The standard free energy change for the formation of ${\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$ from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$ is $∆{\mathbf{G}}^{°}=+237.13\mathbf{KJ}$ The reactions are

$$\begin{gathered} \mathrm{cathode}:2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{H}}_2\left(\mathrm{g}\right)+20{\mathrm{H}}^{-} \\ \mathrm{Anode}:{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-} \end{gathered}$$

Calculate the standard voltage${\mathrm{E}}^{°}$needed to decompose water into its elements by electrolysis. What does the word standard mean in this question?

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