Calculate the equilibrium constant, K, for the following reaction at 25 °C.
Fe3+(aq) + B(s) + 6H 2O(l) → Fe(s)+ H3BO3(s) + 3H3O+(aq)
The balanced reduction half-reactions for the above equation and their respective standard reduction potential values (E degree) are as follows:
Fe3+(aq) + 3e- → Fe(s) E° = - 0.04 V
H3BO3(s) + 3H3O+(aq) + 3e- → B(s) + 6H 2O(l) E° = - 0.8698 V
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