of 287

pm. Based on this information, calculate the density in g/mL of iron.

10. Below is the phase diagram for water. Draw or describe how this phase diagram will

change based on the addition of a weak acid. (Note: This is not really how phase

diagrams of mixtures work, but we’re going to using this to map two ideas together

(phase diagram is of water)

Answer 1

1. Standard Cell Potential

The standard cell potential is calculated as:

E°cell = E°cathode - E°anode

E°cell = 0.000 V - (-0.447 V) = 0.447 V

2. Concentration of H+ in the Hydrogen Compartment

Using the Nernst equation:

Ecell = E°cell - (0.0592/n) * log(Q)

Q = [Fe2+] / ([H+]^2 * PH2)

Rearranging for [H+]:

[H+]^2 = [Fe2+] * PH2 / 10^((E°cell - Ecell) / (0.0592 / n))

Substitute values:

[H+]^2 = (1.00 × 10^-2) * (1.00) / 10^((0.447 - 0.269) / (0.0592 / 2))

[H+]^2 = (1.00 × 10^-2) / 31.62 = 3.16 × 10^-4

[H+] = sqrt(3.16 × 10^-4) = 1.78 × 10^-2 M

3. Ka for the Weak Acid

Ka = [H+][A-] / [HA]

Assume [H+] ≈ [A-] and [HA] ≈ 1.00 - [H+]:

Ka = (1.78 × 10^-2)(1.78 × 10^-2) / (1.00 - 1.78 × 10^-2)

Ka = 3.17 × 10^-4 / 0.982 = 3.23 × 10^-4

4. Standard Free Energy

ΔG° = -RT ln(Ka)

R = 8.314 J/mol·K, T = 298 K, Ka = 3.23 × 10^-4

ΔG° = -(8.314)(298) * ln(3.23 × 10^-4)

ΔG° = -(8.314)(298)(-7.038) = 17,451 J/mol = 17.45 kJ/mol

5. Density of Iron

Iron adopts a body-centered cubic (BCC) structure with 2 atoms per unit cell.

Mass of 2 Fe atoms:

Mass = 2 * (atomic mass of Fe / Avogadro’s number)

Mass = 2 * (55.845 / 6.022 × 10^23) = 1.85 × 10^-22 g

Volume of unit cell:

Volume = a^3 = (287 pm)^3 = (287 × 10^-10 cm)^3 = 2.37 × 10^-23 cm^3

Density:

Density = Mass / Volume = 1.85 × 10^-22 g / 2.37 × 10^-23 cm^3 = 7.81 g/cm^3

6. Phase Diagram for Water with Weak Acid

Adding a weak acid lowers the freezing point and raises the boiling point due to colligative properties. This broadens the liquid-phase region, as the presence of the acid particles affects the thermodynamic behavior of water. The triple point remains relatively stable, but phase boundaries for solid-liquid and liquid-vapor shift slightly.