May I get help on this please

Question

Calculate the equilibrium constant, K for the reaction shown at 25 degrees C

Fe3+(aq)+B(s)+6H2O(l)⟶Fe(s)+H3BO3(s)+3H3O+(aq)

The balanced reduction half‑reactions for the equation and their respective standard reduction potential values Enot are

Fe3+(aq)+3e−⟶Fe(s) 𝐸∘=−0.04 V

H3BO3(s)+3H3O+(aq)+3e−⟶B(s)+6H2O(l) 𝐸∘=−0.8698 V

𝐾=

Answer 1

lnK = -ΔG₀/RT and ΔG = -n_eFξ₀ where n is the mole of electrons transferred, F is Faraday's constant (96485 C/mole e), and ξ₀ is the voltage of the cell under standard conditions.

So K = exp(n_eFξ₀/RT)

Here n_e = 3 ξ₀ = .8698 - ,04 Volts, R = 8.314 J/mole-K , and T is 298.15 K