Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 1 / Section B / 11 / iii

Table of contents
- Solution 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.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version