Question

For $M^{2+}/M$ and $M^{3+}/M^{2+}$ systems the $E^o$ values for some metals are as follows:

$\begin{array}{cccc}C{r}^{2+}/Cr& -0.9V& C{r}^{3+}/C{r}^{2+}& -0.4V\\ M{n}^{2+}/Mn& -1.2V& M{n}^{3+}/M{n}^{2+}& +1.5V\\ F{e}^{2+}/Fe& -0.4V& F{e}^{3+}/F{e}^{2+}& +0.8V\end{array}$

Use this data to comment upon:

A. the stability of $F{e}^{3+}$ in acid solution as compared to that of $C{r}^{3+}$ or $M{n}^{3+}$

B. the ease with which iron can be oxidized as compared to a similar process for either chromium or manganese metal.

Answer 1

$A.\text{Stability of}F{e}^{3+}\text{ion}$

The reduction potentials for the given pairs increase in the following order:

$M{n}^{3+}/M{n}^{2+}>F{e}^{3+}/F{e}^{2+}>C{r}^{3+}/C{r}^{2+}$

Since we know, as the reduction potential increases oxidizing power increases and reducing power decreases.

The $E^{\ominus }$ value for $F{e}^{3+}/F{e}^{2+}$ is greater than that for $C{r}^{3+}/C{r}^{2+}$ and lower than that for $M{n}^{3+}/M{n}^{2+}$. So, the reduction of $F{e}^{3+}$ to $F{e}^{2+}$ is easier than the reduction of $C{r}^{3+}$ to $C{r}^{2+}$, but not as easy as the reduction of $M{n}^{3+}$ to $M{n}^{2+}$. Hence, $F{e}^{3+}$ is more stable than $M{n}^{3+}$, but less stable than $C{r}^{3+}$.

Hence overall stability order follows as:
$M{n}^{3+}<F{e}^{3+}<C{r}^{3+}$

$\text{Final answer}:$ $M{n}^{3+}<F{e}^{3+}<C{r}^{3+}$


$B.\text{Oxidizing ability of metal}$

The reduction potentials for the given pairs increase in the following order:

$F{e}^{2+}/Fe>C{r}^{2+}/Cr>M{n}^{2+}/Mn$

Since we know, as the reduction potential increases oxidizing power increases and reducing power decreases.

The value of reduction potential for $F{e}^{2+}/Fe$ is highest and for $M{n}^{2+}/Mn$ is lowest.
So, the oxidation of $Mn$ to $M{n}^{2+}$ is easiest and oxidation of $Fe$ to $F{e}^{2+}$ is toughest.
Hence, the overall order of their ability to get oxidized is follow as:
$Fe<Cr<Mn$

$\text{Final answer:}Fe<Cr<Mn$