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Lecture 263/30/07. E° F 2 (g) + 2e - ↔ 2F - +2.87 Ag + + e - ↔ Ag (s)+0.80 Cu 2+ + 2e - ↔ Cu (s)+0.34 Zn 2+ + 2e - ↔ Zn (s)-0.76 Quiz 1. Consider these.

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Presentation on theme: "Lecture 263/30/07. E° F 2 (g) + 2e - ↔ 2F - +2.87 Ag + + e - ↔ Ag (s)+0.80 Cu 2+ + 2e - ↔ Cu (s)+0.34 Zn 2+ + 2e - ↔ Zn (s)-0.76 Quiz 1. Consider these."— Presentation transcript:

1 Lecture 263/30/07

2 E° F 2 (g) + 2e - ↔ 2F - +2.87 Ag + + e - ↔ Ag (s)+0.80 Cu 2+ + 2e - ↔ Cu (s)+0.34 Zn 2+ + 2e - ↔ Zn (s)-0.76 Quiz 1. Consider these half-reactions 1. Which is the strongest oxidizing agent? 2. Will F 2 (g) oxidize Ag(s) to Ag + ? 3. Will Zn 2+ reduce Ag + to Ag(s)? 2. Calculate the equilibrium constant for: Ag + (aq) + Cu (s) ⇆ Ag(s) + Cu 2+ (aq)

3 1.50 amps flow through a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited?

4 The anode reaction in a lead storage battery is: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e - If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?

5 Corrosion

6 http://www.gordonengland.co.uk/img/corr5.gif

7 http://awesomegems.com/jewelry/Cleaner-Ionic-tarnish2.jpg

8 Standard Reduction Potentials (25° C) E° F 2 (g) + 2e - ↔ 2F - +2.87 Au 3+ + 3e - ↔ Au (s)+1.50 Cl 2 (g) + 2e- ↔ 2 Cl - +1.36 O 2 (g) + 4 H + + 4e - ↔ 2 H 2 O+1.23 Hg 2+ + 2e - ↔ Hg (l)+0.85 Ag + + e - ↔ Ag (s)+0.80 Fe 3+ + e - ↔ Fe 2+ +0.77 Cu 2+ + 2 e - ↔ Cu (s)+0.34 Sn 4+ + 2e - ↔ Sn 2+ +0.15 2H + + 2 e - ↔ H 2 (g) 0.00 Pb 2+ + 2e - ↔ Pb (s)-0.13 Sn 2+ + 2e - ↔ Sn (s)-0.14 Ni 2+ + 2e - ↔ Ni (s)-0.26 Zn 2+ + 2 e - ↔ Zn (s)-0.76 Cr 2+ + 2 e - ↔ Cr (s)-0.91 Mg 2+ + 2 e - ↔ Mg (s)-2.37 Na + + e - ↔ Na (s)-2.71 Ca 2+ + 2 e - ↔ Ca (s)-2.87 Li + + e - ↔ Li (s)- 3.04

9 Recap Balancing Redox reactions Electrochemical cells Batteries Standard Reduction Potential table Corrosion Nernst equation Electrolysis Molten aqueous Counting electrons

10 www.state.sd.us/denr/DES/mining/P000106.jpg http://www.nelpi.org/tarcreek/slides/5s.html

11 Acid Mine Drainage REDOX in the real world FeS 2 (pyrite) + O 2  SO 4 2- + Fe 2+


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