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JEE MAIN 2026 · Redox Reactions and Electrochemistry

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 56

Updated
Single correct+4 / -1Answer matches NTA’s final key

Given at 298 K: EFe2+/Fe⊖=XE^{\ominus}_{\text{Fe}^{2+}/\text{Fe}} = X Volt and EFe3+/Fe⊖=YE^{\ominus}_{\text{Fe}^{3+}/\text{Fe}} = Y Volt.

The EFe3+/Fe2+⊖E^{\ominus}_{\text{Fe}^{3+}/\text{Fe}^{2+}} in Volt at 298 K is given by:

Choose one

Solution

Solution

Electrode potentials do not add, but free energies do: ΔG⊖=−nFE⊖\Delta G^\ominus = -nFE^\ominus.

  • Fe3+^{3+} + 3e⁻ → Fe: ΔG1=−3FY\Delta G_1 = -3FY
  • Fe2+^{2+} + 2e⁻ → Fe: ΔG2=−2FX\Delta G_2 = -2FX

Subtracting the second from the first gives Fe3+^{3+} + e⁻ → Fe2+^{2+}:

−1⋅FE⊖=−3FY+2FX  ⇒  E⊖=3Y−2X.-1 \cdot F E^\ominus = -3FY + 2FX \;\Rightarrow\; E^\ominus = 3Y - 2X.

Answer: 3Y−2X3Y - 2X.

Answer checked against the official answer key. The solution is our own working.

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