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The general equation for liquid junction potential is given by $$E = {-RT\over F}\sum_i\int_a^b{t_i\over z_i}d ln(a_i)$$ Where a and b are the two phases, $t_i$ is the transport number of the ions and $z_i$ is the charge on the ions. When making the additional assumptions that (1) the concentrations are equivalent to the activities everywhere in the junction and that (2) the concentration of each ion follows a linear transition between the two phases, the above equation becomes the Henderson equation:$$E = {\sum_i{|Z_i| \over z_i}u_i[c_i(b)-c_i(a)]\over \sum_i|z_i|u_i[c_i(b)-c_i(a)]}{RT\over F}{ln{\sum_iz_iu_ic_i(a)\over \sum_iz_iu_ic_i(b)}}$$But when I tried to evaluate the integral by my self by substituting the equation of transfer number $t_i = {z_ic_iu_i\over \sum_iz_ic_iu_i}$ and putting $a_i = c_i$, I'm stuck with the following: $$E = {-RT\over F}\sum_i \int_a^b{{|z_i|\over zi}u_i\over \sum_i|z_i|c_iu_i}dc_i$$ The integral is of the form $\int{1\over {ax + b}}dx$ where $a = |z_i|u_i$ and $b = \sum_{j \ne i}|z_j|c_ju_j$ $$E = {-RT\over F}\sum_i{1\over z_i}ln\left( {|z_i|u_ic_i(b)+\sum_{j \ne i}|z_j|c_ju_j\over |z_i|u_ic_i(a)+\sum_{j \ne i}|z_j|c_ju_j}\right)$$Which is clearly not the Henderson equation given in Bard and Falkner, and I couldn't figure out where to use the second assumption. Where was I wrong and how to proceed?

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