Calculate the standard potential of a cell having the following electrode reactions:
$Cd_{(aq)}^{2+} + 2e^{-} \rightarrow Cd_{(s)}$ $E^{\circ} = -0.403 \ V$
$Zn_{(aq)}^{2+} + 2e^{-} \rightarrow Zn_{(s)}$ $E^{\circ} = -0.763 \ V$ (in $V$)

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 $V$

Reduction of species is dependent on its reduction potential value.

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:

For the reaction $F_2 + 2e^{-} \to 2F^{-}$,$E^{\circ} = 2.8 \, V$. What is the $E^{\circ}$ for the reaction $\frac{1}{2} F_2 + e^{-} \to F^{-}$?

Reduction potential of ions are given below:

$ClO_4^{-}$	$E^{\circ} = 1.19 \ V$
$IO_4^{-}$	$E^{\circ} = 1.65 \ V$
$BrO_4^{-}$	$E^{\circ} = 1.74 \ V$

The correct order of their oxidising power is: