Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25 °C:
$\quad$ $\quad$ 3I₂(s) + 2Fe(s) → 2Fe³⁺(aq) + 6I^- (aq)
E°(I₂/I^-) = + 0.54 V and E°(Fe³⁺/Fe) = -0.036 V

Question

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25 °C:

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3I₂(s) + 2Fe(s) → 2Fe³⁺(aq) + 6I^- (aq)
E°(I₂/I^-) = + 0.54 V and E°(Fe³⁺/Fe) = -0.036 V

Quizplus · Accepted Answer

The standard cell potential (E°cell) is calculated by subtracting the anode potential from the cathode potential: E°cell = E°(cathode) - E°(anode) = -0.036 V - 0.54 V = -0.576 V. The equilibrium constant (K) is related to E°cell by the Nernst equation: E°cell = (RT/nF)lnK. At 25°C, this simplifies to E°cell = (0.0257/n)lnK. For this reaction, n = 6. Solving for K gives K = e^(nFE°cell/RT) = e^(6 * 96485 * -0.576 / 8.314 * 298) ≈ 2.4 × 10^58.

Use the Tabulated Half-Cell Potentials to Calculate the Equilibrium Constant \quad

Use the Tabulated Half-Cell Potentials to Calculate the Equilibrium Constant $\quad$