Druyd:Knowledge

Nernst Potential

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If a potential is applied to a membrane this will lead to a polarization in which the positive charges move to the negative plate and the negative charges to the positive plate. The difference in concentration however leads to a diffusion that tempts to match the distribution. Nernst's potential is the limit potential over which the applied potential exceeds the tendency to diffuse by polarizing the membrane. For potentials smaller than Nernst's potential, diffusion tends to depolarize the membrane.

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ID:(820, 0)

Ohm's law with Conductivity

Equation

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If a potential difference dV of a long conductor dx and section S with a resistivity \rho_e is considered you have with Ohm's law that the current is

I = \displaystyle\frac{S}{\rho_e dx}dV

so with

j=\displaystyle\frac{I}{S}

y

\kappa_e=\displaystyle\frac{1}{\rho_e}

with what

$j =- \kappa \displaystyle\frac{ dV }{ dx }$

ID:(3877, 0)

Conductivity

Equation

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In the case of ion conduction, conductivity must include the sign of the charge, which is entered with the number of charges z divided by the absolute value of said number \mid z \mid. Therefore the conductivity is

$\kappa =\displaystyle\frac{ z }{ \mid z \mid } \mu_e c$

where \mu_e is mobility and c the concentration of ions.

ID:(3876, 0)

Current density

Equation

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The flow density j is understood as the current I by section S, so

$j =\displaystyle\frac{ I }{ S }$

ID:(3221, 0)

Nernst Current

Equation

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The electron current is the dQ charge that passes through a S section in a dt time. If it is assumed that electrons or ions travel at a speed v the volume of these that will pass in time dt through the section S is the same to Svdt. If, on the other hand, the ion concentration is c and its charge is q the current will be

I=\displaystyle\frac{dQ}{dt}=\displaystyle\frac{Svdtc}{dt}=Svc

that is

equation/druyd>

ID:(3222, 0)

Fick's Law for Charged Particles

Equation

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The diffusion leads to the difference in concentrations dc over a distance dx generates a flow of particles j that is calculated by the so-called Fick's law :

$j =- D \displaystyle\frac{ dc }{ dx }$

where D is the diffusion constant.

ID:(3878, 0)

Diffusion Constant for Charged Particles

Equation

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The diffusion constant D was modeled by Einstien and depends on the absolute value of the number of charges \mid z \mid, the mobility \mu_e, the universal gas constant, T the absolute temperature and F the Faraday constant that has a value of 9.649E+4 C/mol:

$D =\displaystyle\frac{ \mu_e R T }{\mid z \mid F }$

ID:(3879, 0)

Concentration of Charges

Equation

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If there is more than one type of ion, the actual concentration of the ions must be estimated, that is, add the concentrations weighted by the number of charges they have

$c_m=\sum_i\mid z_i\mid c_i$

where R is the gas constant, T the temperature, z the number of charges, F the constant Farday and the concentrations between both sides of the membrane c_1 and c_2.

ID:(3883, 0)

Concentration of Charges (1)

Equation

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In case of a type of load

$c_m =\mid z_1 \mid c_1$

where R is the gas constant, T the temperature, z the number of charges, F the constant of Farday and the concentrations between both sides of the c_1 membrane.

ID:(3884, 0)

Concentration of Charges (2)

Equation

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In case of two types of charges

$c_m = \mid z_1\mid c_1 + \mid z_2\mid c_2$

where R is the gas constant, T the temperature, z the number of charges, F the constant Farday and the concentrations between both sides of the membrane c_1 and c_2.

ID:(3885, 0)

Concentration of Charges (3)

Equation

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In case of three types of charges

$c_m = \mid z_1\mid c_1 + \mid z_2\mid c_2 + \mid z_3\mid c_3$

where R is the gas constant, T the temperature, z the number of charges, F the constant Farday and the concentrations between both sides of the membrane c_1, c_2 and c_3.

ID:(3886, 0)

Equilibrium Condition

Equation

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The equilibrium condition occurs when the flow due to the potential difference is equal to the flow due to the diffusion. That is why you have to

-\displaystyle\frac{z\mu_ec}{\mid z\mid}\displaystyle\frac{dV}{dx}=-\displaystyle\frac{\mu_eRT}{\mid z\mid F}\displaystyle\frac{dc}{dx}

for what you have

$dV =\displaystyle\frac{ R T }{ z F }\displaystyle\frac{ dc }{ c }$

ID:(3880, 0)

Nernst Potential

Equation

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If the potential difference is integrated, the relationship of the potential difference corresponding to the limit in which the electric field is compensated with the Diffusion can be established:

$V_m =-\displaystyle\frac{ R T }{ F }\ln\displaystyle\frac{ c_1 }{ c_2 }$

where R is the gas constant, T the temperature, z the number of charges, F the constant Farday and the concentrations between both sides of the membrane c_1 and c_2.

ID:(3881, 0)