Tutorial 8 Solutions

Tutorial 8 Solutions - CHEMISTRY 1AO3 TUTORIAL #8 November...

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CHEMISTRY 1AO3 TUTORIAL #8 November 2-6, 2009 Solutions: Redox Reactions – Ch 5 / Acid-Base Chemistry – Chapter 16 ______________________________________________________________________________ Page 1 of 12 The following problems (1-13) are based on material from Chapters 5, 16 and class notes. 1. Complete and balance the following reaction in basic solution: MnO 2 (s) + ClO 3 - (aq) MnO 4 - (aq) + Cl - (aq) Start with determining the changes in the oxidation numbers for all elements in molecules, as well as identifying oxidizing/reducing agents: Reactants: MnO 2 (Mn: +4; O:-2) and ClO 3 - (Cl: +5; O: -2) Products: MnO 4 - (Mn: +7) and Cl - (Cl: -1). Therefore the MnO 2 is being oxidized (Mn +4 +7; reducing agent) and ClO 3 - is being reduced (Cl +5 -1; oxidizing agent). Then construct balanced half-reaction under basic conditions (balance redox atoms, add electrons, balance charge with OH - . balance H/O with H 2 O, multiple by a factor to eliminate electrons): (NOTE: this is only one possible approach – another would be to balance H with H + , then add H 2 O to balance O, then add OH - to both sides to convert H + to H 2 O) Oxidation Half-reaction: ( MnO 2 (s) + 4OH - (aq) MnO 4 - (aq) + 3e - + 2H 2 O (l) ) X 2 Reduction Half-reaction: ClO 3 - (aq) + 6e - + 3H 2 O (l) Cl - (aq) + 6OH - (aq) Net Redox Reaction: 2MnO 2 (s) + ClO 3 - (aq) + 2OH - (aq) 2MnO 4 - (aq) + Cl - (aq) + H 2 O (l) 2. Complete and balance the following reaction in acidic solution: CH 3 CH 2 OH(l) + Cr 2 O 7 2 (aq) CH 3 COOH(aq) + Cr 3+ (aq) Hint: In order to determine the oxidation number of alcohol and carboxylic acid, consider the carbon species -CH 2 OH and -COOH, respectively, since the CH 3 end group does not change. This is a redox reaction. We are using a strong oxidizing agent with a primary alcohol, so the alcohol (ethanol) will be converted to a carboxylic acid (acetic acid). The orange Cr 2 O 7 2- (aq) ion is reduced to green Cr 3+ (aq) ion. Start with determining the changes in the oxidation numbers for all elements in molecules, as well as identifying oxidizing/reducing agents: Reactants: CH 3 CH 2 OH (C of the CH 2 OH group: -1; O:-2; H: +1) and Cr 2 O 7 2- (Cr: +6; O: -2) Products: CH 3 COOH (C of the COOH group: +3), Cr 3+ (Cr: +3, O: -2). Therefore the Cr 2 O 7 2- is being reduced (Cr +6 +3; oxidizing agent) and CH 2 OH is being oxidized (C -1 +3; reducing agent). Then construct balanced half-reaction under basic conditions (balance redox atoms, add electrons, balance H/O with H 2 O, then balance charge with H + , multiple by a factor to eliminate electrons):
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CHEMISTRY 1AO3 TUTORIAL #8 November 2-6, 2009 Solutions: Redox Reactions – Ch 5 / Acid-Base Chemistry – Chapter 16 ______________________________________________________________________________ Page 2 of 12 Oxidation Half-reaction: ( CH 3 CH 2 OH (aq) + H 2 O (l) CH 3 COOH (aq) + 4H + (aq) + 4e - ) X3 Reduction Half-reaction: ( Cr 2 O 7 2- (aq) + 6e - + 14H + (aq) 2Cr
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This note was uploaded on 12/20/2010 for the course CHEM 1a03 taught by Professor Landry during the Spring '08 term at McMaster University.

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Tutorial 8 Solutions - CHEMISTRY 1AO3 TUTORIAL #8 November...

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