The standard reduction potentials for $Al^{3+} | Al$,$Fe^{2+} | Fe$,and $Br_2 | Br^-$ are $-1.66 \ V$,$-0.45 \ V$,and $1.09 \ V$ respectively. What is the correct order of their reducing power?

$MX$ is a sparingly soluble salt that follows the given solubility equilibrium at $298 \ K$: $MX_{(s)} \rightleftharpoons M^{+}_{(aq)} + X^{-}_{(aq)}$; $K_{sp} = 10^{-10}$. If the standard reduction potential for $M^{+}_{(aq)} + e^- \rightarrow M_{(s)}$ is $(E^{\ominus}_{M^{+}/M}) = 0.79 \ V$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\ominus}_{X^{-}/MX_{(s)}/M}$ is . . . . . . $mV$. (nearest integer) [Given: $\frac{2.303 RT}{F} = 0.059 \ V$]

Given the following standard electrode potentials:
$Cu^{+2} + 2e^{-} \to Cu$,$E^{o} = X_{1}$
$Cu^{+} + e^{-} \to Cu$,$E^{o} = X_{2}$
Calculate the standard electrode potential $E^{o}$ for the reaction: $Cu^{+2} + e^{-} \to Cu^{+}$

$E^o$ values for the half-cell reactions are given below:
$Cu^{2+} + e^- \to Cu^{+} \quad E^o = 0.15 \ V$
$Cu^{2+} + 2e^- \to Cu \quad E^o = 0.34 \ V$
What will be the $E^o$ for the half-cell reaction:
$Cu^{+} + e^- \to Cu$ ? $\dots \ V$

Calculate the cell potential in volts for the cell reaction given below:
$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)

Which of the following has the highest electrode potential?