Electrode potential of hydrogen electrode is 18 mV , then [H^ + ] is

Given : $E_{\mathrm{H}^{+} / \mathrm{H}_2}=18 \times 10^{-3} \mathrm{~V},\left[\mathrm{H}^{+}\right]=?$ Applying Nernst equation, $E_{\mathrm{H}^{+} / \mathrm{H}_2}=E_{\mathrm{H}^{+} / \mathrm{H}_2}^{\circ}-\frac{0.0591}{1} \log \frac{1}{\left[\mathrm{H}^{+}\right]}$18 \times 10^{-3} \mathrm{~V}=0+0.0591 \log \left[\mathrm{H}^{+}\right] 18 \times 10^{-3} \mathrm{~V}=0.0591 \log \left[\mathrm{H}^{+}\right] \log \left[\mathrm{H}^{+}\right]=0.3046 \therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(0.3046)=2.02=2.0