The standard oxidation potential of a calomel electrode is:

Electrode potential data are given below :
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; \, E^o = +0.77 \, V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; \, E^o = -1.66 \, V$
$Br_{2(aq)} + 2e^- \to 2Br^{-}_{(aq)}; \, E^o = +1.08 \, V$
Based on the data given above,the reducing power of $Fe^{2+}$,$Al$ and $Br^{-}$ will increase in the order:

The standard oxidation potentials for the half-reactions are given as $Zn \to Zn^{2+} + 2e^{-}; E^o = +0.76 \ V$ and $Fe \to Fe^{2+} + 2e^{-}; E^o = +0.41 \ V$. The $EMF$ for the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

Given:
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; E^o = +0.77 \ V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; E^o = -1.66 \ V$
$Br_{2(aq)} + 2e^- \to 2Br^-_{(aq)}; E^o = +1.09 \ V$
Considering the electrode potentials,which of the following represents the correct order of reducing power?

The standard electrode potentials $(E^{0}_{Red})$ of four elements $A, B, C,$ and $D$ are $-3.05 \ V, -1.66 \ V, -0.40 \ V,$ and $0.80 \ V$ respectively. Which of the following is the most reactive?

Why does copper not replace hydrogen from acids?