Warning !

Soas is no longer maintained. You are strongly encouraged to switch to its successor, QSoas

sr21l, sr21n, so21l, so21n

This is useful if you believe that the high driving force residual slope in the voltammogram results from a distribution of the orientation of the enzyme molecules on the surface. The model described in ref. [13] can then be used to fit the data after removing a linear baseline extrapolated from the i=0 part of the voltammogram, see the reg command.

For oxidative catalysis, the current equations are:
\begin{subeqnarray}
\frac{i^\star}{i_{\rm lim}}&=
& \frac{1}{a^{\rm ox}} \left( ...
...l{eq.2eo.v2} \\
b_1^{\rm ox}&= & b_2^{\rm ox}
\exp(\beta d_0)
\end{subeqnarray}
(with $ e_{\rm X/Y}=\exp\left[ F/RT\times(E-E_{\rm X/Y}) \right]$) when the limiting current is reached (so2l) and

$\displaystyle \frac{i^\star}{i_{\rm lim}}\approx \frac{1} {\beta d_0  a^{\rm ox}}\ln\frac{a^{\rm ox}+b_2^{\rm ox}}{b_2^{\rm ox}}$ (8)

(with $ a^{\rm ox}$ and $ b_2^{\rm ox}$ as defined in eq 9b and 9c) when the current doesn't level off (so2n).

For reductive catalysis (so2l and sr2n), the current equations read as eqs 9a and 10 but with
\begin{subeqnarray}
a^{\rm red}&= & 1+e_{\rm I/R}(1+e_{\rm O/I})
\\
b_2^{\rm re...
...t(e_{\rm O/I}^{1/2} + e_{\rm
I/R}^{1/2} (1+e_{\rm O/I})\right)
\end{subeqnarray}



Christophe Leger 2009-02-24