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.

5.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?

Answer 1

Answers for parts 1-4 were attempted in another of your posts. Please refer to those.

(5) Standard Cell potential = 0.000 V - (-0.447V)

Eº_cell = +0.447 V

(6) Nernst equation:

E_cell = Eº_cell - RT/nF ln Q and at 25ºC and using log₁₀ we have..

E_cell = Eº_cell - 0.0592/n log Q

0.269 = 0.447 - 0.0592/2 log Q

Solve for Q which is = [Fe²⁺] / [H⁺]

Solve for [H⁺]

(7) HA <==> H⁺ + A^-

Ka = [H⁺][A^-]/[HA]

Plug in value for [H⁺] as determined in #6 which is also [A^-]. Use 1.00 for [HA] and solve for Ka

(8) ∆Gº = -RT ln Q

∆Gº = -(8.314 J/Kmol)(298K) ln Q

Plug in value of Q determined in #6 and solve for ∆Gº

Answer 2

I answered 2 questions for more contact me [email protected]. 1. To solve for the time it takes to capture 1.79 × 10^-2 moles of hydrogen gas, we first need to calculate the amount of charge that flows through the circuit using Faraday's law: Q = nF

where Q is the charge in Coulombs, n is the number of moles of electrons transferred in the reaction, and F is the Faraday constant (96,485 C/mol).

Since each mole of H2 gas produced corresponds to the transfer of 2 moles of electrons in the half-reaction, the number of moles of electrons transferred is:

n = (1.79 × 10^-2 mol H2) x (2 mol e^- / 1 mol H2) = 3.58 × 10^-2 mol e^-

The charge required to transfer this amount of electrons is then:

Q = (3.58 × 10^-2 mol e^-) x (96,485 C/mol) = 3.46 × 10^3 C

The current (I) is given as 1.25 A, so we can use the equation:

Q = It

where t is the time in seconds. Rearranging, we get:

t = Q / I

Substituting the values, we get: t = (3.46 × 10^3 C) / (1.25 A) = 2.77 × 10^3 s

Converting to minutes, we get:

t = 2.77 × 10^3 s x (1 min / 60 s) ≈ 46.2 min

Therefore, it would take approximately 46.2 minutes to capture 1.79 × 10^-2 moles of hydrogen gas.

2. To calculate the size of the round-bottom flask needed to trap the hydrogen gas at a pressure of 0.500 atm, we can use the ideal gas law:

PV = nRT

where P is the pressure in atm, V is the volume in liters, n is the number of moles of gas, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

We know that the pressure of the hydrogen gas is 1.00 atm at 25°C, so we need to calculate the volume of the container required to reduce the pressure to 0.500 atm at 10°C. Converting the temperatures to Kelvin, we have:

P1 = 1.00 atm

P2 = 0.500 atm

T1 = 25°C + 273.15 = 298.15 K

T2 = 10°C + 273.15 = 283.15 K

The number of moles of hydrogen gas produced is equal to the number of moles of electrons transferred in the half-reaction, which we calculated in part 1 to be 1.79 × 10^-2 mol H2.

Substituting the values into the ideal gas law, we get:



V1 = (nRT1) / P1 = [(1.79 × 10^-2 mol) x (0.0821 L·atm/mol·K) x (298.15 K)] / (1.00 atm) ≈ 0.41 L



V2 = (nRT2) / P2 = [(1.79 × 10^-2 mol) x (0.0821 L·atm/mol·K) x (283.15 K)] / (0.500 atm) ≈ 0.68 L



The difference in volume required to reduce the pressure from 1.00 atm to 0.500 atm is