What is E°(Fe³⁺/Fe²⁺) given E°(Fe²⁺/Fe)=X and E°(Fe³⁺/Fe)=Y?

3Y−2X. Using ΔG°=−nFE°: ΔG₃=ΔG₂−ΔG₁=−3FY−(−2FX)=−F(3Y−2X). Since n=1: E°₃=3Y−2X.

Why can't E° values be directly subtracted?

E° is intensive but ΔG°=−nFE° is extensive. Different half-reactions have different n values, so you must work with ΔG° (additive) and then extract E° at the end.

E°(Fe²⁺/Fe)=X E°(Fe³⁺/Fe)=Y find E°(Fe³⁺/Fe²⁺) at 298K

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ChemistryElectrochemistry

QMCQElectrochemistry

Given at 298 K:
$E^\circ(\text{Fe}^{2+}/\text{Fe}) = X$ Volt and $E^\circ(\text{Fe}^{3+}/\text{Fe}) = Y$ Volt

The $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+})$ in Volt at 298 K is given by:

A) $2X-3Y$ B) $3Y-2X$ C) $3Y+2X$ D) $Y+X$

✅ Correct Answer

B) 3Y−2X

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Solution

Write the half-cell reactions with ΔG values

Reaction 1: Fe²⁺ + 2e⁻ → Fe,   $\Delta G_1 = -2FX$
Reaction 2: Fe³⁺ + 3e⁻ → Fe,   $\Delta G_2 = -3FY$
Target Reaction 3: Fe³⁺ + e⁻ → Fe²⁺,   $\Delta G_3 = ?$

Use Hess's law for ΔG

Reaction 3 = Reaction 2 − Reaction 1 (subtract to cancel Fe):

$\Delta G_3 = \Delta G_2 - \Delta G_1 = (-3FY) - (-2FX) = -3FY + 2FX$

Find E° from ΔG₃ = −nFE°

$\Delta G_3 = -1\times F\times E^\circ_3$

$-1\times F\times E^\circ_3 = -3FY + 2FX$

$E^\circ_3 = 3Y - 2X$

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Theory & Key Concepts

Why E° Values Are NOT Simply Additive

A crucial and frequently tested concept: standard electrode potentials (E°) cannot be directly added or subtracted to get another electrode potential. Only Gibbs free energies (ΔG°) are additive through Hess's law.

The fundamental relation: $\Delta G^\circ = -nFE^\circ$, where $n$ is the number of electrons transferred in the half-reaction, $F$ is Faraday's constant (96485 C/mol), and $E^\circ$ is the standard electrode potential.

Why this matters: E° is an intensive property (independent of amount), but ΔG° is extensive (depends on how many moles). When we combine reactions, we must work with ΔG°, not E° directly. This is why E°(Fe³⁺/Fe²⁺) ≠ E°(Fe³⁺/Fe) − E°(Fe²⁺/Fe).

Numerical values: Actual values are E°(Fe²⁺/Fe) = −0.44 V and E°(Fe³⁺/Fe) = −0.04 V. Therefore E°(Fe³⁺/Fe²⁺) = 3(−0.04) − 2(−0.44) = −0.12 + 0.88 = +0.77 V. This confirms that Fe³⁺ is a moderate oxidising agent.

Memory trick: Use $-nFE°$: multiply each E° by its electron count, apply Hess's law for ΔG, then divide by the new n to get the new E°.

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Important Concepts

ΔG° = −nFE° is the Key

E° values cannot be added directly. Use ΔG°=−nFE°: ΔG₁=−2FX (2 electrons), ΔG₂=−3FY (3 electrons). Target reaction = reaction 2 − reaction 1. ΔG₃=ΔG₂−ΔG₁=−3FY+2FX. Since n=1 electron: E₃=(3Y−2X).

Why Not Simple Subtraction?

E°(Fe³⁺/Fe²⁺) ≠ E°(Fe³⁺/Fe) − E°(Fe²⁺/Fe) = Y−X. This would only work if both reactions involved the same number of electrons. Since one involves 2e⁻ and the other 3e⁻, you must use ΔG (which scales with n).

Verification With Real Values

E°(Fe²⁺/Fe)=−0.44V, E°(Fe³⁺/Fe)=−0.04V. Using 3Y−2X: 3(−0.04)−2(−0.44)=−0.12+0.88=+0.77V. Standard value of E°(Fe³⁺/Fe²⁺)=+0.77V ✓

Hess's Law for ΔG

ΔG is extensive (like enthalpy), so it follows Hess's law exactly. Reaction 3 = Reaction 2 − Reaction 1 means ΔG₃=ΔG₂−ΔG₁. This is the correct approach for combining half-reactions.

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FAQs

Why can't we just subtract E° values?

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E° is intensive (per electron, essentially), but ΔG°=−nFE° is extensive. Adding reactions means adding ΔG°, not E°. So 3E°₂−2E°₁ = E°₃ × 1, giving E°₃=3Y−2X.

What is Reaction 3 = Reaction 2 − Reaction 1?

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Reaction 2: Fe³⁺+3e⁻→Fe. Reaction 1: Fe²⁺+2e⁻→Fe. Subtracting: Fe³⁺+3e⁻−Fe²⁺−2e⁻→0, i.e., Fe³⁺+e⁻→Fe²⁺. Correct! n=1.

What is F in the formula?

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F = Faraday's constant = 96485 C/mol ≈ 96500 C/mol. It cancels out in the final answer.

Actual numerical answer?

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X=E°(Fe²⁺/Fe)=−0.44V, Y=E°(Fe³⁺/Fe)=−0.04V. E°(Fe³⁺/Fe²⁺)=3(−0.04)−2(−0.44)=+0.77V.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026 Chemistry Section A.

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