The $E^{0}_{Red}$ values of $A, B,$ and $C$ are $+0.5 \ V, -3.0 \ V,$ and $-1.2 \ V$ respectively. The order of their reducing power is:

Calculate the $E^{\circ}_{cell}$ of the cell $Al | Al^{3+} || H^+ | H_{2(g)}, Pt$ given that $E^{\circ}_{Al^{3+}/Al} = -1.66 \ V$. (in $V$)

The standard reduction potentials of three metals $A, B,$ and $C$ are $+0.5 \, V, -3.0 \, V,$ and $-1.2 \, V$ respectively. What is the order of their reducing power?

Given that the standard reduction potentials for $M^{+}/M$ and $N^{+}/N$ electrodes at $298 \ K$ are $0.52 \ V$ and $0.25 \ V$ respectively. Which of the following is correct in respect of the following electrochemical cell?
$M | M^{+} || N^{+} | N$

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:

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 $V$