$$\mathrm{E=-{2.303RT \over F}pH - {RT \over 2F}\ln {p_{H_2}/p^0}}$$ This is a equation on Wikipedia page of Standard Hydrogen Electrode, but I have no idea where the $pH$ come from. Since $\mathrm{a_{H^+}=f_{H^+} C_{H^+} /C_0}$, how does the $\mathrm{f_{H^+}}$ disappear?


There are a few logarithm rules that you need to apply here. \begin{align} E &= \mathrm{RT \over F}\ln {a_{H^+} \over (p_{H_2}/p^0)^{1/2}}\\ &= \mathrm{RT \over F}\ln a_{H^+} - \mathrm{RT \over F} \ln \left({p_{H_2} \over p^0}\right)^{1/2}\\ &= \mathrm{\ln(10) RT \over F} \lg a_{H^+} - {1 \over 2}\mathrm{RT \over F}\ln \left({p_{H_2} \over p^0}\right)\\ &= -\mathrm{2.303 RT \over F} \mathrm{pH} - {1 \over 2}\mathrm{RT \over F}\ln \left({p_{H_2} \over p^0}\right) \end{align}

You need a few logarithm rules:

  • $\log\left({a \over b}\right) = \log(a) - \log(b)$
  • $\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$
  • $\log_b(x^r) = r \log_b(x)$
  • $\begingroup$ I tried editing but there's a 6 character minimum. You've got a small typo in the derivation. Next to last line from bottom, lg(CH+) should be log(CH+) $\endgroup$
    – Sean Doris
    Feb 29 '16 at 0:15
  • $\begingroup$ I simply don't understand why $f_{H^+} \approx 1$, since the concentration we are dealing is not low. $\endgroup$
    – snsunx
    Feb 29 '16 at 2:55
  • $\begingroup$ You don't need $f_{\ce{H+}} \approx 1$, because the definition of $\mathrm{pH}$ is not $\mathrm{pH} = - \log (c_{\ce{H+}})$ but rather $\mathrm{pH} = - \log (a_{\ce{H+}})$. The definition based on the concentration is just an often used approximation which assumes $f_{\ce{H+}} \approx 1$ and is only valid in cases where this assumption is valid. $\endgroup$
    – Philipp
    Feb 29 '16 at 7:55
  • $\begingroup$ @SeanDoris: Thank you, but this is correct. "... $\log_{10}(x)$ should be written $\lg (x)$ and $\log_e(x)$ should be $\ln (x)$., so it is correct. $\endgroup$ Feb 29 '16 at 7:57
  • $\begingroup$ @Philipp: Thank you, you are right ... changed it. $\endgroup$ Feb 29 '16 at 7:59

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