For $M^{2+}/M$ and $M^{3+}/M^{2+}$ systems the $E^\circ$ values for some metals are as follows: | System | $E^\circ$ | System | $E^\circ$ | |---|---|---|---| | $Cr^{2+}/Cr$ | $-0.9$ V | $Cr^{3+}/Cr^{2+}$ | $-0.4$ V | | $Mn^{2+}/Mn$ | $-1.2$ V | $Mn^{3+}/Mn^{2+}$ | $+1.5$ V | | $Fe^{2+}/Fe$ | $-0.4$ V | $Fe^{3+}/Fe^{2+}$ | $+0.8$ V | Use this data to comment upon: (i) the stability of $Fe^{3+}$ in acid solution as compared to that of $Cr^{3+}$ or $Mn^{3+}$ and (ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

For M²⁺/M and M³⁺/M²⁺ systems the E^° values for s – Class 12 Chemistry

Exercises · 4.17

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

System	$E^\circ$	System	$E^\circ$
$Cr^{2+}/Cr$	$-0.9$ V	$Cr^{3+}/Cr^{2+}$	$-0.4$ V
$Mn^{2+}/Mn$	$-1.2$ V	$Mn^{3+}/Mn^{2+}$	$+1.5$ V
$Fe^{2+}/Fe$	$-0.4$ V	$Fe^{3+}/Fe^{2+}$	$+0.8$ V

Use this data to comment upon:

the stability of $Fe^{3+}$ in acid solution as compared to that of $Cr^{3+}$ or $Mn^{3+}$ and
the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

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Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).

The Core Intuition

An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.

Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.

The Precise Statement

Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.

This depends on three key factors:

Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ in some contexts).
Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ is more stable for tin.

Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.

The Core Idea: Energy Minimisation

An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:

Ionisation energy (energy needed to remove electrons)
Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
Electronic configuration (half-filled / fully-filled subshells)

There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.

Key Principle 1: Inert Pair Effect (for p-block elements)

Why it holds:

For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.

Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
Example: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

Derivation logic:

The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Why it holds:

A half-filled ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

Derivation logic:

The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.

Key Principle 3: Lattice Energy / Hydration Energy Compensation

For transition metals, stability of a particular oxidation state in aqueous solution depends on:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

Why it holds:

Higher oxidation states have higher ionisation energy (harder to remove electrons).
But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
The balance determines which state is stable.

Example:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds: …

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The key idea is to use the standard electrode potentials to compare the tendency of $M^{3+}$ to get reduced (stability in acid) and the ease of oxidation of $M$ to $M^{2+}$ . For (i), $Fe^{3+}$ is more stable than $Mn^{3+}$ but less stable than $Cr^{3+}$ in acid. For (ii), iron is harder to oxidise than chromium but easier than manganese.

Understanding the Concept: What the Potentials Tell Us

The two sets of potentials given — $E^\circ$ for $M^{2+}/M$ and $E^\circ$ for $M^{3+}/M^{2+}$ — are like two different thermometers for the same metal. The first tells us how easily the metal loses electrons to become $M^{2+}$ (oxidation of the metal). The second tells us how easily $M^{2+}$ loses another electron to become $M^{3+}$ (oxidation of the +2 ion), or equivalently, how stable the $M^{3+}$ ion is against being reduced back to $M^{2+}$ .

A more positive $E^\circ$ for $M^{3+}/M^{2+}$ means the half-reaction $M^{3+} + e^- \rightarrow M^{2+}$ is more spontaneous — so $M^{3+}$ is a stronger oxidising agent and is less stable in solution (it wants to grab an electron and become $M^{2+}$ ). Conversely, a more negative $E^\circ$ for $M^{3+}/M^{2+}$ means $M^{3+}$ is harder to reduce, hence more stable.

For the metal itself, a more negative $E^\circ$ for $M^{2+}/M$ means the oxidation $M \rightarrow M^{2+} + 2e^-$ is more spontaneous (since the reverse reduction is less favourable). So the metal is easier to oxidise.

Let's apply this logic step by step.

(i) Stability of $Fe^{3+}$ in acid solution compared to $Cr^{3+}$ and $Mn^{3+}$

1. Compare the $M^{3+}/M^{2+}$ potentials.

The relevant values are:

$Cr^{3+}/Cr^{2+}$ : $E^\circ = -0.4$ V
$Fe^{3+}/Fe^{2+}$ : $E^\circ = +0.8$ V
$Mn^{3+}/Mn^{2+}$ : $E^\circ = +1.5$ V

2. Interpret the sign and magnitude.

A more positive $E^\circ$ means the reduction $M^{3+} + e^- \rightarrow M^{2+}$ is more favourable. So $Mn^{3+}$ (with $+1.5$ V) is the strongest oxidising agent among the three — it is most eager to accept an electron and become $Mn^{2+}$ . That makes $Mn^{3+}$ the least stable in solution.

$Fe^{3+}$ ( $+0.8$ V) is less eager than $Mn^{3+}$ but still quite willing to be reduced.

$Cr^{3+}$ ( $-0.4$ V) has a negative potential — the reduction is non-spontaneous under standard conditions. So $Cr^{3+}$ is the most stable; it does not want to become $Cr^{2+}$ .

3. Order of stability.

Stability in acid solution: $Cr^{3+} > Fe^{3+} > Mn^{3+}$ .

Watch out

A common mistake is to think a more positive $E^\circ$ for $M^{3+}/M^{2+}$ means $M^{3+}$ is more stable. Actually, it means the opposite — it is more easily reduced, hence less stable.

4. Conclusion for (i).

$Fe^{3+}$ is more stable than $Mn^{3+}$ but less stable than $Cr^{3+}$ in acid solution.

--- …

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Method: Standard Electrode Potential Comparison for Redox Stability

This method uses the sign and magnitude of $E^\circ$ values to determine:

Stability of a higher oxidation state (from $M^{3+}/M^{2+}$ data)
Ease of oxidation of the metal (from $M^{2+}/M$ data)

Steps

Step 1: Understand what each $E^\circ$ tells you

System	Meaning	More positive $E^\circ$ →
$M^{2+}/M$	Tendency of $M^{2+}$ to get reduced to $M$	$M^{2+}$ is easier to reduce (less stable as ion)
$M^{3+}/M^{2+}$	Tendency of $M^{3+}$ to get reduced to $M^{2+}$	$M^{3+}$ is easier to reduce (less stable as $M^{3+}$ )

Step 2: Analyse stability of $M^{3+}$ (part i)

$Cr^{3+}/Cr^{2+}$ : $E^\circ = -0.4$ V

→ Negative $E^\circ$ means $Cr^{3+}$ is not easily reduced to $Cr^{2+}$

→ $Cr^{3+}$ is stable in acid
$Mn^{3+}/Mn^{2+}$ : $E^\circ = +1.5$ V

→ Large positive $E^\circ$ means $Mn^{3+}$ is very easily reduced to $Mn^{2+}$

→ $Mn^{3+}$ is unstable (it will readily accept electrons)
$Fe^{3+}/Fe^{2+}$ : $E^\circ = +0.8$ V

→ Positive $E^\circ$ , but smaller than +1.5 V

→ $Fe^{3+}$ is moderately stable — less stable than $Cr^{3+}$ , but more stable than $Mn^{3+}$

Conclusion for (i):

Stability order in acid: $\boxed{Cr^{3+} > Fe^{3+} > Mn^{3+}}$

$Fe^{3+}$ is less stable than $Cr^{3+}$ but much more stable than $Mn^{3+}$ .

Step 3: Analyse ease of oxidation of the metal (part ii)

Oxidation of metal: $M \rightarrow M^{2+} + 2e^-$

This is the reverse of the given $M^{2+}/M$ reduction. …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing "Stability" with "Reducing Power"

The error: Students often think a more negative $E^\circ$ for $M^{3+}/M^{2+}$ means the ion is more stable. They see $Cr^{3+}/Cr^{2+} = -0.4$ V and think $Cr^{3+}$ is "unstable" because the value is negative.

Why it's wrong: Stability of a higher oxidation state (like $+3$ ) in acid solution depends on how easily it gets reduced to the $+2$ state. A more positive $E^\circ$ for $M^{3+}/M^{2+}$ means the $M^{3+}$ ion is a stronger oxidising agent — it readily accepts electrons and gets reduced. This makes the $+3$ state less stable in solution.

How to avoid: Remember the rule:

More positive $E^\circ$ for $M^{3+}/M^{2+}$ → $M^{3+}$ is easily reduced → less stable
More negative $E^\circ$ for $M^{3+}/M^{2+}$ → $M^{3+}$ is hard to reduce → more stable

For the given data:

$Cr^{3+}/Cr^{2+} = -0.4$ V → $Cr^{3+}$ is most stable
$Fe^{3+}/Fe^{2+} = +0.8$ V → $Fe^{3+}$ is moderately stable
$Mn^{3+}/Mn^{2+} = +1.5$ V → $Mn^{3+}$ is least stable (strongest oxidiser)

Mistake 2: Misinterpreting "Ease of Oxidation" for Metals

The error: Students look at $M^{2+}/M$ values and think a more positive value means easier oxidation. They see $Fe^{2+}/Fe = -0.4$ V and $Cr^{2+}/Cr = -0.9$ V and conclude iron oxidises more easily.

Why it's wrong: Oxidation of a metal $M$ to $M^{2+}$ is the reverse of the reduction reaction $M^{2+} + 2e^- \rightarrow M$ . A more negative $E^\circ$ for $M^{2+}/M$ means the reduction is less favourable — so the reverse (oxidation) is more favourable.

How to avoid: Flip your thinking:

More negative $E^\circ$ for $M^{2+}/M$ → metal is easier to oxidise (stronger reducing agent)
Less negative $E^\circ$ for $M^{2+}/M$ → metal is harder to oxidise

For the given data:

$Mn^{2+}/Mn = -1.2$ V → easiest to oxidise
$Cr^{2+}/Cr = -0.9$ V → moderate
$Fe^{2+}/Fe = -0.4$ V → hardest to oxidise among the three

Mistake 3: Forgetting to Compare Both Systems Together

The error: Students answer part (i) using only $M^{3+}/M^{2+}$ data and part (ii) using only $M^{2+}/M$ data, without linking the two.

Why it's wrong: The question asks for comparative statements. For part (ii), "ease with which iron can be oxidised" refers to $Fe \rightarrow Fe^{2+}$ , which uses $Fe^{2+}/Fe$ data. But the comparison with Cr and Mn must use their respective $M^{2+}/M$ values.

How to avoid: Write your answer in clear comparative sentences:

Part (i):

$Cr^{3+}$ is most stable (most negative $E^\circ$ for $Cr^{3+}/Cr^{2+}$ )
$Fe^{3+}$ is moderately stable
$Mn^{3+}$ is least stable (most positive $E^\circ$ )

Part (ii):

$Mn$ is easiest to oxidise (most negative $E^\circ$ for $Mn^{2+}/Mn$ )
$Cr$ is next …

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Understanding the Concept: What the Potentials Tell Us

(i) Stability of Fe3+Fe^{3+}Fe3+ in acid solution compared to Cr3+Cr^{3+}Cr3+ and Mn3+Mn^{3+}Mn3+

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(i) Stability of $Fe^{3+}$ in acid solution compared to $Cr^{3+}$ and $Mn^{3+}$

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