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chapter19_part1

# chapter19_part1 - Chapter 19 Electrochemistry and Its...

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Unformatted text preview: Chapter 19: Electrochemistry and Its Applications 872 End-of-Chapter Solutions for Chapter 19 Summary Problem Answer: (a) 4.055 V; see equations, below; no, cation does not remain near anode (b) 1.935 V; see equations below; no; no, cation does not remain near anode; add an anion to precipitate the cation (c) 5.360 10 4 A, 2037 g PbO 2 (d) 11.9 h (e) Exposure to nickel compounds have been linked to cancer. Strategy and Explanation: (a) Look up the standard reduction potential for the two half-reactions in Table 19.1: PbO 2 (s) + SO 4 2– (aq) + 4 H 3 O + (aq) + 2 e – (aq) PbSO 4 (s) + 6 H 2 O( l ) E ° cathode = + 1.685 V Mg(s) Mg 2+ (aq) + 2 e – E ° anode = – 2.37 V PbO 2 (s) + SO 4 2– (aq) + Mg(s) Mg 2+ (aq) + PbSO 4 (s)+ 6 H 2 O( l ) Calculate the E ° cell using the equation from Sections 19.4, on page 934. E ° cell = E ° cathode – E ° anode = (+1.685 V) – (– 2.37 V) = 4.055 V This cell could not be the basis of a secondary battery. Nothing can prevent the magnesium ion produced at anode from floating away. The recharging of this battery would likely produce a lead-acid battery since the PbSO 4 would react like it does in the lead-acid battery. (b) Look up the Standard Reduction Potential for the two half-reactions in Table 19.1: PbO 2 (s) + SO 4 2– (aq) + 4 H 3 O + (aq) + 2 e – (aq) PbSO 4 (s) + 6 H 2 O( l ) E ° cathode = + 1.685 V Ni(s) Ni 2+ (aq) + 2 e – E ° anode = – 0.25 V PbO 2 (s) + SO 4 2– (aq) + Ni(s) Ni 2+ (aq) + PbSO 4 (s)+ 6 H 2 O( l ) E ° cell = E ° cathode – E ° anode = (+1.685 V) – (– 0.25 V) = 1.935 V The lead-acid battery is described in Section 19.9. The E ° cell for it is given on page 955 as 2.041 V. The Ni/PbO 2 cell potential is less than that of a single sell of the lead-acid battery. This cell could not be the basis of a secondary battery. Nothing can prevent the nickel (II) ion produced at anode from floating away. The recharging of this battery would likely produce a lead-acid battery since the PbSO 4 would react like it does in the lead-acid battery. If an anion were added that could precipitate the Ni 2+ ion, such as OH – or S 2– , that could prevent the cation from floating away from the proximity of the anode and make it possible to build a secondary battery from this cell. (c) Calculate the moles of Ni reacting. 500.0 g Ni 1 mol Ni 58.6934 g Ni = 8.519 mol Ni Use stoichiometry to determine the number of electrons passed, then Faraday’s constant to determine the coulombs of charge. 8.519 mol Ni 2 mol e 1 mol Ni 96485 C 1 mol e =1.644 10 6 C Divide charge by time, to calculate amperes. 1.644 10 6 C 30.000 min 1 min 60 s 1A 1s 1C = 5.360 10 4 A Use stoichiometry and molar mass to calculate the mass of PbO 2 ....
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chapter19_part1 - Chapter 19 Electrochemistry and Its...

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