Question

For $M^{2+}/M\text{and}{M}^{3+}/M^{2+}$ systems, $E^0$ values for some metals are as follows:

$C{r}^{2+}\left(aq\right)+2e^{-}\to Cr\left(s\right);E^0=-0.91VC{r}^{3+}\left(aq\right)+e^{-}\to C{r}^{2+}\left(aq\right);E^0=-0.41V$
$M{n}^{2+}\left(aq\right)+2e^{-}\to Mn\left(s\right);E^0=-1.2VM{n}^{3+}\left(aq\right)+e^{-}\to M{n}^{2+}\left(aq\right);E^0=+1.51V$
$F{e}^{2+}\left(aq\right)+2e^{-}\to Fe\left(s\right);E^0=-0.44VF{e}^{3+}\left(aq\right)+e^{-}\to F{e}^{2+}\left(aq\right);E^0=+0.77V$

Which of the following statement is/are correct regarding the given standard reduction potentials?

• A. $Mn$ is the powerful reducing agent
• B. $Mn$ is more vulnerable to oxidation than $Cr$
  
• C. $F{e}^{3+}$ has more oxidising power than $F{e}^{2+}$
• D. $M{n}^{3+}$ is less stable than $M{n}^{2+}$

Answer 1

Answer: (A), (B), (C), (D)

$M{n}^{3+}$ is less stable than $M{n}^{2+}$

$M{n}^{2+}/Mn$ couple has the least $E^0$ value than redox couples.
Thus, $Mn$ will readily go oxidation than others and hence, it is the powerful reducing agent.

Therefore, statement (a) and (b) are correct.

$F{e}^{3+}/F{e}^{2+}$ has higher $E^0$ value than $F{e}^{2+}/Fe$ couple. Thus, $F{e}^{3+}$ will have more tendency to reduce to $F{e}^{2+}$ than $F{e}^{2+}$ to $Fe$
Hence, $F{e}^{3+}$ will have more oxidising power than $F{e}^{2+}$.

Comparing $Mn$ couples,
$M{n}^{2+}/Mn$ has lower $E^0$ value than $M{n}^{3+}/M{n}^{2+}$ couple. Thus, $Mn$ will readily get oxidised to $M{n}^{2+}$ ions.
Whereas, $M{n}^{3+}$ will readily get reduced to $M{n}^{2+}$.
Hence, $M{n}^{3+}$ ion is least stable than $M{n}^{2+}$ ion.

Hence, all the given statements are correct.