The potential for the given half cell at $298 \ K$ is $(-) \ldots \ldots \ldots \times 10^{-2} \ V.$
$2 H^{+}_{(aq)} + 2 e^- \rightarrow H_{2(g)}$
$[H^{+}] = 1 \ M, P_{H_2} = 2 \ atm$
(Given: $2.303 RT / F = 0.06 \ V, \log 2 = 0.3$)

For the cell $Zn_{(s)} | Zn^{2+}_{(1\,M)} || Cu^{2+}_{(1\,M)} | Cu_{(s)}$,the $E^o_{cell} = 1.10 \, V$. When the cell is fully discharged at $298 \, K$,the ratio of the concentrations $[Zn^{2+}] / [Cu^{2+}]$ is:

$A$ copper electrode is dipped in a $0.1 \, M$ copper sulfate solution at $25 \, ^\circ C$. Calculate the reduction potential of the copper electrode. $(E^o_{Cu^{2+}/Cu} = 0.34 \, V)$ (in $, V$)

In the electrochemical cell $:$
$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\,M)\,|\,Cu$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\,M$ and that of $CuSO_4$ changed to $0.01\,M,$ the $emf$ changes to $E_2.$ From the followings,which one is the relationship between $E_1$ and $E_2$ $?$ (Given,$RT/F = 0.059$)

If $E^{\circ}(Ag^{+}_{(aq)} \mid Ag_{(s)}) = +0.80 \ V$,what is the potential developed for $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} (0.01 \ M) + e^{-}$ at $298 \ K$?

The hydrogen electrode is dipped in a solution of $pH = 3$ at $25\,^oC$. The potential of the cell would be ............. $V$ (the value of $2.303\,RT/F$ is $0.059\,V$).