Which is a stronger reducing agent $Cr ^{2+}$ or $Fe ^{2+}$ and why?

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The following reactions are involved when $Cr ^{2+}$ and $Fe ^{2+}$ act as reducing agents.

$\longrightarrow Cr\,\, Cr _{2+}$    $3+ Fe _{2+} \longrightarrow Fe _{3+}$

${E^\circ }_{C{r^3}/C{r^{2 + }}}$ The value is $-0.41\, V$ and ${E^\circ }_{F{e^3}/F{e^{2 + }}}$ is $+0.77$ $V$. This means that $C r^{2+}$ can be easily oxidized to $Cr ^{3+},$ but $Fe ^{2+}$ does not get oxidized to $Fe ^{3+}$ easily. Therefore, $Cr ^{2+}$ is a better reducing agent that $Fe ^{3+}$

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