The standard electrode potential (E°) for the half-cell reaction Fe3+ + e- -> Fe2+ at 298 K is (Given: E°(Fe3+/Fe) = -0.04 V and E°(Fe2+/Fe) = -0.44 V at 298 K)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q56 – NEET-UG 2026

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Home Chemistry ElectrochemistryQuestion #409

The standard electrode potential (

E^{\circ}

) for the half-cell reaction

{Fe}^{3 +} + e^{-} \to {Fe}^{2 +}

298 K

is (Given:

E^{\circ} ({Fe}^{3 +} / Fe) = - 0.04 V

and

E^{\circ} ({Fe}^{2 +} / Fe) = - 0.44 V

298 K

)

Clue 1 of 3: Standard Electrode Potentials

Standard electrode potentials are intensive properties and cannot be directly added or subtracted.

🧠 Full Solution Path

Step 1: Calculate Gibbs Free Energy for Given Reactions✦ Active

The relationship between standard Gibbs free energy ( $Δ G^{\circ}$ ) and standard electrode potential ( $E^{\circ}$ ) is $Δ G^{\circ} = - n F E^{\circ}$ , where $n$ is the number of electrons transferred and $F$ is Faraday's constant. We calculate $Δ G^{\circ}$ for the given reactions:

{Fe}^{3 +} + 3 e^{-} \to Fe (n_{1} = 3, E_{1}^{\circ} = - 0.04 V)

Δ G_{1}^{\circ} = - 3 \times F \times (- 0.04 V) = + 0.12 F

{Fe}^{2 +} + 2 e^{-} \to Fe (n_{2} = 2, E_{2}^{\circ} = - 0.44 V)

Δ G_{2}^{\circ} = - 2 \times F \times (- 0.44 V) = + 0.88 F

💡 Teacher's Secret Hint

Remember that standard electrode potentials are intensive properties, so they cannot be directly added or subtracted. Convert them to extensive Gibbs free energy values first.

Step 2: Combine Reactions to Find Target

Δ G^{\circ}

○ Expand

The target reaction is ${Fe}^{3 +} + e^{-} \to {Fe}^{2 +}$ . This reaction can be obtained by subtracting the second given reaction from the first (or adding the first reaction to the reverse of the second reaction):

({Fe}^{3 +} + 3 e^{-} \to Fe) - ({Fe}^{2 +} + 2 e^{-} \to Fe)

{Fe}^{3 +} + 3 e^{-} - {Fe}^{2 +} - 2 e^{-} \to Fe - Fe

{Fe}^{3 +} + e^{-} \to {Fe}^{2 +}

Δ G_{3}^{\circ} = Δ G_{1}^{\circ} - Δ G_{2}^{\circ}

Δ G_{3}^{\circ} = (+ 0.12 F) - (+ 0.88 F) = - 0.76 F

💡 Teacher's Secret Hint

Ensure the number of electrons ( $n$ ) is correctly identified for each half-reaction when calculating $Δ G^{\circ}$ .

Step 3: Calculate Standard Electrode Potential for Target Reaction○ Expand

For the target reaction ${Fe}^{3 +} + e^{-} \to {Fe}^{2 +}$ , the number of electrons transferred is $n_{3} = 1$ . Using the relationship $Δ G_{3}^{\circ} = - n_{3} F E_{3}^{\circ}$ :

- 0.76 F = - 1 \times F \times E_{3}^{\circ}

E_{3}^{\circ} = \frac{- 0.76 F}{- 1 F} = + 0.76 V

💡 Teacher's Secret Hint

Pay close attention to the signs during calculations, especially when subtracting $Δ G^{\circ}$ values.

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Chemistry Question 56 – NEET-UG 2026

Clue 1 of 3: Standard Electrode Potentials

🧠 Full Solution Path