Given at 298 K : E°Fe2+/Fe = X Volt E°Fe3+/Fe = Y Volt The E°Fe3+/Fe2+ in Volt at 298 K is given by:

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

Chemistry Q56 – JEE-MAIN 2026

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Home Chemistry ElectrochemistryQuestion #403

Given at 298 K :

E_{{Fe}^{2 +} / Fe}^{\circ}

= X Volt

E_{{Fe}^{3 +} / Fe}^{\circ}

= Y Volt The

E_{{Fe}^{3 +} / {Fe}^{2 +}}^{\circ}

in Volt at 298 K is given by:

Clue 1 of 3: Gibbs Free Energy

Standard electrode potentials are not directly additive. Instead, their corresponding standard Gibbs free energies are additive.

🧠 Full Solution Path

Step 1: Write down the given half-reactions and their corresponding standard Gibbs free energy changes.✦ Active

The given standard electrode potentials correspond to the following half-reactions and their standard Gibbs free energy changes:

{Fe}^{2 +} (a q) + 2 e^{-} \to Fe (s); E_{1}^{\circ} = X V ⟹ Δ G_{1}^{\circ} = - 2 F X

{Fe}^{3 +} (a q) + 3 e^{-} \to Fe (s); E_{2}^{\circ} = Y V ⟹ Δ G_{2}^{\circ} = - 3 F Y

We need to find $E^{\circ}$ for the reaction:

{Fe}^{3 +} (a q) + e^{-} \to {Fe}^{2 +} (a q); E_{3}^{\circ} = ? ⟹ Δ G_{3}^{\circ} = - 1 F E_{3}^{\circ}

Step 2: Manipulate the half-reactions to obtain the desired reaction and sum their Gibbs free energies.○ Expand

To obtain the desired reaction ( ${Fe}^{3 +} \to {Fe}^{2 +}$ ), we can subtract the first reaction from the second reaction, or equivalently, add the reverse of the first reaction to the second reaction. Reversing the first reaction gives:

Fe (s) \to {Fe}^{2 +} (a q) + 2 e^{-}; Δ G_{1, r e v}^{\circ} = - (- 2 F X) = 2 F X

Adding this reversed reaction to the second reaction:

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

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

Therefore, the total Gibbs free energy change for the desired reaction is:

Δ G_{3}^{\circ} = Δ G_{2}^{\circ} + Δ G_{1, r e v}^{\circ} = - 3 F Y + 2 F X

Step 3: Calculate the standard electrode potential for the desired reaction.○ Expand

Substitute $Δ G_{3}^{\circ} = - 1 F E_{3}^{\circ}$ into the equation:

- 1 F E_{3}^{\circ} = 2 F X - 3 F Y

Dividing by $- F$ gives:

E_{3}^{\circ} = - (2 X - 3 Y) = 3 Y - 2 X

Thus, the standard electrode potential $E_{{Fe}^{3 +} / {Fe}^{2 +}}^{\circ}$ is $3 Y - 2 X$ Volt.

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Chemistry Question 56 – JEE-MAIN 2026

Clue 1 of 3: Gibbs Free Energy

🧠 Full Solution Path