find reduction potential using standard hydrogen electrode

To determine the reduction potential of any species, we can use a standard hydrogen electrode. Since $2\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}\to \mathrm{H}{}_2$ has ${\xi }^{\circ }=0$, the overall cell potential is equal but opposite in sign to the reduction potential of the other reaction.

galvanic cell with standard hydrogen electrode

Imagine the following setup of galvanic cell: anode: $\mathrm{Zn}(\mathrm{s})\to \mathrm{Zn}{}^{2+}(\mathrm{aq})+2\mathrm{e}{}^{-}$ cathode: $2\mathrm{H}{}^{+}(\mathrm{aq})+2\mathrm{e}{}^{-}\to \mathrm{H}{}_2(\mathrm{g})$ (standard hydrogen electrode / SHE) overall: $2\mathrm{H}{}^{+}+\mathrm{Zn}(\mathrm{s})\to \mathrm{Zn}{}^{2+}+\mathrm{H}{}_2(\mathrm{g})$, $E_{cell}^{\circ }=0.76\mathrm{V}$

Since we have $E_{cell}^{\circ }=E_{cathode}^{\circ }+E_{anode}^{\circ }$ and cathode has $\mathrm{H}{}^{+}\to \mathrm{H}{}_2$ ($E^{\circ }=0\mathrm{V}$) We know that for anode: $\mathrm{Zn}\to \mathrm{Zn}{}^{2+}$ ($E^{\circ }=0.76\mathrm{V}$, $E_{reduction}=-0.76\mathrm{V}$)

By setting $2\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}\to \mathrm{H}{}_2$ to zero we can find values for all other half-reactions.