What is to be done to determine the potential of any half cell?

At $25^\circ C$,$1 \, atm$ of gas $X$ is bubbled through a solution containing $1 \, M$ $Y^-$ and $1 \, M$ $Z^-$. If the reduction potential order is $Z > Y > X$,then:

The cell potential of the given cell is $0.34 \, V$. Write the cell reaction and calculate $E^o_{Cu^{2+}|Cu}$.
$Pt_{(s)} \mid H_{2(g)} (1 \, bar) \mid H^+_{(1 \, M)} \parallel Cu^{2+}_{(1 \, M)} \mid Cu_{(s)}$

What is electrode potential?

Given standard electrode potentials:
$Fe^{2+} + 2e^- \to Fe; E^o = -0.440 \ V$
$Fe^{3+} + 3e^- \to Fe; E^o = -0.036 \ V$
The standard electrode potential $(E^o)$ for $Fe^{3+} + e^- \to Fe^{2+}$ is .............. $V$.

For the reduction of $NO_3^{-}$ ion in an aqueous solution,$E^0$ is $+0.96 \ V$. Values of $E^0$ for some metal ions are given below:
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V \ \ \ \ E^0 = -1.19 \ V$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe \ \ \ \ E^0 = -0.04 \ V$
$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au \ \ \ \ E^0 = +1.40 \ V$
$Hg^{2+}_{(aq)} + 2e^{-} \rightarrow Hg \ \ \ \ E^0 = +0.86 \ V$
The pair$(s)$ of metals that is(are) oxidized by $NO_3^{-}$ in aqueous solution is(are):
$(A) V$ and $Hg$
$(B) Hg$ and $Fe$
$(C) Fe$ and $Au$
$(D) Fe$ and $V$