In the process of electrolysis, electrical power is used to separate water into oxygen and hydrogen molecules via the reaction:
H2O --> H2 + ½O2
This is very much like running a hydrogen fuel cell in reverse. We assume that only the activation potential for the hydrogen reaction is non-negligible. All other potentials are negligible. Hence the relevant parameters are :
PO2 PH2 Temp j0(H2) α(H2)
1 atm 1 atm 350 K 0.10 A/cm2 0.50
What is the minimum voltage needed to drive this reaction at these conditions, in volts?
What is the current density in A/cm2 at a voltage of 1.5 V?
What area of the cell, in cm2, do we need in order to get a rate of H2 production of 1 mol/sec?
To determine the minimum voltage required to drive the electrolysis of water under these conditions, we'll need to calculate the potential difference using the Nernst equation and the given parameters for hydrogen.
The Nernst equation is given by:
E = Eº - (RT/nF) * ln(Q)
Where:
E is the cell potential
Eº is the standard cell potential (which is 0V for this reaction)
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (350K)
n is the number of electrons transferred in the reaction (2 for this reaction)
F is Faraday's constant (96485 C/mol)
Q is the reaction quotient, which can be calculated using the concentrations of the species involved in the reaction.
For the hydrogen reaction:
2H+ + 2e- --> H2
Since the activation potential for the hydrogen reaction is non-negligible, the overpotential (η) needs to be considered, which affects the cell potential. The overpotential can be calculated using the Tafel equation:
η = (RT/αF) * ln(j/j0)
Where:
η is the overpotential
R, T, and F have the same definitions as before
α is the transfer coefficient for the reaction (given as 0.50)
j0 is the exchange current density (given as 0.10 A/cm2)
j is the actual current density
Now let's calculate the minimum voltage needed to drive the reaction:
1. Calculate the reaction quotient Q:
Q = (PH2)^2 / (PO2)^(1/2) = (1 atm)^2 / (1 atm)^(1/2) = 1 atm
2. Calculate the overpotential η at the given current density of 0.10 A/cm2:
η = (RT/αF) * ln(j/j0) = (8.314 J/(mol·K) * 350 K) / (0.50 * 96485 C/mol) * ln(0.10 A/cm2 / 0.10 A/cm2) = 0 V
Since the overpotential is 0V, it does not contribute to the minimum voltage required. Therefore, the minimum voltage needed to drive the reaction at these conditions is 0V.
Now let's calculate the current density at a voltage of 1.5V:
1. Calculate the overpotential η at the given voltage of 1.5V:
η = (RT/αF) * ln(j/j0) = (8.314 J/(mol·K) * 350 K) / (0.50 * 96485 C/mol) * ln(j / 0.10 A/cm2) = 1.5 V
Solving for j:
ln(j / 0.10 A/cm2) = (1.5 V) * (0.50 * 96485 C/mol) / (8.314 J/(mol·K) * 350 K)
j = e^[(1.5 V) * (0.50 * 96485 C/mol) / (8.314 J/(mol·K) * 350 K)] * 0.10 A/cm2 ≈ 0.447 A/cm2
Therefore, the current density at a voltage of 1.5V is approximately 0.447 A/cm2.
To calculate the area of the cell needed to produce 1 mol/sec of H2, we'll need to use Faraday's law.
Faraday's law states that the amount of substance produced or consumed in an electrolytic reaction is directly proportional to the amount of charge passed through the cell. The equation is given by:
n = (it) / (F * A)
Where:
n is the number of moles of H2 produced per unit time (1 mol/sec)
i is the current passing through the cell (in amperes)
t is the time (in seconds)
F is Faraday's constant (96485 C/mol)
A is the area of the cell (in cm2)
Rearranging the equation to solve for A:
A = (it) / (F * n)
Substituting the given values:
A = (1 A * 1 s) / (96485 C/mol * 1 mol/sec) ≈ 1.04 x 10^-5 cm2
Therefore, the area of the cell needed to produce 1 mol/sec of H2 is approximately 1.04 x 10^-5 cm2.