Quiz 1 Scores. Chapter 4: Chemical Reactions F D C- C C+ B- B B+ A- A A+ Chem 6A Michael J. Sailor, UC San Diego

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1 Chapter 4: Chemical Reactions Quiz 1 Scores F D C- C C+ B- B B+ A- A A+ 2

2 Grade Projection Quiz 1 (data from 2010) score on quiz #1 F D C- C C+ B- B B+ A- A + final grade in course 3 Announcements: Thurs ct 6 quiz (#2) will cover chapter 3 Bring student ID or we cannot accept your quiz! No notes, no calculators You need to know your name, PID, and section # 4

3 Types of Reactions xidation-reduction Precipitation Acid-Base 5 Problem: balancing equations Balance the following reaction by placing the appropriate numbers in front of each reactant and product. Indicate the states [for example, (aq), (s), etc.] of each reactant and product: C 12 H H 2 S 4 C + H 2 S 4 + H 2 6

4 Solution: balancing equations C 12 H 22 11(s) + H 2 S 4(l) 12 C (s) + H 2 S 4(aq) + 11 H 2 (l) How do you know the states? You need to memorize the states of the elements at normal (room temperature and atmospheric pressure conditions (e.g., C is a solid, He is a gas, Br 2 is a liquid) For compounds, you will be given their state if needed. 7 Problem: balancing difficult redox Balance the following reaction by placing the appropriate numbers in front of each reactant and product: Fe 2+ + Mn H + Fe 3+ + Mn 2+ + H 2 This is an example of a REDX reaction. Many REDX reactions are difficult to balance by inspection 8

5 Solution: balancing difficult redox (1) Assign oxidation states to all elements: Fe 2+ + Mn H + Fe 3+ + Mn 2+ + H 2 9 Rules for Assigning xidation States 1. For single elements, oxidation state = charge on the ion: 2. In single-element compounds, oxidation state = 0 In compounds, assign using the following rules: 1. xidation states must sum to equal the charge on the molecule. 2. H is always +1 when combined with nonmetals, -1 when combined with metals 3. is usually -2 (Exception: Peroxide 2 2- ) 4. Halogens are -1 unless combined with or other halogens (the halogen on top in the periodic table wins--f is always -1). 10

6 Solution: balancing difficult redox (1) Assign oxidation states to all elements: Fe 2+ + Mn H + Fe 3+ + Mn 2+ + H 2 11 Solution: balancing difficult redox (2) Assign who gets oxidized, who gets reduced xidation: oxidation state becomes more positive Reduction: oxidation state becomes more negative Fe 2+ loses 1 electron Fe 2+ + Mn H + Fe 3+ + Mn 2+ + H 2 Mn gains 5 electrons 12

7 (2) Assign who gets oxidized, who gets reduced xidation: oxidation state becomes more positive Reduction: oxidation state becomes more negative Fe 2+ loses 1 electron Fe 2+ + Mn H + Fe 3+ + Mn 2+ + H 2 Mn gains 5 electrons Mn 4 - is reduced in the reaction; it is the oxidizing agent Fe 2+ is oxidized in the reaction; it is the reducing agent 13 Solution: balancing difficult redox (3) Split into half-reactions, balance atoms being reduced or oxidized Fe 2+ + Mn H + Fe 3+ + Mn 2+ + H 2 xidation half-reaction: Reduction half-reaction: Fe 2+ Fe 3+ Mn 4 - Mn 2+ 14

8 What is a half-reaction? Example: Electrolysis of water (water splitting) xidation half-reaction: H 2 H 2 + ½ 2 H 2 ½ 2 + 2H + + 2e - Reduction half-reaction:2h 2 + 2e - H 2 + 2H - 1 3H 2 H 2 + ½ 2 + 2H 2 15 Solution: balancing difficult redox (4) Balance oxidation numbers with electrons xidation half-reaction: Reduction half-reaction: Fe 2+ Fe 3+ + e - Mn e - Mn 2+ 16

9 Solution: balancing difficult redox (5) Multiply one of the half-reactions to get the same number of electrons in both equations xidation half-reaction: Reduction half-reaction: 5Fe 2+ 5Fe e - Mn e - Mn Solution: balancing difficult redox (6) Add half-reactions together xidation half-reaction: Reduction half-reaction: 5Fe 2+ 5Fe e - Mn e - Mn 2+ 5Fe 2+ + Mn e - 5Fe e - + Mn 2+ 18

10 Solution: balancing difficult redox (7) Cancel electrons, balance charges with H + xidation half-reaction: Reduction half-reaction: 5Fe 2+ 5Fe e - Mn e - Mn 2+ 8H + + 5Fe 2+ + Mn 4-5Fe 3+ + Mn on this side 17+ on this side 19 Solution: balancing difficult redox (8) Balance H + with H 2 8H + + 5Fe 2+ + Mn 4-5Fe 3+ + Mn H 2 20

11 Solution: balancing difficult redox (9) Simplify, check to see if charge and atoms balance 8H + + 5Fe 2+ + Mn 4-5Fe 3+ + Mn H 2 Left side Right side Number of H: 8 4x2 = 8 Number of Fe: 5 5 Number of Mn: 1 1 Number of : 4 4 Total charge: 8 + 5(2) -1 = 17 5(3) + 2 = Aqueous solubility rules (Table 4.1) Green compounds are not in Table 4.1 but you should know them 20 22

12 Problem: precipitation and quantitative analysis ml of a contaminated water sample contains Cd 2+ ions at a concentration of x 10-3 M. If 50.0 ml of a solution containing excess Na 2 S is added to the solution, the precipitate that is formed will weigh: a) g b) 6.28 x 10-2 g c) g d) 6.28 x 10-3 g e) none of the above 23 Solution: precipitation and quantitative analysis Balanced equation: Cd 2+ (aq) + 2 Na+ (aq) + S2- (aq) CdS (s) + 2 Na+ (aq) Net ionic equation: Cd 2+ (aq) + S2- (aq) CdS (s) 24

13 Solution: precipitation and quantitative analysis How many moles of Cd 2+ are in that sample? x 10-3 mol CdL L solution = 8.03 x 10-4 mol Cd 2+ How many grams of CdS are in that sample? Molecular weight of CdS: = = g/mol CdS 25 Solution: precipitation and quantitative analysis So the number of grams of CdS (s) produced is x 10-3 mol Cd 2+ L L solution 1 mol CdS 1 mol Cd g CdS mol CdS = g CdS Since there is excess S 2-, you didn t need to know how much of it was there 26

14 Problem: balancing difficult redox Balance the following reaction by placing the appropriate numbers in front of each reactant and product. Identify the oxidant and the reductant: N 2 + H 2 N + HN 3 This is another example of a REDX reaction that is difficult to balance by inspection. 27 Problem: balancing difficult redox Balance the following reaction by placing the appropriate numbers in front of each reactant and product. Identify the oxidant and the reductant: 3N 2 + H 2 N + 2HN 3 28

15 Sections SECTIN TIME LCATIN D01 M 2-2:50 pm WLH 2115 D02 M 3-3:50 pm WLH 2115 D03 M 4-4:50 pm WLH 2115 D04 W 2-2:50 pm WLH 2115 D05 W 3-3:50 pm WLH 2115 D06 W 4-4:50 pm WLH 2115 D07 F 2-2:50 pm WLH 2115 D08 F 3-3:50 pm WLH 2115 D09 F 4-4:50 pm WLH Extras 30

S 8 + F 2 SF 6 4/9/2014. iclicker Participation Question: Balance the following equation by inspection: H + + Cr 2 O 7 + C 2 H 5 OH

S 8 + F 2 SF 6 4/9/2014. iclicker Participation Question: Balance the following equation by inspection: H + + Cr 2 O 7 + C 2 H 5 OH Today: Redox Reactions Oxidations Reductions Oxidation Numbers Half Reactions Balancing in Acidic Solution Balancing in Basic Solution QUIZ 3 & EXAM 3 moved up by one day: Quiz 3 Wednesday/Thursday next