Electrochemistry

At 298 K , if emf of the cell corresponding to the reaction $\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(0.01 \mathrm{M})+\mathrm{H}_2(g) (1 \mathrm{~atm})$ is 0.28 V , then the pH of the solution at the hydrogen electrode is $\left(\frac{2.303 R T}{F}=0.06 \mathrm{~V}\right)$, $\left(E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\right)$