Which of the following conditions will increase the voltage of the cell,represented by the equation $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$?

The following reaction takes place at $298 \, K$ in an electrochemical cell involving two metals $A$ and $B$,
$A^{2+}_{(aq)} + B_{(s)} \rightarrow B^{2+}_{(aq)} + A_{(s)}$
with $[A^{2+}] = 4 \times 10^{-3} \, M$ and $[B^{2+}] = 2 \times 10^{-3} \, M$ in the respective half-cells,the cell $EMF$ is $1.091 \, V$.
The equilibrium constant of the reaction is closest to

Calculate the $E.M.F.$ of the following cell at $298 \ K$: $Zn_{(s)} | ZnSO_4(0.01 \ M) || CuSO_4(1.0 \ M) | Cu_{(s)}$ if $E^o_{cell} = 2.0 \ V$. (in $V$)

For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

If $E^{\circ}(Zn_{(aq)}^{+2} \mid Zn_{(s)}) = -0.76 \ V$,calculate the potential for $Zn_{(s)} \rightarrow Zn_{(0.01 \ M)}^{+2} + 2e^{-}$ at $298 \ K$.

$A$ hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $pH=9$ and passing hydrogen gas around the platinum wire at $1.2 \ atm$ pressure. The oxidation potential of such an electrode equals $V$.