What is the reduction potential of a silver wire dipped in a $0.1 \ M \ AgNO_3$ solution at $25^\circ C$?

At a temperature of $298 \ K$, the $emf$ of the following electrochemical cell: $Ag_{(s)} | Ag^{+}(0.1 \ M) || Zn^{2+}(0.1 \ M) | Zn_{(s)}$ will be (Given, $E^{\circ}_{cell} = -1.562 \ V$) (in $V$)

Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):
$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$
(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^o$ for the cell at $298 \ K$.
$2 X_2(s)+3 Y^{2+}(a q) \rightarrow 2 X_2^{3+}(a q)+3 Y(s)$

$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$:

Calculate $pH$ of $HCl$ solution at $298\,K$ temperature for the following cell: $Pt_{(s)} \mid H_2 \,(1\,bar) \mid HCl\,(xM) \parallel Cu^{2+}\,(0.02\,M) \mid Cu_{(s)}$. Given that the standard cell potential $E^{\circ}_{cell} = 0.34\,V$ and the measured cell potential $E_{cell} = 0.45\,V$.