Consider a fuel cell that uses the reaction of ethanol with oxygen to produce el

Question

Consider a fuel cell that uses the reaction of ethanol with oxygen to produce electricity,

Calculate the equilibrium constant, K, for the following reaction at 25 degreeC. 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 (Edegree) are as follows: Fe^3+(aq)+3e^- rightarrow Fe(s) Edegree = -0.04 v H_3BO_3(s)+3H_3O^+(aq)+3e^- rightarrow B(s)+6H_2O(l) Edegree = -0.8698 v K = 0.4698

Explanation / Answer

Solution :-

Lets calculate the Eo cell using the standard reuction potentials

Fe^3+ is reduced at cathode and B is oxidized at anode

Eo cell = Eo cathode – Eo anode

            = -0.04 V – (-0.8698 V)

            = 0.8298 V

Now lets calculate the equilibrium constant

Eo cell = (0.0592 / n ) log K

n= 3 electrons

0.8298 V = (0.0592 / 3 ) log K

0.8298 V / (0.0592 / 3 ) = log K

42.05 = log K

Antilog [42.05] = K

1.12*10^42 = K

So the value of the equilibrium constant K is 1.12*10^42

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