ussername
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I'm trying to understand the concept of Butler-Volmer equation and its kinetic derivation. What I don't know and didn't find it anywhere is related to the charge transfer coefficient.
Let's have a reaction coordinate during electrode reaction with a transfer of electrons:
Can anybody explain why the derivation of free activation energy of oxidation ##\Delta G*_{ox}## with respect to the electrode potential ##E## is:
$$\left( \frac{\partial \Delta G*_{ox}}{\partial E} \right)_{T,p,E_{eq}}=-F\cdot \alpha_{ox}$$
where ##\alpha_{ox}## is the charge transfer coefficient of oxidation - dimensionless number with value from 0 to 1.
Why is there Faraday constant and not any other number?
The explanation could be related to the change of free energy during transport of electrons through the electrode potential - it is ##\Delta G_{m}=-F\cdot E## for reversible case and ##\Delta G_{m}>-F\cdot E## for irreversible case (with heat dissipation). In that case ##\alpha_{ox}=1## would stand for reversible charge transport and ##\alpha_{ox}=0## would stand for totally irreversible charge transport.
Note: I know there are more definitions of charge transfer coefficient but please let's work with this difinition:
$$\alpha_{ox}=-\frac{1}{F}\cdot \left (\frac{\partial \Delta G*_{ox}}{\partial E} \right )_{T,p,E_{eq}}$$ $$\alpha_{red}=\frac{1}{F}\cdot \left (\frac{\partial \Delta G*_{red}}{\partial E} \right )_{T,p,E_{eq}}$$
Let's have a reaction coordinate during electrode reaction with a transfer of electrons:
Can anybody explain why the derivation of free activation energy of oxidation ##\Delta G*_{ox}## with respect to the electrode potential ##E## is:
$$\left( \frac{\partial \Delta G*_{ox}}{\partial E} \right)_{T,p,E_{eq}}=-F\cdot \alpha_{ox}$$
where ##\alpha_{ox}## is the charge transfer coefficient of oxidation - dimensionless number with value from 0 to 1.
Why is there Faraday constant and not any other number?
The explanation could be related to the change of free energy during transport of electrons through the electrode potential - it is ##\Delta G_{m}=-F\cdot E## for reversible case and ##\Delta G_{m}>-F\cdot E## for irreversible case (with heat dissipation). In that case ##\alpha_{ox}=1## would stand for reversible charge transport and ##\alpha_{ox}=0## would stand for totally irreversible charge transport.
Note: I know there are more definitions of charge transfer coefficient but please let's work with this difinition:
$$\alpha_{ox}=-\frac{1}{F}\cdot \left (\frac{\partial \Delta G*_{ox}}{\partial E} \right )_{T,p,E_{eq}}$$ $$\alpha_{red}=\frac{1}{F}\cdot \left (\frac{\partial \Delta G*_{red}}{\partial E} \right )_{T,p,E_{eq}}$$