ChiefRunningPhist
Well-Known Member
There are some positive contributions being made here. Good work!
V = I·Ω
Ω = ρ·(L/A)
V = voltage in volts
I = current in amps
Ω = resistance in ohms
ρ = 1/EC
L = distance between electrodes
A = electrode surface area
EC = 1/ρ = [(submerged electrode Surface Area)/(distance between electrodes)]·Ω
As Ag concentration increases so does the EC. lf a voltage source were implemented, the reaction rate would increase with time. If a current source were utilized the reaction rate would maintain at a constant rate.
EDIT:
Temperature (& pH) also effects EC. Utilizing a current source mitigates this as well.
The electrical potential is dependent on the electrode spacing and surface area as well as the solution EC. The farther the electrode spacing, smaller the electrode surface area, lower the EC, the greater the voltage needed to pass a set current.That is a surprisingly high voltage difference. The actual electrochemistry can be driven by three volts or maybe less. Can you tell me what the benefit of the higher voltage is ? I may be missing something obvious for which my med-chem bias hasn’t set me up.
V = I·Ω
Ω = ρ·(L/A)
V = voltage in volts
I = current in amps
Ω = resistance in ohms
ρ = 1/EC
L = distance between electrodes
A = electrode surface area
EC = 1/ρ = [(submerged electrode Surface Area)/(distance between electrodes)]·Ω
As Ag concentration increases so does the EC. lf a voltage source were implemented, the reaction rate would increase with time. If a current source were utilized the reaction rate would maintain at a constant rate.
EDIT:
Temperature (& pH) also effects EC. Utilizing a current source mitigates this as well.
Last edited: