Question bank/NEET UG/Chemistry

NEET UGChemistryModerateRe-NEET 2026

The standard electrode potential (E∘E^{\circ}) for the half-cell reaction Fe3++e−→Fe2+Fe^{3+} + e^- \to Fe^{2+} at 298 K is

(Given : E∘(Fe3+/Fe)=−0.04E^{\circ}(Fe^{3+}/Fe) = -0.04 V and E∘(Fe2+/Fe)=−0.44E^{\circ}(Fe^{2+}/Fe) = -0.44 V at 298 K)

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ΔGFe3+/Fe∘=ΔGFe3+/Fe2+∘+ΔGFe2+/Fe∘\Delta G^{\circ}_{Fe^{3+}/Fe} = \Delta G^{\circ}_{Fe^{3+}/Fe^{2+}} + \Delta G^{\circ}_{Fe^{2+}/Fe}
−nFE∘=−nFE∘+−nFE∘-nFE^{\circ} = -nFE^{\circ} + -nFE^{\circ}
−3F(−0.04)=−1FE∘+2F(0.44)-3F(-0.04) = -1FE^{\circ} + 2F(0.44)
3F(0.04)=−Fe∘+0.88F3F(0.04) = -Fe^{\circ} + 0.88F
0.12=−E∘+0.880.12 = -E^{\circ} + 0.88
0.12−0.88=−E∘0.12 - 0.88 = -E^{\circ}
−0.76=−E∘-0.76 = -E^{\circ}
E∘=0.76E^{\circ} = 0.76 V

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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