Question

Calculate the equilibrium constant, k, for the reaction shown at 25 °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 (e∘) are fe3 (aq) 3e−⟶fe(s)h3bo3(s) 3h3o (aq) 3e−⟶b(s) 6h2o(l)e∘e∘

Answer 1

The complete question is ;

Calculate the equilibrium constant, K, for the reaction shown at 25 °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 (E?) are

Fe3+(aq)+3e------>Fe(s) E°=-0.04V

H3BO3(s)+3H3O+(aq)+3e------->B(s)+6H2O(l) E°=0.8698 V

Answer:

1.05

Explanation:

E° cell= E°cathode- E°anode

E°cathode = 0.8698 V

E°anode= -0.04V

E°cell=0.8698 V - (-0.04V)

E°cell= 0.9098 V

E°cell= 0.0592/n logK

From the balanced redox reaction equations above, n=3

0.9098 V= 0.0592/3 logK

logK= 0.0217

K= Antilog (0.0217)

K= 1.05