$H _{2} \rightarrow 2 H ^{+}+2 e ^{-}$
At $1\,atm$ pressure, $pH$ is equal to $10$ .
Therefore,
$\left[ H ^{+}\right]=10^{-10}$
The standard reduction potential of $S.H.E$ is equal to zero.
Apply Nernst equation,
$E_{ H _{2} / H ^{+}}=0-\frac{0.059}{2} \log \frac{\left(10^{-10}\right)^{2}}{1}$
$=0.0295 \times 20$
$=0.59 \,V$
$Zn^{2+}(aq) + 2e^{-} $ $\rightleftharpoons$ $ Zn (s)$ , $E^o_{RP}= -0.762\, V$,
$Cr^{3+}(aq) + 3e^{-} $ $\rightleftharpoons$$ Cr(s)$, $E^o_{RP} = -0.740\, V$
$2H^{+}(aq) + 2e^{-} $$\rightleftharpoons$$ H_2(g)$, $E^o_{RP} = 0.00\,V$,
$Fe^{3+}_{(aq)} + 2e^{-} $$\rightleftharpoons$$ Fe^{2+}(aq)$ , $E^o_{RP} = 0.77 \,V$
$M{g^{2 + }} + 2{e^ - } \to Mg(s);\,\,E = - 2.37\,V$
$C{u^{2 + }} + 2{e^ - } \to Cu(s);\,\,\,E = + 0.33\,V$

$2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g})$
${\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}}$
(Given: $2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3$ )