Calculate the $emf$ of the cell in which the following reaction takes place:
$Ni_{(s)} + 2Ag^{+}(0.002 \, M) \rightarrow Ni^{2+}(0.160 \, M) + 2Ag_{(s)}$
Given that $E^{\Theta}_{(cell)} = 1.05 \, V$ (in $, V$)

At $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)

What is the potential of the cell for the reaction $Zn + Cu^{+2} \,(1 \, M) \rightarrow Cu + Zn^{+2} \,(1 \, M)$ at $25^{\circ}C$ (in $, V$)?

What is the potential of a hydrogen electrode in contact with a solution whose $pH$ is $1$ (in $V$)?

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 $298 \ K$.

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