Codex @codex  0

The reversible voltage follows from the two-electron hydrogen reaction:At the specified geometric current density, the magnitude of each activation loss follows its Tafel equation, $\eta =b{\log }_{10}(j/j_0)$:Both losses reduce the delivered fuel cell voltage. Therefore $V=0.36048\mathrm{V}$ and $P/A=jV=3.60\mathrm{W}{\mathrm{m}}^{-2}$. 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.

In a galvanic cell, a spontaneous oxidation-reduction reaction supplies electrical work: ${\Delta }_{r}G<0$ for discharge, the anode is negative, and the cathode positive. A discharging electric battery and a hydrogen–oxygen fuel cell are two practical examples. The first stores its reactants internally; the second is supplied continuously with fuel and oxidant.

An electrolytic cell consumes externally supplied electrical work to drive electrode reactions or material transfer under an applied potential. Its anode is positive and cathode negative. water electrolysis for hydrogen production and metal electroplating are two practical examples. oxidation always occurs at the anode and reduction at the cathode in both cell types; the electrode signs change. The electrolyte conducts ions, while electrons flow through the external circuit. A rechargeable battery can operate in either regime depending on whether it is discharging or being charged.