The hydrogen electrode is dipped in a solution of pH =3 at 25^° C. The potential of the electrode will be - ____ × 10^-2 V. ((2.303 RT)/F=0.059 V )

The hydrogen electrode is dipped in a solution of $\displaystyle \mathrm{pH}=3$ at $\displaystyle 25^{\circ} \mathrm{C}$. The potential of the electrode will be - $\displaystyle \_\_\_\_$ $\displaystyle \times 10^{-2} \mathrm{~V}$. $$\left(\frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.059 \mathrm{~V}\right) $$