\newcommand{\cout}{\frac{[C]_{\text{out}}}{[C]_{\text{in}}}}
$\begin{array}[b]{rl}
E &= \frac{RT}{zF}\ln\cout = \frac{1.987\operatorname{cal}\operatorname{K}^{-1}\operatorname{mol}^{-1}\cdot 37^{\circ}\operatorname{C}}{z\cdot 96480\operatorname{C}\operatorname{mol}^{-1}}\ln\cout\\
&= \frac{1.987\operatorname{cal}\operatorname{K}^{-1}\cdot 310.15\operatorname{K}}{z\cdot 96480\operatorname{C}} \ln\cout = 
\frac{1.987\cdot 4.2 \operatorname{J} \cdot 310.15}{z\cdot 96480\operatorname{C}} \ln\cout\\
&= \frac{1.987\cdot 4.2\cdot 310.15 \operatorname{VC}}{z\cdot 96480\operatorname{C}}\ln\cout = \frac{1.987\cdot 4.2\cdot 310.15 \operatorname{V}}{z\cdot 96480}\cdot\frac{\log_{10}\cout}{\log_{10}e}\\
&=^* \frac{1.987\cdot 4.2\cdot 310.15 \operatorname{V}}{z\cdot 96480\cdot\log_{10}e}\log_{10}\cout \approx \frac{0.062\operatorname{V}}{z}\cdot\log_{10}\cout\\
&= \frac{62\operatorname{mV}}{z}\log_{10}\cout
\end{array}\qed$\\
Um das selbe für $T=20^{\circ}\operatorname{C}$ zu tun, nehmen wir den Term bei $*$ und ersetzen 310.15 (Grad Kelvin) durch die geänderte Temperatur (293.15):\\
\[
\frac{1.987\cdot 4.2\cdot 293.15 \operatorname{V}}{z\cdot 96480\cdot\log_{10}e}\log_{10}\cout \approx \frac{0.058\operatorname{V}}{z}\cdot\log_{10}\cout = \frac{58\operatorname{mV}}{z}\log_{10}\cout
\]