Calculate the equilibrium constant, 𝐾, for the reaction shown at 25 °C. Fe3+(aq)+B(s)+6H2O(l)⟶Fe(s)+H3BO3(s)+3H3O+(aq)

Question

The balanced reduction half‑reactions for the equation and their respective standard reduction potential values (𝐸∘) are

Fe3+(aq)+3e−⟶Fe(s) 𝐸∘=−0.04 V

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

Answer 1

To calculate the equilibrium constant (K) for the given reaction, we need to use the Nernst equation:

E = E° - (RT/nF)ln(Q)

where E is the cell potential at nonstandard conditions, E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred, F is Faraday's constant, and Q is the reaction quotient.

At equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (K). Therefore, we can rewrite the Nernst equation as:

E = E° - (RT/nF)ln(K)

We can start by writing the overall balanced equation for the reaction:

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

This reaction can be split into two half-reactions:

Fe3+(aq) + 3e- ⟶ Fe(s)

H3BO3(s) + 3H3O+(aq) + 3e- ⟶ B(s) + 6H2O(l)

The standard cell potential (E°) for the overall reaction is the difference between the standard reduction potentials (E°) for the two half-reactions:

E°cell = E°cathode - E°anode

E°cell = 0 V - (-0.04 V) = 0.04 V

Now, we can use the Nernst equation to calculate the equilibrium constant (K) at 25°C (298.15 K):

E = E° - (RT/nF)ln(K)

0.04 V = -0.8698 V - (8.314 J/mol K / (3 mol e- * 96485 C/mol e-))(298.15 K)ln(K)

Simplifying and solving for K:

ln(K) = (-0.04 V + 0.8698 V) / [(8.314 J/mol K / (3 mol e- * 96485 C/mol e-))(298.15 K)]

ln(K) = 41.276

K = e^(41.276)

K ≈ 6.0 x 10^17

Therefore, the equilibrium constant (K) for the given reaction at 25°C is approximately 6.0 x 10^17.