Animated Solution for Chemistry - d and f-Block Elements: Four successive members of the first row transition elements listed below with atomic numbers. Which one of them is expected to have the highest $E_{M^{3+}/M^{2+}}^{\circ }$ value?

When we encounter a question asking for the highest $E_{M^{3+}/M^{2+}}^{\circ }$ value, we are essentially being asked a thermodynamic question: Which $M^{3+}$ ion is the most desperate to gain an electron and become $M^{2+}$?

A high positive standard reduction potential indicates that the reactant ($M^{3+}$) is highly unstable relative to the product ($M^{2+}$). To solve this, we must analyze the electronic configurations and the underlying thermodynamic factors like ionization enthalpy and hydration energy for each element.

Let's break down the transition from $M^{3+}$ to $M^{2+}$ for each given metal:

1. Chromium (Cr): The electronic configuration of $C{r}^{3+}$ is $3d^3$. In an aqueous octahedral field, these three electrons perfectly half-fill the lower energy $t_{2g}$ orbitals. This imparts immense stability to the $C{r}^{3+}$ ion. Because it is already so stable, it strongly resists gaining an electron to become $C{r}^{2+}$ ($3d^4$). Consequently, its reduction potential is negative: $E^{\circ }=-0.41\text{ V}$.

2. Manganese (Mn): Manganese presents a classic case of half-filled stability. The $M{n}^{2+}$ ion has a $3d^5$ configuration, which is an exactly half-filled $d$-subshell—a state of exceptional stability. Therefore, the $M{n}^{3+}$ ion ($3d^4$) acts as a strong oxidizing agent, eagerly accepting an electron to achieve that coveted $3d^5$ state. This strong driving force results in a high positive potential: $E^{\circ }=+1.57\text{ V}$.

3. Iron (Fe): For iron, the situation is reversed compared to manganese. Here, it is the $F{e}^{3+}$ ion that possesses the stable $3d^5$ configuration. Since it is already in a highly stable state, it has a much lower tendency to accept an electron and become $F{e}^{2+}$ ($3d^6$). Thus, its reduction potential drops significantly compared to manganese: $E^{\circ }=+0.77\text{ V}$.

4. Cobalt (Co): This is where many students fall into a trap. Neither $C{o}^{3+}$ ($3d^6$) nor $C{o}^{2+}$ ($3d^7$) boasts a half-filled $d^5$ configuration. So, why would its reduction potential be high?

The secret lies in the third ionization enthalpy ($I{E}_3$). The energy required to remove the third electron from cobalt is exceptionally high. While the hydration enthalpy of $C{o}^{3+}$ is large, it is not sufficient to compensate for this massive ionization energy. As a result, $C{o}^{3+}$ is thermodynamically highly unstable in aqueous solutions. It acts as a ferocious oxidizing agent, desperately pulling an electron back to revert to the $C{o}^{2+}$ state.

This overwhelming thermodynamic instability gives cobalt the highest standard reduction potential among the first-row transition metals: $E^{\circ }=+1.97\text{ V}$.

Comparing the values, we get the clear order: $Co(+1.97\text{ V})>Mn(+1.57\text{ V})>Fe(+0.77\text{ V})>Cr(-0.41\text{ V})$

Therefore, Cobalt (Co) is expected to have the highest $E_{M^{3+}/M^{2+}}^{\circ }$ value.