Does anyone know what the intermediate calculations are to obtain equation 1.4.14? Taken from Fundamentals and Applications (Bard) page 31Fundamentals and Applications (Bard) pg. 31

To be more precise, how to obtain (1.4.14) from $\frac{nFAm_O}{nFAm_R}=\frac{i_l-i}{i}$

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    $\begingroup$ You have to use equation 1.4.12, i.e. $E = E^{0^{\prime}} + \frac{RT}{nF} \ln \frac{C_{\mathrm{O}} (x=0)}{C_{\mathrm{R}} (x=0)}$, as basis and substitute equations 1.4.11 and 1.4.13 into it, then you arrive at 1.4.14 by using the basic property of logarithms that $\ln (ab) = \ln (a) + \ln(b)$. $\endgroup$
    – Philipp
    Commented Feb 21, 2014 at 16:21

1 Answer 1


Thank you for the comment Philipp. With this clarification, the derivation is simple:

$E=E_0+\frac{RT}{nF}ln\frac{C_O(x=0)}{C_R(x=0)}$ (1.4.12) where

$C_O(x=0)=\frac{i_l-i}{nFAm_O}$ (1.4.11) and $C_R(x=0)=\frac{i}{nFAm_R}$ (1.4.13).

Than, combining equations (1.4.11) and (1.4.13) to (1.4.12), we obtain:

$E=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{nFAm_O}}{\frac{i}{nFAm_R}}=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{i}}{\frac{nFAm_O}{nFAm_R}}=E_0+\frac{RT}{nF}(ln\frac{i_l-i}{i}- ln\frac{nFAm_O}{nFAm_R})$=



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