= Solution
Take unit activity of each dissolved metal ion, unit activity of each metal, $p_{\mathrm H_2}=1\,\mathrm{bar}$ and $T=298\,\mathrm K$. The hydrogen-electrode <Nernst equation> gives
$$
E_{\mathrm H^+/\mathrm H_2}=-\frac{RT\ln10}{F}\mathrm{pH}
=-0.059117\,\mathrm{pH}\quad\mathrm V.
$$
Thus $E_H=-0.05912\,\mathrm V$ at pH 1 and $-0.82764\,\mathrm V$ at pH 14. For corrosion coupled to hydrogen evolution,
$$
\mathrm M+z\mathrm H^+\longrightarrow\mathrm M^{z+}+\frac z2\mathrm H_2,
\qquad E_{\rm cell}=E_H-E_M^\circ,\qquad\Delta G=-zFE_{\rm cell}.
$$
For alkaline solution the same electron balance can be written $\mathrm M+z\mathrm H_2O\to\mathrm M^{z+}+z\mathrm{OH^-}+(z/2)\mathrm H_2$, before allowing for metal-ion hydrolysis or complexation. Using $z=2$ for zinc and lead, and $z=1$ for lithium, the results per mole of metal are
$$
\begin{array}{c|rr|rr}
&\mathrm{pH}=1&\mathrm{pH}=1&\mathrm{pH}=14&\mathrm{pH}=14\\
\text{metal}&E_{\rm cell}\ (\mathrm V)&\Delta G\ (\mathrm{kJ\,mol^{-1}})&E_{\rm cell}\ (\mathrm V)&\Delta G\ (\mathrm{kJ\,mol^{-1}})\\\hline
\mathrm{Zn}&0.7009&-135.3&-0.06764&+13.05\\
\mathrm{Pb}&0.07088&-13.68&-0.6976&+134.6\\
\mathrm{Li}&3.0409&-293.4&2.2724&-219.3
\end{array}
$$
\b[Lithium is strongly unstable against hydrogen evolution in both media in this model.] The positive alkaline values for zinc and lead apply to the stipulated unit-activity free-ion products; they are not complete predictions of alkaline corrosion once hydroxide precipitation and complexation change the product chemical potentials.
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