Q.Why is the $E^\circ$ value for the $Mn^{3+}/Mn^{2+}$ couple much more positive than that for $Cr^{3+}/Cr^{2+}$ or $Fe^{3+}/Fe^{2+}$? Explain.

Concept: Stability of Oxidation States — The $E^\circ$ value reflects the ease of reduction; a more positive value means $Mn^{3+}$ is much more easily reduced (i.e., less stable) than $Cr^{3+}$ or $Fe^{3+}$.

Reasoning:

1. $Mn^{3+}$ has a $d^4$ configuration. In an octahedral field, the fourth electron occupies the higher-energy $e_g$ orbital (high-spin), making the ion highly unstable and prone to gain an electron to reach the stable half-filled $d^5$ ($Mn^{2+}$).
2. $Cr^{3+}$ is $d^3$ — a half-filled $t_{2g}$ set — giving it extra stability (exchange energy). $Fe^{3+}$ is $d^5$ — a half-filled $d$ subshell — also very stable. Both resist reduction. …

The Core Idea: Stability of Oxidation States and Electronic Configuration

The standard electrode potential $E^\circ$ for a redox couple $M^{3+}/M^{2+}$ tells us how easily $M^{2+}$ gets oxidised to $M^{3+}$. A more positive $E^\circ$ means the $M^{2+}$ state is more stable relative to $M^{3+}$ — it resists oxidation. Conversely, a less positive (or negative) $E^\circ$ means $M^{2+}$ is easily oxidised.

The key lies in the electronic configurations of the ions, specifically the stability associated with half-filled and fully-filled $d$ subshells. In aqueous solution, these are high-spin complexes for first-row transition metals.

Let's examine each case.

Step-by-Step Reasoning

1. The $Mn^{3+}/Mn^{2+}$ couple: The half-filled $d^5$ fortress

• $Mn^{2+}$ has the electronic configuration $[Ar]\,3d^5$. In an octahedral field (high-spin), this is $t_{2g}^3 e_g^2$ — each of the five $d$ orbitals is singly occupied.
• This is a half-filled $d$ subshell, which confers exceptional stability due to:
  • Maximum exchange energy (Hund's rule).
  • Symmetrical distribution of electron density.
• To oxidise $Mn^{2+}$ to $Mn^{3+}$, you must remove an electron from this stable $3d^5$ arrangement. $Mn^{3+}$ has $3d^4$ ($t_{2g}^3 e_g^1$), which is less stable (Jahn-Teller distortion also adds instability).
• Therefore, the oxidation $Mn^{2+} \rightarrow Mn^{3+} + e^-$ is energetically very difficult. This means $Mn^{2+}$ strongly resists being oxidised, so the equilibrium $Mn^{3+} + e^- \rightleftharpoons Mn^{2+}$ lies far to the right. A large positive potential is needed to drive the reduction of $Mn^{3+}$.

2. The $Fe^{3+}/Fe^{2+}$ couple: The other half-filled story

• $Fe^{2+}$ is $[Ar]\,3d^6$ ($t_{2g}^4 e_g^2$). It does not have a half-filled subshell.
• $Fe^{3+}$ is $[Ar]\,3d^5$ ($t_{2g}^3 e_g^2$) — the same half-filled $3d^5$ configuration that made $Mn^{2+}$ so stable.
• Here, oxidation of $Fe^{2+}$ to $Fe^{3+}$ produces the stable half-filled configuration. This is energetically favourable.
• Hence, $Fe^{2+}$ is more easily oxidised than $Mn^{2+}$, and $Fe^{3+}$ is a weaker oxidising agent than $Mn^{3+}$. The $E^\circ$ is therefore much less positive.

3. The $Cr^{3+}/Cr^{2+}$ couple: Stability from the $t_{2g}$ subshell

• $Cr^{2+}$ is $[Ar]\,3d^4$ ($t_{2g}^3 e_g^1$). This is not particularly stable.
• $Cr^{3+}$ is $[Ar]\,3d^3$ ($t_{2g}^3 e_g^0$). This is a half-filled $t_{2g}$ subshell — a very stable arrangement in an octahedral field (maximum exchange energy within the $t_{2g}$ set).
• Oxidation of $Cr^{2+}$ to $Cr^{3+}$ again produces a stable configuration. So $Cr^{2+}$ is easily oxidised, and $Cr^{3+}$ is a weak oxidising agent.

Method: Electronic Configuration & Stability Analysis

This method uses electronic configuration and exchange energy to explain why certain oxidation states are more stable than others.

Step 1: Write the electronic configurations

Step 2: Identify the stability factor

The key is exchange energy — the energy released when electrons with parallel spins exchange positions. More unpaired electrons → higher exchange energy → greater stability.

• Mn²⁺ has 5 unpaired electrons (d⁵, half-filled) → maximum exchange energy → very stable.
• Mn³⁺ has only 4 unpaired electrons → less stable.

So, Mn²⁺ is unusually stable compared to Mn³⁺. This makes the reduction:

highly favourable → large positive $E^\circ$ value.

Step 3: Compare with Cr and Fe

Here are the most common mistakes students make when answering this question, along with how to avoid each one.

Mistake 1: Only Mentioning the Electronic Configuration (Without Linking to Stability)

The Mistake:

Students often stop at stating the configurations:

• $Mn^{2+}$ is $3d^5$ (half-filled).
• $Cr^{2+}$ is $3d^4$.
• $Fe^{2+}$ is $3d^6$.

They then conclude that $Mn^{2+}$ is "more stable" without explaining why this makes the $E^\circ$ value more positive.

Why it’s wrong:

A more positive $E^\circ$ means the reduction ($Mn^{3+} + e^- \rightarrow Mn^{2+}$) is more spontaneous. You must connect the stability of the product ($Mn^{2+}$) to the driving force for the reaction.

How to Avoid It:

Always link the electronic configuration to extra stabilization energy.

• Correct logic: $Mn^{2+}$ ($3d^5$) has zero exchange energy loss upon reduction because all spins are parallel. $Mn^{3+}$ ($3d^4$) has one paired electron, so it is less stable. The jump from an unstable $Mn^{3+}$ to a highly stable $Mn^{2+}$ releases a lot of energy, making the reduction potential very positive.
• Compare: $Fe^{3+}$ ($3d^5$) is also half-filled, so $Fe^{3+}$ is already very stable. Reducing it to $Fe^{2+}$ ($3d^6$) actually loses that extra stability, so the $E^\circ$ is less positive.

Mistake 2: Confusing the Direction of the Reaction

The Mistake:

Students think a positive $E^\circ$ means the oxidation is easy. They might say "Mn is easily oxidized to Mn³⁺" which is the opposite of what the data shows.

The Fact:

The given couple is $Mn^{3+}/Mn^{2+}$. A high positive $E^\circ$ means $Mn^{3+}$ is a strong oxidizing agent (it wants to get reduced to $Mn^{2+}$). It does not mean Mn metal is easily oxidized.

How to Avoid It:

• Memorize the sign convention: $E^\circ > 0$ means the reduction is spontaneous relative to SHE.
• Use a mnemonic: "Positive potential = reduction is potent."
• Check the species: $Mn^{3+}$ is the reactant (oxidizing agent). A high $E^\circ$ tells you $Mn^{3+}$ is unstable and wants to become $Mn^{2+}$.

Mistake 3: Ignoring the Role of Hydration Enthalpy

The Mistake:

Students only discuss the $d^5$ configuration and forget that hydration enthalpy also plays a role, especially for $Cr^{3+}/Cr^{2+}$.

Why it matters:

$Cr^{3+}$ has a $t_{2g}^3$ configuration (half-filled in the $t_{2g}$ set) and a high charge (+3) with a small ionic radius. This gives it an exceptionally high hydration enthalpy. This extra stabilization of $Cr^{3+}$ makes it harder to reduce it to $Cr^{2+}$, resulting in a less positive $E^\circ$ for $Cr^{3+}/Cr^{2+}$.

How to Avoid It:

• Always check for CFSE effects: For $d^3$ and $d^8$ configurations in octahedral fields, the CFSE is very high.
• Compare the two factors:
  • For Mn: Electronic configuration (half-filled stability of $Mn^{2+}$) dominates.
  • For Cr: High hydration enthalpy of $Cr^{3+}$ (due to high charge and CFSE) dominates, making $Cr^{3+}$ very stable and the reduction potential low.

Mistake 4: Forgetting the Exact $E^\circ$ Values

The Mistake:

Students give a vague answer like "Mn has a higher value" without quoting the numbers.

Why it’s a problem: …