Question

Consider a hydrogen-oxygen fuel cell, an electrochemical cell that generates electricity from the chemical reaction

2 H₂(g) + O₂(g) → 2 H₂O(l) .
Yes, this is the same reaction as in additional question 1, but now we are looking at electrochemical aspects of it. On one side of the cell, H2 is pumped in, and the half-cell reaction is

2 H₂(g) + 4 OH^–(aq) → 4 H₂O(l) + 4 e– ; at the other side, O₂ is pumped in:

O₂(g) + 2 H₂O(l) + 4 e- → 4 OH^–(aq) .

a) Which half-cell is oxidation? Which is reduction?

b) Compute the standard cell potential, E°, from the standard half cell potentials given in your textbook.

c) Suppose the cell operates at room temperature (25°C) and with H₂ partial pressure of 0.10 bar and O₂ partial pressure 0.20 bar. Use the Nernst equation to compute the cell potential under these conditions.

Answer 1

2 H₂(g) + 4 OH^–(aq) → 4 H₂O(l) + 4 e– oxidation half reaction

O₂(g) + 2 H₂O(l) + 4 e- → 4 OH^–(aq) . reduction half reaction

b.2 H₂(g) + 4 OH^–(aq) → 4 H₂O(l) + 4 e–      E0 = 0.83v

O₂(g) + 2 H₂O(l) + 4 e- → 4 OH^–(aq) .       E0 = 0.4v

------------------------------------------------------------------------

2 H₂(g) + O₂(g)----------------------> 2 H₂O(l) E0 cell = 1.23v

c.

n = 4

Ecell   = E0 cell - 0.0592/n logQ

           = 1.23- 0.0592/4 log1/P²H2 PO2

        = 1.23- 0.0592/4 log1/(0.1)^2*0.2

      = 1.23- 0.0148*-2.6989    = 1.2699v >>>>answer