Problems for Chapter

10.1 The solid-state electrochemical cell: NbO | Nb(Ru) || electrolyte || Ta2 O5| Ta consists on the right of a Ta/Ta2O5 couple that produces a fixed electrode potential and on the left a half cell containing a mixture of NbO and Nb dissolved in ruthenium. Ru is inert electrochemically and serves only to dilute the active niobium metal component. The cell operates at 1000 K with various mole fractions of Nb dissolved in ruthenium. For the overall cell reaction

NbO + ^Ta = Nb + 1^Ta2O5 , the standard free energy change is AG° = 3.6 kJ/mole.

(a) What is the cell potential if the Nb is pure (i.e., no ruthenium solvent)?

(b) When the Nb is dissolved in ruthenium so that xNb = 0.6, the observed cell voltage is 61mV. What is the activity coefficient of Nb in the alloy?

10.2 Using Table 10.1, determine the standard electrode potential for the following half-cell reaction:

10.3 A mixture of powdered Fe and FeO is stirred into a solution of uranyl nitrate maintained at a pH of 5. At equilibrium, what is the ratio of the UO2+ and U4+ concentrations?

10.4 Determine the EMF of the cell Fe\Fe2+Helectrolytel\Fe2+\Fe3+ when the ferrous ion concentration in the left-hand electrode is 0.5 M and the ratio of Fe3+ to Fe2+ concentrations in the right hand half cell is 0.1.

10.5 The sketch shows the internals of a common lead-acid battery. The abbreviated designation is: Pb|PbSO4||H2SO4||PbO2|PbSO4

The unit is constructed of alternating plates of lead and lead dioxide. The lead electrodes are spongy to provide a high surface area to contact the electrolyte, which is 6 M sulfuric acid. The lead dioxide is packed into plates to form the second electrode. Operation of the battery depletes both electrodes by converting both to lead sulphate, which is insoluble in the electrolyte.

The standard free energies of formation of the species in solution are given in the table below

Species

AG°, kJ/mole

PbSO4(s)

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