For the half‑reaction Fe^3++e^-→ Fe^2+, E° = +0.77 V at 25°C (n=1). Calculate the equilibrium constant K for this half‑reaction. (Take RT/F = 0.0257 V.)

For the half‑reaction $F e^{3 +} + e^{-} \to F e^{2 +}$ , $E^{\circ} = + 0.77 V$ at $2 5^{\circ} C$ (n=1). Calculate the equilibrium constant $K$ for this half‑reaction. (Take $R T / F = 0.0257 V$ .)

medium MCQ 2 marks Class 12 · Chemistry · Electrochemistry

Answer

K = 1.0 \times 1 0^{13}

The correct answer is $K = 1.0 \times 1 0^{13}$ — Convert to $K$ : $K = e^{29.96} \approx 1.0 \times 1 0^{13}$ .

Use the relation $Δ G^{\circ} = - n F E^{\circ}$ and $Δ G^{\circ} = - R T ln K$ . Equate: $- n F E^{\circ} = - R T ln K$ .
Solve for $ln K$ : $ln K = \frac{n E ^{\circ}}{R T / F} = \frac{E ^{\circ}}{0.0257}$ .
Substitute $E^{\circ} = 0.77 V$ : $ln K = \frac{0.77}{0.0257} = 29.96$ .
Convert to $K$ : $K = e^{29.96} \approx 1.0 \times 1 0^{13}$ .

Common mistake: Using

R T

instead of

R T / F

in the exponent, giving a much smaller

K

Concept tested: Link between electrode potential and equilibrium constant