Which one of the following has a potential more than zero?

For Pt, \(\frac{1}{2} \mathrm{H}_{2}(1 \mathrm{~atm}) \mid \mathrm{HCl}(\mathrm{x})\)
According to Nernst equation, electrode potential is given by
\[
\mathrm{E}_{\text {cell }}=\frac{0.0591}{2} \log \mathrm{x}
\]
(a) For Pt, \(\frac{1}{2} \mathrm{H}_{2}(1 \mathrm{~atm}) \mid \mathrm{HCl}(2 \mathrm{M})\)
\[
\begin{aligned}
\text { E cell }_{\text {cell }} &=\frac{0.0591}{2} \log 2 \\
&=\frac{0.0591}{2} \times 0.3010=0.0089
\end{aligned}
\]
(b) For Pt, \(\frac{1}{2} \mathrm{H}_{2}(1\) atm \() \mid \mathrm{HCl}(0.1) \mathrm{M}\)
\[
\begin{aligned}
\mathrm{E}_{\text {cell }} &=\frac{0.0591}{2} \log (0.1) \\
&=\frac{0.0591}{2} \times(-1)=-0.0295
\end{aligned}
\]
(c) For Pt, \(\frac{1}{2} \mathrm{H}_{2}(1 \mathrm{~atm}) \mid \mathrm{HCl}(0.5 \mathrm{M})\)
\[
\begin{aligned}
\mathrm{E}_{\text {cell }} &=\frac{0.0591}{2} \log (0.5) \\
&=\frac{0.0591}{2}(-0.3010) \\
&=-0.0089
\end{aligned}
\]
(d) For \(\mathrm{Pt}, \frac{1}{2} \mathrm{H}_{2}\) (1 atm) \(\mid \mathrm{HCl}\) (1 M)
\[
\mathrm{F}_{\text {cell }}=\frac{0.0591}{2} \log 1=0
\]
Hence, the cell given in option (a) has potential more than zero.