Calculate the equilibrium constant, K, for the following reaction at 25 degree C. Fe^3 + (aq) + B (s) + 6H_2O(l) rightarrow Fe(s) + H_3BO_3(s) + 3H_3O^+ (aq) The balanced reduction half-reactions for the above equation and their respective standard reduction potential values (E degree) are as follows: Fe^3 + (aq) +3e^- rightarrow Fe(s) E degree = -0.04 V H_3BO_3(s) + 3H_3O^+(aq) + 3e^- rightarrow B(s) + 6H_2O(l) E degree = -0.8698 V K =