chapter19_part1

chapter19_part1 - Chapter 19: Electrochemistry and Its...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 Faradays 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 ....
View Full Document

This note was uploaded on 07/15/2009 for the course CHE 131 taught by Professor Kerber during the Fall '08 term at SUNY Stony Brook.

Page1 / 14

chapter19_part1 - Chapter 19: Electrochemistry and Its...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online