Given at $298\mathrm{~K}$: $\mathrm{E}_{\mathrm{Fe}^{2+}/\mathrm{Fe}}^{\ominus}=\mathrm{X}$ Volt $\mathrm{E}_{\mathrm{Fe}^{3+}/\mathrm{Fe}}^{\ominus}=\mathrm{Y}$ Volt The $\mathrm{E}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\ominus}$ in Volt at 298 K is given by :

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Redox Reactions and Electrochemistry

Given at

298 K

E_{Fe^{2 +} / Fe}^{⊖} = X

Volt

E_{Fe^{3 +} / Fe}^{⊖} = Y

Volt

The

E_{Fe^{3 +} / Fe^{2 +}}^{⊖}

in Volt at 298 K is given by :

Answer: (B)

Express each reduction through its free energy,

Δ G^{⊖} = - n F E^{⊖}

. For

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

Δ G_{1}^{⊖} = - 2 F X

; for

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

Δ G_{2}^{⊖} = - 3 F Y

. The couple

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

is obtained by subtracting the first reaction from the second, so

Δ G_{3}^{⊖} = Δ G_{2}^{⊖} - Δ G_{1}^{⊖} = - 3 F Y + 2 F X

. Since this couple transfers one electron,

Δ G_{3}^{⊖} = - F E_{F e^{3 +} /F e^{2 +}}^{⊖}

, and dividing by

- F

gives

E_{F e^{3 +} /F e^{2 +}}^{⊖} = 3 Y - 2 X

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option