HW9-ex3 - dsj294 hw09 Holcombe(53570 This print-out should...

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dsj294 – hw09 – Holcombe – (53570) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. You must use the Table of Stan- dard Reduction Potentials on Dr. Mc- Cord’s course website for all prob- lems in which values are not given. http://courses.cm.utexas.edu/pmccord/ ch302/help/red-table.html 001 10.0 points In the half-reaction Cr 2 O 2 7 Cr 3+ what is the change in the oxidation state of Cr? (Note that oxygen retains an oxidation state of - 2). 1. - 8 to +3 2. - 4 to +3 3. +6 to +3 correct 4. +8 to +3 5. +4 to +3 6. - 6 to +3 Explanation: Oxidation numbers are per atom. The oxi- dation number of a monatomic ion is equal to the charge on the ion. Chromium in Cr 3+ has an oxidation number of +3. The sum of ox- idation numbers in a polyatomic ion is equal to the charge on the ion. We set the sum of the oxidation numbers in Cr 2 O 2 7 equal to - 2 and solve for the oxidation number of Cr: 2 x + 7( - 2) = - 2 Cr 2 O 2 7 : x = +6 002 10.0 points Balance the reaction SnCl 2 + K 2 Cr 2 O 7 + HCl CrCl 3 + KCl + SnCl 4 + H 2 O . What is the coefficient of CrCl 3 in the bal- anced equation? Correct answer: 2. Explanation: The balanced equation is 3 SnCl 2 + K 2 Cr 2 O 7 + 14 HCl 2 CrCl 3 + 2 KCl + 3 SnCl 4 + 7 H 2 O . 003 (part 1 of 2) 10.0 points Using oxidation and reduction half- reactions, balance the skeletal equation O 3 (aq) + Br (aq) O 2 (g) + BrO 3 (aq) of the action of ozone on bromide ions. The reaction takes place in a basic solution. What is the smallest possible integer coefficient of BrO 3 in the combined balanced equation? Correct answer: 1. Explanation: Reduction half-reaction: O 3 (g) O 2 (g) Balance O: O 3 (g) O 2 (g) + H 2 O( ) Balance H: 2 H 2 O( ) + O 3 (g) O 2 (g) + H 2 O( ) + 2 OH (aq) Cancel H 2 O and balance the charge: H 2 O( ) + O 3 (g)+2 e O 2 (g) + 2 OH (aq) Oxidation half-reaction: Br (aq) BrO 3 (aq) Balance O: 3 H 2 O( ) + Br (aq) BrO 3 (aq) Balance H: 6 OH (aq) + 3 H 2 O( ) + Br (aq) BrO 3 (aq) + 6 H 2 O( ) Cancel H 2 O and balance the charge: 6 OH (aq) + Br (aq) BrO 3 (aq) + 3 H 2 O( ) + 6 e Balance the e : 3 [H 2 O( ) + O 3 (g) + 2 e O 2 (g) + 2 OH (aq)] 6 OH (aq) + Br (aq) BrO 3 (aq) + 3 H 2 O( ) + 6 e Add the half-reactions: 3 O 3 (g) + Br (aq) 3 O 2 (g) + BrO 3 (aq)
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dsj294 – hw09 – Holcombe – (53570) 2 004 (part 2 of 2) 10.0 points Identify the oxidizing agent in this reaction.
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