Calculate the equilibrium constant, K, for the following reaction at 25 °C.
Fe3+(aq)+B(s)+6H2O(l)→Fe(s)+H3BO3(s)+3H3O+(aq)Fe3+(aq)+B(s)+6H2O(l)→Fe(s)+H3BO3(s)+3H3O+(aq)
The balanced reduction half-reactions for the above equation and their respective standard reduction potential values (E°) are as follows:
Fe3+(aq)+3e−→Fe(s)Fe3+(aq)+3e−→Fe(s) E°=-0.04V
H3BO3(s)+3H3O+(aq)+3e−→B(s)+6H2O(l)H3BO3(s)+3H3O+(aq)+3e−→B(s)+6H2O(l) E°=-0.8698V