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

The oxygen-pressure exponents can be explained by a change in the dominant charge state of metal vacancies, with holes carrying the p-type current. For a divalent metal oxide, a doubly negative metal vacancy is $V_{\mathrm{M}}^{\prime \prime }$ in Kröger–Vink notation. A conventional oxygen-incorporation reaction isTreating regular oxygen-site activity as constant, the law of mass action gives $K_2=[V_{\mathrm{M}}^{\prime \prime }]p_h^2/p_{{\mathrm{O}}_2}^{1/2}$. If these are the compensating defects, charge neutrality gives $p_h=2[V_{\mathrm{M}}^{\prime \prime }]$, hence $p_h^3\propto p_{{\mathrm{O}}_2}^{1/2}$ and $\sigma \propto p_{{\mathrm{O}}_2}^{1/6}$, assuming nearly constant hole mobility.

For singly negative vacancies, useNow $p_h=[V_{\mathrm{M}}^{\prime }]$ and $p_h^2\propto p_{{\mathrm{O}}_2}^{1/2}$, so $\sigma \propto p_{{\mathrm{O}}_2}^{1/4}$. The charge-state equilibrium $V_{\mathrm{M}}^{\prime }\rightleftharpoons V_{\mathrm{M}}^{\prime \prime }+h^{\bullet }$ gives $[V_{\mathrm{M}}^{\prime \prime }]/[V_{\mathrm{M}}^{\prime }]=K_{\mathrm{ion}}/p_h$: the singly charged fraction can therefore dominate as oxygen pressure and hole concentration increase. This is a consistent dilute-defect model, not a unique experimental identification of every defect. Changes in mobility, defect association or other compensating defects require independent tests.

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.

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.

At unchanged intrinsic catalyst kinetics, increasing the electrochemically accessible true area by $100$ multiplies exchange current per geometric area by $100$. Equivalently, the local current density at a fixed total geometric current falls by $100$. Each Tafel equation activation loss decreases by $b{\log }_{10}100=2b$:Thus $V_{\mathrm{new}}=0.59948\mathrm{V}$ and $P_{\mathrm{new}}/A=5.99\mathrm{W}{\mathrm{m}}^{-2}$, an increase of $2.39\mathrm{W}{\mathrm{m}}^{-2}$, or about $66.3\%$. Power does not increase a hundredfold, because the voltage improvement is logarithmic and the specified geometric current is unchanged. This assumes both electrodes gain accessible active area, the original exchange currents use the original area basis, and the porous structure does not introduce additional transport or Ohmic losses.