Multiple rate-determining steps and rate expression

Question

I am trying to understand the concept of the rate laws governing hydrogen evolution of platinum. More specifically, I am trying to understand the multiple occurrence of a rds to the overall rate law of a reaction.

In the lecture, we talked about how there are three possible steps; the Volmer-step, the Heyrovsky step and the Tafel step:

For simplicity, lets talk about the Volmer-Tafel mechanism. We suppose the Volmer step to be very slow and rate-determining, and the Tafel step to be really fast; therefore, the Volmer reaction has to happen twice in order for the Tafel step to take place.

My professor derived the following rate expression:

This rate expression seems to agree with literature. However, I am not able to understand how neither the reaction order in protons nor the "Tafel slope" (exponential term, similar to Arrhenius) is not affected by the fact that the rate-determining step has to happen twice. I guess this is also a question in more general terms.

Why would multiple occurrence of an RDS not alter the rate expression?

Answer 1

In case I you only consider the electrochemical hydrogen reduction and its inverse reaction

$$ \ce{H^+ + M + e <=>[k_1][k_{-1}] MH} \hspace{3ex}[1] $$

The rate of reaction [1] is given by

$$ \ce{ v_1 = k_1 c_{H^+} (1 -\theta_H)} $$

where $\ce{c_{H^+}}$ is the surface concentration of hydrogen, and $\theta_H$ is the surface coverage of adsorbed hydrogen.

Since reaction [1] is an electron tranfer reaction on an electrode surface, it rate depends on the electrode potential (E) - or overpotential ($\ce{\eta\,=\,E -E_0}$). If the oxidation step at the applied potential, $\ce{k_1(\eta)}$ is given by

$$ \rm{k_1 = k^0 e^{-\alpha f \eta} }$$

where $\rm{\alpha}$ is the symmetry factor and $\rm{f = RT/F}$. It is natural that this expression reminds you of the Arrhenius equation. Actually, you can interpret as describing the effect of the electrode potential of the height of the reaction barrier.

Thus, $$ \ce{ v_1 = k^0 e^{-\alpha f \eta} c_{H^+} (1 -\theta_H)} $$

For low coverage this expression simplifies to

$$ \ce{ v_1 \approx k^0 e^{-\alpha f \eta} c_{H^+} } $$

Since reaction [1] is the RDS, the overall rate depends on the rate of this step, and

$$ \ce{ v_{HER} = \frac{v_1}{\color{red}{2}} = \frac{k^0}{2} e^{-\alpha f \eta} c_{H^+}} $$

Guess the 2 factor is subsumed into the experimental $\ce{k^0_{exp}}$. Maybe I am missing something.