Understanding the Concept

The question asks which of the given first-row transition elements has the highest standard electrode potential ($E_{M^{3+}/M^{2+}}^0$) for the $M^{3+}/M^{2+}$ couple. This potential indicates the tendency of the $M^{3+}$ ion to get reduced to the $M^{2+}$ ion. A higher (more positive) $E_{M^{3+}/M^{2+}}^0$ value means the $M^{3+}$ ion is a stronger oxidizing agent and is more stable.

The stability of the +3 oxidation state relative to the +2 state is governed by several factors, with the most critical one being the third ionization energy. This is the energy required to remove an electron from a gaseous $M^{2+}$ ion to form a gaseous $M^{3+}$ ion. A high third ionization energy makes it difficult to form the $M^{3+}$ ion, making it a poor oxidizing agent and resulting in a low (negative) $E_{M^{3+}/M^{2+}}^0$ value. Conversely, a lower third ionization energy favors the formation of the M3+ ion, making it a strong oxidizing agent with a high (positive) $E_{M^{3+}/M^{2+}}^0$ value.

Step-by-Step Analysis

Step 1: Identify the Elements and Their Electron Configurations

• Cr (Z=24): [Ar] 3d⁵ 4s¹ → Cr²⁺ is [Ar] 3d⁴
• Mn (Z=25): [Ar] 3d⁵ 4s² → Mn²⁺ is [Ar] 3d⁵ (stable half-filled d-subshell)
• Fe (Z=26): [Ar] 3d⁶ 4s² → Fe²⁺ is [Ar] 3d⁶
• Co (Z=27): [Ar] 3d⁷ 4s² → Co²⁺ is [Ar] 3d⁷

Step 2: Analyze the Third Ionization Energy

The third ionization energy is exceptionally high for Manganese (Mn). This is because an electron must be removed from the stable, half-filled 3d⁵ configuration of Mn²⁺ to form Mn³⁺ (3d⁴). Removing an electron from this stable configuration requires a large amount of energy.

In contrast, for Cobalt (Co), the third ionization energy is relatively lower. An electron is removed from the 3d⁷ configuration of Co²⁺ to form the more stable 3d⁶ configuration of Co³⁺.

Step 3: Relate Ionization Energy to E°

A high third ionization energy (like for Mn) means the $M^{3+}$ ion is very difficult to form. Therefore, the $M^{3+}/M^{2+}$ couple will have a very low (highly negative) $E_{M^{3+}/M^{2+}}^0$ value, as the reaction $M^{3+}+e^{-}\to M^{2+}$ is not favored.

A lower third ionization energy (like for Co) means the $M^{3+}$ ion is easier to form and is more stable. This makes the $M^{3+}$ ion a strong oxidizing agent, resulting in a high (positive) $E_{M^{3+}/M^{2+}}^0$ value.

Step 4: Compare the Given Elements

Among the options:

• Mn has the highest third ionization energy, so it has the lowest (most negative) $E_{M^{3+}/M^{2+}}^0$.
• Co has a relatively low third ionization energy, so it has the highest (most positive) $E_{M^{3+}/M^{2+}}^0$.
• Fe and Cr have values in between.

Final Answer

Cobalt (Co, Z = 27) is expected to have the highest $E_{M^{3+}/M^{2+}}^0$ value.

Related Topics & Formulae

Key Concepts:

• Ionization Energy: The energy required to remove an electron from an isolated gaseous atom or ion. Successive ionization energies increase, but the magnitude of the increase depends on electronic configuration.
• Stability of Oxidation States: Oxidation states are stabilized by half-filled (d⁵) or fully-filled (d¹⁰) subshells. Removing an electron from these stable configurations requires more energy.
• Standard Electrode Potential (E°): A measure of the tendency for a chemical species to be reduced. It is directly related to the free energy change of the reduction reaction ($\mathrm{\Delta G}=-\mathrm{nFE}°$).

Important Formulae & Theory:

The standard electrode potential for a half-cell reaction is related to the standard Gibbs free energy change by the equation:

$\Delta G^0=-\mathrm{nF}{E}^0$

Where:

• $\Delta G^0$ is the standard Gibbs free energy change
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