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20 Electrochemistry Solutions to Exercises 20.72 (a) E° = -0.136 V (-0.126 V) = -0.010 V; = 2 0.22 = -0.010 0.0592 2 = 0.010 0.0592 2 1.00 log [Pb²⁺] = 0.0592 = = = = 10⁻⁸ M (b) For PbSO₄(s), = [Pb²⁺] = = 10⁻⁸ Batteries and Fuel Cells (section 20.7) 20.73 (a) The emf of a battery decreases as it is used. This happens because the concentrations of products increase and the concentrations of reactants decrease. According to the Nernst equation, these changes increase Q and decrease (b) The major difference between AA- and D-size batteries is the amount of reactants present. The additional reactants in a D-size battery enable it to provide power for a longer time. 20.74 (a) First, H₂O is a reactant in the cathodic half-reaction, so it must be present in some form. Additionally, liquid water enhances mobility of the hydroxide ion in the alkaline battery. OH- is produced in the cathode compartment and consumed in the anode compartment. It must be available at all points where Zn(s) is being oxidized. If the Zn(s) near the separator is mostly reacted, OH- must diffuse through the gel until it reaches fresh Zn(s). A small amount of mobilizes OH- so that redox can continue until reactants throughout the battery are depleted. (b) Highly concentrated or solid reactants offer a large amount of reactant in a small sample volume. The more available reactant, the longer the cell a voltage. A voltaic cell with solid or highly concentrated reactants has the advantages of small size and long operational lifetime. 20.75 Analyze/Plan. Given mass of a reactant (Pb), calculate mass of product (PbO₂), and coulombs of charge transferred. This is a stoichiometry problem; we need the balanced equation for the chemical reaction that occurs in the lead-acid battery. The overall cell reaction is: Pb(s) + + (aq) + (aq) Solve. (a) g Pb mol Pb mol PbO₂ g PbO₂ (b) From the half-reactions for the lead-acid battery, 2 mol electrons are transferred for each mol of Pb reacted. From section 20.5, 96,485 C/mol 632