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Unformatted text preview: obennoskey (blo242) – Postlab 6 – Lyon – (52635) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Give the oxidation numbers for copper in the compound Cu 2 O and for Mo in the compound Mo 2 O 2 7 , where oxygen maintains an oxida tion number of 2. 1. 0; 0 2. +1; +12 3. 0; 2 4. +2; +6 5. +1; +1 6. +2; +7 7. +1; +6 correct Explanation: Oxidation numbers are per atom. The sum of oxidation numbers in a neutral molecule is zero and in a polyatomic ion is equal to the charge on the ion. We know the oxidation number of O is 2. We use these values and set the sum of the oxidation numbers equal to the appropriate value. The oxidation number of Cu and Mo is symbolized by x : 2 x + ( 2) = 0 Cu 2 O : 2 x = +2 x = +1 2 x + 7( 2) = 2 Mo 2 O 2 6 : 2 x 14 = 2 2 x = 12 x = +6 002 10.0 points If the oxidation state of H is +1 and O is 2, what is the oxidation state of C in C 2 H 4 O? 1. +2 2. +1 3. 1 correct 4. 1 2 5. 2 Explanation: To calculate oxidation numbers, remember that the the sum of the numbers must equal the total charge on the molecule or in the case of neutral species, zero. Let x be the oxidation number of C. 2 x + 4(+1) + ( 2) = 0 C 2 H 4 O : 2 x + 2 = 0 2 x = 2 x = 1 003 10.0 points Consider the process in which H 2 SO 3 is con verted to H 2 SO 4 . The oxidation state of oxy gen remains 2, and the oxidation state of H remains +1. Select the correct statement. 1. S is reduced and gains electrons. 2. S is oxidized and loses electrons. correct 3. S is reduced and loses electrons. 4. S is oxidized and gains electrons. 5. S neither gains nor loses electrons in this reaction. Explanation: Oxidation numbers are per atom. The sum of oxidation numbers in a neutral molecule is zero and in a polyatomic ion is equal to the charge on the ion. We know the oxidation number of O is 2. We use these values and set the sum of the oxidation numbers equal to the appropriate value. The oxidation number of S is symbolized by x : 2(+1) + x + 3( 2) = 0 H 2 SO 3 : obennoskey (blo242) – Postlab 6 – Lyon – (52635) 2 2 + x 6 = 0 x = +4 2(+1) + x + 4( 2) = 0 H 2 SO 4 : 2 + x 8 = 0 x = +6 Sulfur goes from +4 to +6 for a loss of two electrons....
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This note was uploaded on 04/14/2010 for the course CH CH302 taught by Professor Mccord during the Spring '10 term at University of Texas at Dallas, Richardson.
 Spring '10
 MCCORD

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