Where does the pH come from in the equation of Standard Hydrogen Electrode?
There are a few logarithm rules that you need to apply here.
\begin{align}
E &amp;= \mathrm{RT \over F}\ln {a_{H^+} \over (p_{H_2}/p^0)^{1/2}}\\
  &amp;= \mathrm{RT \over F}\ln a_{H^+} - \mathrm{RT \over F} \ln \left({p_{H_2} \over p^0}\right)^{1/2}\\
  &amp;= \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)\\
  &amp;= -\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:


$\log\left({a \over b}\right) = \log(a) - \log(b)$
$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$
$\log_b(x^r) = r \log_b(x)$