Q.Use the data given in Q.8 ($E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$) and find out the most stable ion in its reduced form.

Concept: Standard Electrode Potentials – A more positive reduction potential means the oxidised form is a stronger oxidising agent, and the corresponding reduced form is weaker (less stable) as a reducing agent. Conversely, a more negative reduction potential means the reduced form is a stronger reducing agent (less stable in its reduced state). The most stable reduced form is the one that is hardest to oxidise — i.e., the one with the least negative (or most positive) reduction potential for its own oxidation.

Step 1: For each species in its reduced form, we need the potential for the reverse reaction (oxidation). That is simply the negative of the given reduction potential.

Step 2:

• For $Cl^-$: oxidation to $Cl_2$ has $E^\circ = -1.36\ V$
• For $Cr^{3+}$: oxidation to $Cr_2O_7^{2-}$ has $E^\circ = -1.33\ V$ …

The question asks: among $Cl^-$, $Cr^{3+}$, $Cr$ (metal), and $Mn^{2+}$, which is the most stable in its reduced form?

“Stable in reduced form” means the species has little tendency to get oxidised back — it prefers to stay as it is. In electrochemistry, that tendency is measured by the standard reduction potential $E^\circ$ of the corresponding half‑reaction.

But careful: we want the reduced form to be stable. That means the reduced form should have a low tendency to get oxidised. The tendency to get oxidised is the reverse of the reduction potential. So we look at the reduction potential of the oxidised form — if that potential is very positive, the oxidised form is a strong oxidiser and the reduced form is weak (unstable as a reductant). Conversely, if the reduction potential is very negative, the reduced form is a strong reductant (easily oxidised, hence unstable in its reduced state).

Therefore, the most stable reduced form corresponds to the most positive reduction potential of the oxidised species. Let’s list the given data properly.

1. For $Cl^-$: The half‑reaction is

The reduced form is $Cl^-$. The potential is quite positive, so $Cl_2$ is a strong oxidiser and $Cl^-$ is a weak reductant — fairly stable.

2. For $Cr^{3+}$: There are two relevant half‑reactions:

and

The reduced form $Cr^{3+}$ appears on the right of the first reaction and on the left of the second. To judge its stability, we need the potential for the half‑reaction that produces $Cr^{3+}$ from a higher oxidation state — that’s the $+1.33\ \text{V}$ one. But also, $Cr^{3+}$ can be further reduced to $Cr$ metal at $-0.74\ \text{V}$, which tells us $Cr^{3+}$ is a weak oxidiser (hard to reduce). However, the question asks about stability of the reduced form — here $Cr^{3+}$ is the reduced form relative to dichromate. Its $E^\circ = +1.33\ \text{V}$ is slightly less positive than $Cl_2/Cl^-$.

3. For $Cr$ (metal): The half‑reaction is $Cr^{3+} + 3e^- \rightarrow Cr \quad E^\circ = -0.74\ \text{V}$ …

Method: Comparing Standard Reduction Potentials ($E^\circ$)

Concept: The most stable species in its reduced form is the one that is hardest to oxidise back — i.e. the one whose corresponding couple has the highest (most positive) standard reduction potential. A high $E^\circ$ means the oxidised form has a strong pull for electrons, so once reduced, that species has little desire to give the electrons back — it is stable in its reduced state. (Do not confuse this with reducing-agent strength, which runs the opposite way: a very negative $E^\circ$, like $Cr^{3+}/Cr$, means the reduced form — metallic Cr — is a strong reducing agent, i.e. UNSTABLE, because it readily gives up electrons again.)

Steps

1. List the given half-reactions with their $E^\circ$ values

   • $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ $E^\circ = +1.33\ V$
   • $Cl_2 + 2e^- \rightarrow 2Cl^-$ $E^\circ = +1.36\ V$
   • $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ $E^\circ = +1.51\ V$
   • $Cr^{3+} + 3e^- \rightarrow Cr$ $E^\circ = -0.74\ V$

2. Identify the reduced form in each case

   • $Cl_2$ (oxidised) → $Cl^-$ (reduced)
   • $Cr_2O_7^{2-}$ (oxidised) → $Cr^{3+}$ (reduced)
   • $MnO_4^-$ (oxidised) → $Mn^{2+}$ (reduced)
   • $Cr^{3+}$ (oxidised) → $Cr$ (reduced)

3. Compare the $E^\circ$ values of the half-reactions …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "most stable reduced form" with "strongest reducing agent"

The error: Students often think the species with the most negative $E^\circ$ is the most stable reduced form. They pick Cr ($-0.74\ V$) because it's the most negative.

Why it's wrong:

• A more negative $E^\circ$ means the reduced form is a stronger reducing agent (easily gets oxidised) — meaning it is less stable in its reduced state.
• A more positive $E^\circ$ means the reduced form is harder to oxidise — hence more stable.

How to avoid:

• Remember: Higher $E^\circ$ → Reduced form is more stable (it "wants" to stay reduced).
• Think of it like a tug-of-war: a high positive $E^\circ$ means the oxidised form pulls electrons strongly, so the reduced form is content and stable.

Mistake 2: Comparing potentials of half-cells with different numbers of electrons

The error: Students directly compare $E^\circ$ values without noticing that some half-reactions involve different $n$ (number of electrons). For example:

• $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ ($n=6$)
• $Cl_2 + 2e^- \rightarrow 2Cl^-$ ($n=2$)
• $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ ($n=5$)
• $Cr^{3+} + 3e^- \rightarrow Cr$ ($n=3$)

Why it's wrong:

• $E^\circ$ is an intensive property — it does not depend on $n$. So comparing $E^\circ$ directly is actually correct here.
• However, some students mistakenly try to "scale" $E^\circ$ by multiplying with $n$ (like $\Delta G^\circ = -nFE^\circ$). That would be wrong for stability comparison.

How to avoid:

• For stability of the reduced form, compare $E^\circ$ values directly — not $\Delta G^\circ$ or $n \times E^\circ$.
• Only use $\Delta G^\circ$ if the question asks for spontaneity or thermodynamic favourability of a specific reaction.

Mistake 3: Forgetting to check which species is the "reduced form"

The error: Students look at the given $E^\circ$ values and pick the oxidised form (like $MnO_4^-$ or $Cr_2O_7^{2-}$) instead of the reduced form.

Why it's wrong:

• The question asks for the most stable ion in its reduced form.
• For each half-cell, the reduced form is:
  • $Cl^-$ (from $Cl_2/Cl^-$)
  • $Cr^{3+}$ (from $Cr_2O_7^{2-}/Cr^{3+}$)
  • $Mn^{2+}$ (from $MnO_4^-/Mn^{2+}$)
  • $Cr$ (from $Cr^{3+}/Cr$)

How to avoid:

• Always write the half-reaction in the reduction direction:

• The species on the right is the reduced form — that's what you compare.

Mistake 4: Misinterpreting the sign of $E^\circ$ for $Cr^{3+}/Cr$ …