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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Sign up to join this communityDoes 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 $\frac{nFAm_O}{nFAm_R}=\frac{i_l-i}{i}$
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:
$E=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{nFAm_O}}{\frac{i}{nFAm_R}}=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{i}}{\frac{nFAm_O}{nFAm_R}}=E_0+\frac{RT}{nF}(ln\frac{i_l-i}{i}- ln\frac{nFAm_O}{nFAm_R})$=
$E_0-\frac{RT}{nF}ln\frac{m_O}{m_R}+\frac{RT}{nF}ln\frac{i_l-i}{i}$