Chapter_14_Redox_Part_4_

Chapter_14_Redox_Part_4_ - Chapter 14 Part 4 Oxidation...

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Chapter 14 Part 4 Oxidation reduction reactions Skip section 14.5 (balancing complex redox reactions) Skip section 14.7 (corrosion prevention) Fix Figure 14.15 Cathode should be marked as negative Anode should be marked as positive NOTE: quiz Thursday Redox part 1 , 2, and 3 -----
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Brief review of redox reactions Loss and gain of electrons Electron donor Electron acceptor Determination of oxidation states of chemical elements helps to identify redox reactions We follow a set of rules for determining oxidation states Example Determine oxidation states of elements in H 2 SO 4 1+ ? 2- 0 = 2(+1) + X + 4 (-2) X = 6+ --
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Is the following a redox reaction? Pb + O 2 -> PbO 2 Oxidation states 0 0 ? 2- Determine oxidation state of Pb in PbO 2 0 = x +2 (-2) X = 4 Yes redox reaction -------------
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Is the following a redox reaction? Pb(s) + PbO 2 (s) + 2 H 2 SO 4 (aq) -> 2 PbSO 4 (s) + 2 H 2 O 0 4+ 2- 1+ 6+ 2- 2+ 6+ 2- 1+ 2- Yes, redox reaction --
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Electrochemical Cells (1) Voltaic cell Constructed to use a favorable redox reaction to generate electricity (2) Electrolytic cell Constructed to use electricity to drive an unfavorable redox reaction (It has industrial applications) --
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Zn/Cu battery Redox reaction Zn(s) + Cu(NO 3 ) 2 (aq) - > Cu (s) + Zn(NO 3 ) 2 (aq) Ionic equation Zn(s) + Cu 2+ (aq) + 2 NO 3 - (aq) - > Cu(s) + Zn 2+ (aq) + 2 NO 3 - (aq) Half reactions: Zn -> Zn +2 + 2 e - (Zn loses two electrons) Cu +2 +2 e - -> Cu (Cu gains two electrons) _____________________________ net ionic reaction
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Chapter_14_Redox_Part_4_ - Chapter 14 Part 4 Oxidation...

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