Does anyone know what the intermediate calculations are to obtain equation 1.4.14? Taken from Fundamentals and Applications (Bard) page 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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asked Feb 21 '14 at 15:06
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2$\begingroup$ 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 Feb 21 '14 at 16:21
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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:
$E=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{nFAm_O}}{\frac{i}{nFAm_R}}=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{i}}{\frac{nFAm_O}{nFAm_R}}=E_0+\frac{RT}{nF}(ln\frac{i_l-i}{i}- ln\frac{nFAm_O}{nFAm_R})$=
$E_0-\frac{RT}{nF}ln\frac{m_O}{m_R}+\frac{RT}{nF}ln\frac{i_l-i}{i}$
answered Feb 22 '14 at 10:37
