 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2019/paper-1/section-b/11/iii/solution

The reversible voltage follows from the two-electron hydrogen reaction:

E_{rev} = - \frac{Δ _{r} G ^{\circ}}{2 F} = \frac{237000}{193000} = 1.22798 V .

At the specified geometric current density, the magnitude of each activation loss follows its Tafel equation,

η = b lo g_{10} (j / j_{0})

η_{H} = 0.0295 lo g_{10} (10/1 0^{- 4}) = 0.1475 V, η_{O} = 0.09 lo g_{10} (10/1 0^{- 7}) = 0.7200 V .

Both losses reduce the delivered fuel cell voltage. Therefore $V = 0.36048 V$ and $P / A = jV = 3.60 W m^{- 2}$ . The very small oxygen exchange current is responsible for most of the activation loss. The calculation uses current densities on a common area basis, unit water activity and the stated gas conditions, and neglects the losses the problem excludes.

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