The $E^{\circ}$ of $Ce^{4+} / Ce^{3+} = 1.6 \ V$ and $Fe^{3+} / Fe^{2+} = 0.77 \ V$. The $E^{\circ}$ of the reaction where $Fe^{3+}$ oxidises $Ce^{3+}$ is:

Reduction of species is dependent on its reduction potential value.

The standard electrode potential $(E^{\circ})$ values of $Al^{3+}/Al, Ag^{+}/Ag, K^{+}/K$ and $Cr^{3+}/Cr$ are $-1.66 \ V, 0.80 \ V, -2.93 \ V$ and $-0.74 \ V,$ respectively. The correct decreasing order of reducing power of the metal is:

$FeO_4^{2-}$ $\xrightarrow{2.2 \ V} Fe^{3+}$ $\xrightarrow{0.70 \ V} Fe^{2+}$ $\xrightarrow{-0.45 \ V} Fe^0$
$E_{FeO_4^{2-} / Fe^{2+}}^{\theta}$ is $x \times 10^{-3} \ V$. The value of $x$ is $.........$.

What does the negative sign in the expression $E_{Zn^{2+}|Zn}^o = -0.76 \ V$ mean?

Electrode potentials $(E^o)$ are given below:
$Cu^{+}/Cu = +0.52 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$
$Ag^{+}/Ag = +0.88 \ V$
Based on the above potentials,the strongest oxidizing agent will be: