Gouy chapmann through of liberating of ions from a sheet parallel to the electrode. However, the free ions become exposed to the thermal forces from the particles of the solution. The behaviour of ions the vicinity of the electrode is affected by the electrical forces arising from the charge on the electrode and by thermal motion in solution. Thus in this model, the excess charge density on OHP is not equivalent to that on the metal, but is less. The excess charge density in the solution less with distance from the electrode.
The electrode has a sort of ionic atmosphere near the metal it's charge attracts the solvated ions to the second row. Further thermal forces have influence comparable to the forces from the electrode. Sufficiently into the solution, the net charge density is zero, because of positively (+) and negatively (-) ions are equal to any region.
with the ionic cloud on the electrode, the Gouy- chapmann mode resembles ion - ion interaction in solution. The excess charge density on OHP in smaller in magnitude that the on the metal. The remaining charge is distributed in solutions. The diffuse qm, qd placed at a distance k⁻¹ from x =0 plane.
An equation for the total diffuse charge density scattered in the solution under the influence of electrical and thermal forces.
Let us consider a lamina in the electrolyte ∣∣ᵉ to the electrode a distance 'x' from it. The charge density (ρχ) for the χ dimension in rectangular coordinates can obtained using the poisson equation.
ρχ = - ∈i d²ψ .......................(1)
4π dx²
Where ψ be the potential difference between the lamina and the bulk of the solution (ψ→0, x →∞)
The charge density in terms of Blotzmann-Distribution ,
ρχ = ∑ᵢηᵢ Ziᵉ = ∑ni zie₀ e⁻ zie₀ψx ......(2)
kT
Where ηi and ni are the concentration of the species 'i' in the lamina and in the bulk of the solution respectively.
Zi - valence of the species 'i'
e - electronic charge.
Zie₀ψχ - ratio of the electrical and
ΚΤ
thermal energies of a ion at the distance 'x' from the electrode.
From equation (1) and (2)
d²ψ = - 4π ∑iniZie₀ ψx /kT................ (3)
dx² ∈i
The solution for the ulinearised differential equation is to the found,
(ʐi = ʐ− = ʐ ) (ni = n⁰- = n⁰)
dψ = - [32πkTn⁰ ]¹/₂ sin ʐe₀ψx .........(4)
dx RT RT
dψ gives the field at distance x from the electrode according to diffuse charge of Gouy chapmann. Hence this equation give the relation between the electric field and potential at any distance x from the electrode.
The total diffuse charge (qd) in the solution in terms of potential :-
This can be written from electrostates. H is the charge contain in a closed surface (or) Gaussion box which is equal to ∈/4π times the area of the field normal to the surface.
q = ∈i dψ ....... (5)
4π dx (gauss's law from electrostatics) by knowing.. dy
dx
from equation (4) q can be calculated. This 'q' is the charge enclosed with in the closed volume at the surface of which the field is dψ
dx
When we consider qd, we have to take,
χ= 0 →∞ and consider electrode a point charge.
Now, qd = ∈i dψ ........(5)
4π dχ
qd = ∈i dψ − [ 32πn⁰kT ]¹/₂ sinh ʐeψ0
4π dχ ∈i 2kT
qd = - 2 [∈ikT]¹/₂ sinh ʐeψ0
2π 2kT
qd = - 2[∈in⁰kT]¹/₂ sinh ʐe₀ψχ .......(7)
2π 2kT
ψ₀ - potential at x =0 relative to the bulk of the solution where the potential is taken zero.
According to the above equation, the total diffuse charge varies with the to total potential charge varies with the total potential drop in the solution according to a hyperbolic -sine relation. Equation (7) shows that the total diffuse charge varies with potential drop the solution according to a hyperbolic - sine relation.
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