$K,$ $Ca,$ and $Li$ metals may be arranged in the increasing order of their standard electrode potentials as

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:

Standard electrode potential of $SHE$ at $298 \, K$ is ............. $V$.

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 ............ $V$.

What will be the $EMF$ of the cell formed by the following half-cells in $V$?
$Mg^{2+} + 2e^- \to Mg_{(s)}; E = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E = +0.33 \ V$

The standard reduction potential at $298 \ K$ for the following half-cell reactions is given as:
$Zn^{2+}_{(aq)} + 2e^{-} \rightarrow Zn_{(s)} ; \quad E^{\circ} = -0.762 \ V$
$Cr^{3+}_{(aq)} + 3e^{-} \rightarrow Cr_{(s)} ; \quad E^{\circ} = -0.740 \ V$
$2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)} ; \quad E^{\circ} = 0.0 \ V$
$F_{2(g)} + 2e^{-} \rightarrow 2F^{-}_{(aq)} ; \quad E^{\circ} = 2.87 \ V$
Which of the following is the strongest reducing agent?