Chemistry problems help

Question

Consider a galvanic cell based on the half reactions:

Fe2+ (aq) + 2e^- -> Fe (s) Ered = - 0.447 V

2H+ (aq) + 2e^- -> H2 (g) Ered = 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.

1. If the cell runs at a current of 1.25 A, how many minutes will it take to capture 1.79 × 10^-2 moles of hydrogen gas?

2. If you trapped that gas at 10 °C, and you wanted the pressure to be as close to 0.500 atm as possible. What size round-bottom flask in mL would you need? Consider that these flasks tend to come in increments of 25.0 mL.

3. Write the balanced electrochemical reaction that is occurring in this system. Include phases and states of all species.

4. Predict the sign of the standard entropy of this reaction. Explain your reasoning.

Answer 1

(1) 1.79x10^-2 mols H₂ x 2 mols e-/mol H₂ x 96500 C/mol e- = 3455C

3455 C x 1 sec / 1.25 C = 2764 sec x 1 min / 60 sec = 46.1 minutes

(2) PV = nRT

P = 0.500 atm; V = ?; n = 1.79x10^-2; R = 0.0821; T = 283K

V = nRT/P = (1.79x10^-2)(0.0821)(283)/0.5

V = 0.83 L = ~830 mls … use a 900 ml flask if available (or 1 L)

(3) Fe(s) + 2H⁺(aq) ==> Fe²⁺(aq) + H₂(g)

(4) Sign of entropy change (∆S) would be positive since solid Fe is going to aqueous Fe (+∆S) and aqueous H⁺ is going to gas H₂ (+∆S).