Calculate the reversible potential of oxygen electrode in a solution of pH = 1, when the partial pressure of ${\mathrm{O}}_2$ is ${10}^{-2}$atm $[\mathrm{Assume} {\mathrm{E}}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}^0=1.23 \mathrm{V},\mathrm{vs} \mathrm{SHE} \mathrm{at} 1 \mathrm{atm} \mathrm{and} 25°\mathrm{C} ]$

$$\begin{gathered} {\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{e}}^{-}\to 4{\mathrm{OH}}^{-} \\ \begin{array}{l}\mathrm{pH}=1\\ \mathrm{pOH}=14-\mathrm{pH}\\ \mathrm{pOH}=14-1\\ \mathrm{pOH}=13\\ \left[{\mathrm{OH}}^{-}\right]={10}^{-\mathrm{pOH}}\\ \left[{\mathrm{OH}}^{-}\right]={10}^{-13}\mathrm{M}\end{array} \\ \begin{array}{l}{\mathrm{E}}_{{\mathrm{O}}_2,{\mathrm{H}}_2\mathrm{O}}={\mathrm{E}}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}^0-\frac{0.0591}{\mathrm{n}}\log \frac{{\left[{\mathrm{OH}}^{-}\right]}^4}{{\mathrm{P}}_{{\mathrm{O}}_2}}\\ {\mathrm{E}}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}=1.23 \mathrm{V}-\frac{0.0591}{4}\log \frac{{\left({10}^{-13}\right)}^4}{{10}^{-2}}\\ {\mathrm{E}}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}=1.23 \mathrm{V}-\frac{0.0591}{4}\log \frac{{\left({10}^{-13}\right)}^4}{{10}^{-2}}\\ {\mathrm{E}}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}=1.97 \mathrm{V}\end{array} \end{gathered}$$

Therefore, the correct option is C