The chemical reaction \\(2H_2(g)+O_2(g)<->2H_2O(l)\\) is an oxidation-reduction

Question

The chemical reaction      (2H_2(g)+O_2(g)<->2H_2O(l))      is an oxidation-reduction reaction, and therefore can be used as an electrochemical cell. Such a cell is known as a fuel cell, and is one of the foundations on which the so-called hydrogen economy is based. The fuel cell runs under basic conditions according to the following half reactions.

      (2H_2O(l)+2e^-<->H_2(g)+2OH^-(aq))          (E^0=-0.8277V)     

       (O_2(g)+2H_2O(l)+4e^-<->4OH^-(aq))          (E^0=+0.40V)     

a. What is oxidized in the fuel cell?

b. What is reduced in the fuel cell reaction?

c. If this were an irreversible reaction, what would be the standard potential, in V, of the fuel cell?

d. Calculate the actual standard potential, in V, for the cell.

e. Calculate the actual equilibrium constant for the cell at 25 degrees Celcius.

f. Calculate the maximal electrical work, in J, obtainable from the fuel cell reaction under standard conditions. [consider the fuel cell as the system in assisging a sign to the work]

g. Predict whether the maximal work obtainable from the fuel cell would increase, decrease, or remain the same if the temperature were increased. Explain you reasoning.

Explanation / Answer

Anode Reaction: 2H2 + 2O2? ? 2H2O + 4e? Cathode Reaction: O2 + 4e

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