Lec-35-Review - Exam 3 Room Assignments Wednesday Apr 18th,...

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1 Exam 3 Room Assignments Wednesday Apr 18 th , 8:30 PM – 10:00 PM Javits 100 105816591 000000000 Javits 102 105930059 105816592 Old Chemistry 116 999999999 105930060 Room To ID From ID • As seating is very tight, it is essential that you take the exam in the room to which you are assigned below. • Students in the wrong room may be redirected with the result that they may have less time on the exam. Workshops have been cancelled for Thursday, the day after the exam . Identifying Atoms Involved in Electron Transfer Consider the reaction: Cu(s) + HNO 3 (aq) Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O Oxidation: Cu Cu 2+ + 2e Reduction: Where do these electrons go? A method of accounting for electrons is needed for such cases. Oxidation number (ON): The charge an atom would have if bonding electrons are assigned to the more electronegative atom. Consider the oxidation numbers for N and O in NO. Draw the Lewis structure: Note: formal charges are 0 for both N and O Assign e to most electronegative atom: Determine the oxidation numbers: ON(O) = 6 valence e 8 assigned e = 2 ON(N) = 5 valence e 3 assigned e = +2 NO All bonding e are assigned to O as it is more electronegative
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2 Identifying Atoms Involved in Electron Transfer What are the oxidation numbers for N and O in NO 3 ? Draw the Lewis structure (resonance can be ignored): NO O O H Assign bonding e to the more electronegative atom Determine the oxidation numbers: ON(O) = 6 valence e 8 assigned e = 2 ON(H) = 1 valence e 0 assigned e = +1 ON(N) = 5 valence e 0 assigned e = +5 So where did the electrons go? N went from an oxidation state of +5 in NO 3 to +2 in NO N was reduced All bonding e assigned to O as its electronegativity is higher than N or H Assigning Oxidation Numbers By Rules Oxidation numbers can often be determined through a set of rules that circumvent the need to draw Lewis structures. F: 1always O: 2 almost always Cl: 1 usually 4. Highly electronegative elements tend to have ON = standard charge Example Rule For CO, ON(C) = +2 as ON(O) = 2 5. The sum of oxidation numbers equals the charge on the formula H 2 O: ON(H) = +1 LiH: ON(H) = 1 3. ON(H) = +1 except in H & H 2 (0) and metal hydrides ( 1) O 2 : 2 Na + :+1 2. ON = charge for monatomic ions H, H 2 , O 2 , Na, Fe, P 4 1. ON = 0 for a neutral atom or atoms in elemental form
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3 Examples of Determining Oxidation Numbers H 2 SO 4 : O is most electronegative, ON(O) = 2; ON(H) = +1 2 × ON(H) + ON(S) + 4 × ON(O) = ON(S) + 2(1) + 4( 2) = 0 ON(S) = 0 2 + 8 = +6 ClF 3 : F is most electronegative, ON(F) = 1 ON(Cl) + 3 × ON(F) = ON(Cl) + 3( 1) = 0 ON(Cl) = +3 CH 4 : H is given its standard ON of +1 ON(C) + 4 × ON(H) = ON(C) + 4 = 0 ON(C) = 4 CH 3 OH: O is most electronegative, ON(O) = 2; ON(H) = +1 ON(C) + 4 2 = 0 ON(C) = 2 Note: C is reduced by adding H, removing O C is oxidized by adding O, removing H H 2 O 2 : O is most electronegative, ON(O) = 2; ON(H) = +1 2 × ON(H) + 2 × ON(O) = 2 4 = 2 ????
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This note was uploaded on 03/06/2010 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.

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Lec-35-Review - Exam 3 Room Assignments Wednesday Apr 18th,...

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