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What are E°cell and ∆G° of a redox reaction at 25 °C for which n =...

What are E°cell and ∆G° of a redox reaction at 25 °C for which n = 1 and K = 5.0 x 10^4?
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Answer #1

We know that ΔG0 = -R*T*ln K where T = absolute temperature of the reaction = 25°C = (25 + 273) K = 298 K; plug in values.

ΔG0 = -(8.314 J/mol.K)*(298 K)*ln (5.0*104) = -26806.77972 J/mol = -(26806.77972 J/mol)*(1 kJ/1000 J) = -26.80677972 kJ/mol ≈ -26.81 kJ/mol (ans).

Again, we know that ΔG0 = -n*F*E0cell where n = 1 mole and F = 96485 J/V.mol. Plug in values.

-26806.77972 J/mol = -(1 mole)*(96485 J/V.mol)*E0cell

====> E0cell = 0.27787 V ≈ 0.278 V (ans).

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