Lecture of 19 February 2020Outline of today's lecture

Midterm exam - average = 71•Electrochemistry and oxidation-reduction equilibria•

Midterm Exam discussion

problem 7 charge balanceKt t C A 0

ret Ai O I ly

C kaCA xa

Ht KaC 0.3M Kai 1 8 0 4

CA 1 0 1M

htt 3.6 0 414 pH 3 45

problem 7 O 05M malonic acidO l M Naz Melchett

calculate the pHNat t HH CO A t 2CA

Cpa 0.1514 Carat O 2Ml

Nat A x 2 CA

0.2M CTtacCIHt tka.CI tkaTka A

CtkaiKa AtHt t ka Itt c Kaikaz

htt 404 0 6 M pLt 5.39

problem 13 Hasoy Hsoj Ht Kaos

HSE Isaf DHT KaHo

2

problem 14 Ksp ApbaAsap

2E S2 go28

In O M e I 78 f 26 0414

in 0.01M I I 6265S 2.01 0 414

TKsplot

Z

Oxidation Reduction Equilibriaenergy of a charged species depends on OIelectrical electrostatic potential v

related to work required to bring G chargefrom

infinity to potentialOI with 1 J 7 C x 7 V

n moles of electrons throughpotentialdifference0

work ng Na o of Jchange potentce drop

of 1.602 10 19 C single Ct charge1 C G 24 X 018 changes

Coulombs mole e g Na 96487 C meE Faraday F

current flow 1 Amp I C s I

6 24 10 8charggs I

0 Gee n f DOIw un

B

Respiration Example

basal rate of Ozconsumption I met Oz hi

02 4 e 4 Ht 2 Hz 0

I mol Oz hi 4 mole hi

4F Chr l 100 Amps

power 100 A X IV COO W 100 J s t

104 KJ day I

2500 nutritionalcalories 1 day

4

electron transfer oxidation reduction reaction

involves change in oxidation state number

assigning oxidaton Statesi elements by themselves 20

2 H generally t 1

37 0 generally 24 for wins oxidation state chargeFe Fett festoxidation ferricthan metal

reductionferrous

in general oxidation stats are assigned basedon electronegativities of bonded atoms

bonding to a more electroneg atom_oxidations

bondingto a less electroneg atom reduction

CHL Cµ

OIC O

c HY co HY o cut ofmore reduced C more oxidized C

formal oxidation state 5

Oxidation loss of electronsreduction gain of electrons

balancing oxidation reduction reactions

assign oxidation States

Cpus tguys of t.fi Npg

Zt 2 b t a It 6 2 2 2

Gtoxidation Cris SOY a Instance

5T 2Treduction NOI NO reduction

half reacts

balance egs wrt atoms being redufqdiaffeybalance Wrt electrons

ccu s soy ta t 8e

N05 3e NO

6

balance 0 by adding H2O

a St 4h20 Cu t Sai Se

Ng 3e NO t 2H2O

balance Lt by adding Itt

Cust 4H2O Cult 5042 85 8 Ht

Nos t 3 e t 4 Itt NO t 2420

multiply each reaction to get samenumber of electrons and add

3 Cus 8N05 8Ht 3 Cu't x 3Soy8 Not 4h20

Voila the equation is balanced