The reduction potential of a hydrogen electrode at $pH = 1$ is ........ $V$.

If $E^{\circ}(Cu^{2+}_{(aq)} \mid Cu_{(s)}) = +0.34 \ V$. What is the potential for $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} (0.1 \ M) + 2e^-$ at $298 \ K$?

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

Which graph correctly correlates $E_{cell}$ as a function of concentrations for the cell (for different values of $M$ and $M'$):-
$Zn_{(s)} + Cu^{2+}(M) \to Zn^{2+}(M') + Cu_{(s)};$ $E^o_{cell} = 1.10 \, V$
$X$-axis : $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$Y$-axis : $E_{cell}$

For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]

Consider the single electrode process $2H^{+} + 2e^{-} \rightleftharpoons H_2$ catalyzed by a platinum black electrode in $HCl$ electrolyte. The potential of the electrode is $-0.059 \ V$. What is the concentration of the acid in the hydrogen half-cell if the $H_2$ pressure is $1 \ bar$ (in $M$)?