Question

Which is a stronger reducing agent $C{r}^{2+}$ or $F{e}^{2+}$ and why ?

Answer 1

$\text{Stronger reducing agent out of}C{r}^{2+}\text{and}F{e}^{2+}$

The $E_{C{r}^{3+}/Cr2+}^0$ and $E_{F{e}^{3+}/Fe2+}^0$ values are $-0.41V$ and $0.77V$ respectively.
Hence, $C{r}^{2+}$ can easily oxidize to $C{r}^{3+}$, but $F{e}^{2+}$ does not get oxidized to $F{e}^{3+}$ easily.
Therefore, $C{r}^{2+}$ is a better reducing agent that $F{e}^{3+}$.

$\text{Final answer:}$

$C{r}^{2+}$ is a stronger reducing agent.