Copyright © 2020 Dan Dill [email protected] Lecture 16 CH102...

Lecture 16 CH102 Summer 1 2020 6/24/2020 4:06 PM

Copyright © 2020 Dan Dill [email protected] 1

Lecture 16 CH102 Summer 1 2020Wednesday, June 24, 2020

Chapter 24: Oxidation-reduction reactions• Oxidation states• Transition metal oxidation states and Lewis structures• Balancing oxidation reduction reactions• Oxidation-reduction reactions and chemical analysisChapter 25: Electrochemistry • Electrochemical cells harness spontaneous electron flow• Cell diagrams line notation Next: Continue chapter 25

Oxidation statesIn redox reactions electrons rearrange from one place to another.

A way to assess movement of electrons in redox processes is to monitor changes in oxidation states of elements.

Oxidation states are assigned in one of two ways, which give the same result:

by rules equivalent to the second way ;

by partitioning electrons in Lewis structures equivalent to the first way so that the more electronegative atom gets all of shared electrons.

Lecture 16 CH102 Summer 1 2020

7

Copyright © 2020 Dan Dill [email protected]

Rules for assigning oxidation states1. Free elements are 0

2. Sum is the net charge

3. Alkali metals group 1 are always 1

4. F atoms in compounds is always 1

5. Alkaline-earth atoms group 2 , Zn, and Cd in compounds are always 2

6. H atoms in compounds are always 1

7. O atoms in compounds are almost always 2


8


Rules for assigning oxidation statesUse rules 1 to 7 for Mn ClO 𝑠


9


2 7

8 9



Lewis structure oxidation states8. Partition electrons around each atom in Lewis structure, assigned shared

electrons to the more electronegative atom.


10




Use rules 1 7 to ind the oxidation states in acetic acid, C H C OOH


11


TP What is the oxidation number of the carbon labelled 𝑎 in acetic acid, C H C OOH?

1. 42. 33. 24. 15. 06. 17. 28. 39. 4

Copyright © 2020 Dan Dill [email protected] 16 CH102 Summer 1 2020

12



Use rules 1 7 to ind the oxidation states in acetic acid, C H C OOH

Rules 1 7 give O 2 , H 1 , C 𝑥 𝑦 0


13


10 11

12 13





Use rule 8 to find the oxidation states in acetic acid, C H C OOH


14




Use rules 1 7 and rule 8 to ind the oxidation states in acetic acid, C H C OOH


Rule 8 is necessary to give C 3


15


TP What is the oxidation number of the carbon labelled 𝑏 acetic acid, C H C OOH?

1. 42. 33. 24. 15. 06. 17. 28. 39. 4


16



Use rules 1 7 and rule 8 to ind the oxidation states in acetic acid, C H C OOH





17


14 15

16 17



Transition metal valence electronsRule 8 is used to determine the number of valence electrons in transition-metal atoms in covalent compounds:

The number of valence electrons is defined as the oxidation state.


18


Transition metal valence electronsThereby the number of valence electrons in transition-metal covalent compounds is determined and so the Lewis structure can be determined.

Construct the Lewis structure of MnO .

Mn oxidation state 𝑥 satisfies 𝑥 4 2 2, and so 𝑥 6, the number of valence electrons contributed by Mn

The total number of valence electrons is therefore 6 4 6 2 32.


19


Oxidation reduction reactionsOxidation reduction reactions involve transfer of electrons from one reactant to another.

H S 𝑎𝑞 ClO 𝑎𝑞 → S 𝑠 Cl 𝑎𝑞 H O 𝑙


20


Oxidation reduction reactionsH S 𝑎𝑞 ClO 𝑎𝑞 → S 𝑠 Cl 𝑎𝑞 H O 𝑙

S 2 → S 0 oxidized, gives off electrons, reducing agent

Cl 1 → Cl 1 reduced, takes on electrons, oxidizing agent


21


18 19

20 21



TP In the redox process P 𝑠 S O 𝑎𝑞 → H PO 𝑎𝑞 HSO 𝑎𝑞 , the oxidizing agent is …

1. P2. S O3. H PO4. HSO


22

Balancing oxidation reduction equationsFor each oxidation reduction pair half-reaction1. Balance elements other than O and H2. Balance O with H O 𝑙3. Balance H with H 𝑎𝑞 not H O 𝑎𝑞4. Balance charge by adding e to the side that is most positive


23


Balancing oxidation reduction equationsBalance P 𝑠 → H PO 𝑎𝑞

Answer: 16 H O 𝑙 P 𝑠 → 4 H PO 𝑎𝑞 20 H 𝑎𝑞 20 e


24


Balancing oxidation reduction equationsBalance S O 𝑎𝑞 → HSO 𝑎𝑞

Answer: 2 e 2 H 𝑎𝑞 S O 𝑎𝑞 → 2 HSO 𝑎𝑞


25


22 23

24 25



Balancing oxidation reduction equationsCombine balanced half-reactions1. Adjusts e to be the same2. Combine3. Cancel where possible


26


Balancing oxidation reduction equationsCombine balanced half-reactions

16 H O 𝑙 P 𝑠 → 4 H PO 𝑎𝑞 20 H 𝑎𝑞 20 e

2 e 2 H 𝑎𝑞 S O 𝑎𝑞 → 2 HSO 𝑎𝑞

16 H O 𝑙 P 𝑠 10 S O 𝑎𝑞 → 4 H PO 𝑎𝑞 20 HSO 𝑎𝑞


27


Balancing oxidation reduction equationsIf in acidic solution, convert H 𝑎𝑞 to H O 𝑎𝑞1. Add as many H O 𝑙 as H 𝑎𝑞 to both sides2. Combine H 𝑎𝑞 and H O 𝑙 into H O 𝑎𝑞3. Cancel H O 𝑙 where possible


28


Balancing oxidation reduction equationsIf in basic solution, convert H 𝑎𝑞 to OH 𝑎𝑞1. Add as many OH 𝑎𝑞 as H 𝑎𝑞 to both sides2. Combine H 𝑎𝑞 and OH 𝑎𝑞 into H O 𝑙3. Cancel H O 𝑙 where possible


29


26 27

28 29



Balancing oxidation reduction equationsIf in basic solution, convert H 𝑎𝑞 to OH 𝑎𝑞Balance IO 𝑎𝑞 I 𝑎𝑞 → IO 𝑎𝑞 I 𝑎𝑞Answer: H O 𝑙 IO 𝑎𝑞 3 I 𝑎𝑞 → IO 𝑎𝑞 I 𝑎𝑞 2 OH 𝑎𝑞


30


Oxidation-reduction titrationProblem 24-45: A 10.0-gram sample containing Sn 𝑎𝑞 is completely oxidized by 34.6 mL of a 0.556-M solution of NaI 𝑎𝑞 . Calculate the mass of tin in the sample and its mass percentage on the rock. The unbalanced equation for the reaction is I 𝑎𝑞 Sn 𝑎𝑞 → Sn 𝑎𝑞 I 𝑎𝑞 .


31


TP A sample containing Sn is completely oxidized to Sn by a solution of I , which is reduced to I . How many moles of Sn are formed by each mole of I ?

1. 1/32. 1/23. 14. 25. 36. More information needed


32


2 e I 𝑎𝑞 → 3 I 𝑎𝑞


33


30 31

32 33





Sn 𝑎𝑞 → Sn 𝑎𝑞 2 e


34





Sn 𝑎𝑞 I 𝑎𝑞 : 0.0346 L 0.556 M 0.0192 mol


35





Sn 𝑎𝑞 I 𝑎𝑞 : 0.0346 L 0.556 M 0.0192 mol

Sn : 0.0192 mol . 2.28 g


36





Sn 𝑎𝑞 I 𝑎𝑞 : 0.0346 L 0.556 M 0.0192 mol

Sn : 0.0192 mol . 2.28 g

mass % . .

100% 22.8%


37


34 35

36 37



Cu2+(aq) oxidizes Zn(s)Spontaneous flow of electrons from Zn to Cu

Cu2 𝑎𝑞 2 e Cu 𝑠Zn 𝑠 Zn2 𝑎𝑞 2 e


39

Cu2+(aq) oxidizes Zn(s) spontaneouslyLet’s sketch an electrochemical cell to harness the spontaneity of

Cu2 𝑎𝑞 Zn 𝑠 Cu 𝑠 Zn2 𝑎𝑞


40

Galvanic (Voltaic) CellsLecture 16 CH102 Summer 1 2020

41


Cell diagram (line notation)Zn 𝑠 | Zn2 𝑎𝑞 ||Cu2 𝑎𝑞 | Cu 𝑠

• Oxidation on left, “||” is salt bridge, reduction on right

• Phases separated by “|”, same phases separated by “,”

• If no solid, inert electrode Pt or graphite

• Left to right order matches flow of electrons


42

39 40

41 42



Cell diagram (line notation)Fe 𝑠 | Fe2 𝑎𝑞 ||MnO4 𝑎𝑞 , Mn2 𝑎𝑞 | Pt 𝑠

Write the balanced half-reactions in acidic solution


44


Write the balanced half-reactions in acidic solutionFe 𝑠 Fe2 𝑎𝑞 2 eMnO4

𝑎𝑞 5 e 8 H 𝑎𝑞 Mn2 𝑎𝑞 4 H2O 𝑙


45


Convert the balanced half-reactions in acidic solution to basic solutionFe 𝑠 Fe2 𝑎𝑞 2 𝑒MnO4

𝑎𝑞 5 𝑒 8 H 𝑎𝑞 Mn2 𝑎𝑞 4 H2O 𝑙


46


Convert the balanced half-reactions in acidic solution to basic solutionFe 𝑠 Fe2 𝑎𝑞 2 𝑒MnO4

𝑎𝑞 5 𝑒 8 H 𝑎𝑞 Mn2 𝑎𝑞 4 H2O 𝑙MnO4

𝑎𝑞 5 𝑒 4 H2O 𝑙 Mn2 𝑎𝑞 8 OH 𝑙


47

44 45

46 47

Copyright © 2020 Dan Dill [email protected] Lecture 16 CH102...

Documents

Transcript of Copyright © 2020 Dan Dill [email protected] Lecture 16 CH102...