Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How...

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Elechtrochemistry 17

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Transcript of Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How...

Page 1: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Elechtrochemistry

17

Page 2: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Reduction

3+ 2+

Oxidation

3+2+

Page 3: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

F = (6.022 x 1023 mol-1) x (1.602192 x 10-19 C) = 96,484 C mol-1

How much? – Faraday’s constant.

Page 4: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Faraday

Page 5: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.
Page 6: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Electricity

Current = Charge/time - I = Q/t [Ampere]=[Coulomb]/[sec]

Potential = Current x Resistance - E = I R (Ohms law)[Volt]=[Ampere] [Ohm]

Page 7: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

So what is resistance?

NB: Potential is relative!

Page 8: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Work = Potential x Charge - W = E q [Joule] = [Volt] [Coulomb]

Page 9: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.
Page 10: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.
Page 11: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.
Page 12: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.
Page 13: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.
Page 14: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.
Page 15: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.
Page 16: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

NERNST EQUATION

Page 17: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

NERNST EQUATION

At 25 oC

Page 18: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

NERNST EQUATION

Walther Hermann Nernst 1864–1941

Page 19: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

NERNST EQUATION

Page 20: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

NERNST EQUATION

Page 21: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

THERMODYNAMICS and equilibrium

Page 22: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

THERMODYNAMICS and equilibrium

Page 23: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

NERNST EQUATION

Page 24: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

0.50 M AgNO3(aq) 0.010 M Cd(NO3)2(aq)

Page 25: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

0.50 M AgNO3(aq)

0.010 M Cd(NO3)2(aq)

Page 26: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

0.50 M AgNO3(aq)

0.010 M Cd(NO3)2(aq)

My way:

Page 27: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

Page 28: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

Page 29: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

Page 30: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

Page 31: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

Page 32: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

Page 33: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

Page 34: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

BUN: 7 to 20 mg/dL CO2 (carbon dioxide): 20 to 29 mmol/L

Creatinine: 0.8 to 1.4 mg/dL Glucose: 64 to 128 mg/dL

Serum chloride: 101 to 111 mmol/L Serum potassium: 3.7 to 5.2 mEq/L Serum sodium: 136 to 144 mEq/L

CHEM-7 is a group of blood tests that provides information about your body's metabolism. The test is commonly called a basic metabolic panel.

Page 35: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Ion selective electrode

SelectivityAnd

interference

Page 36: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

NERNST EQUATION

Measure E to determine 1 unknown concentration(…..so fix the other concentrations)

Page 37: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Reference electrodes

Page 38: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Reference electrodes

My way:

Page 39: Elechtrochemistry 17. F = (6.022 x 10 23 mol -1 ) x (1.602192 x 10 -19 C) = 96,484 C mol -1 How much? – Faraday’s constant.

Reference electrodes


S TOICHIOMETRY. L ETS R EVIEW - M OLES AND P ARTICLES 1 mole = 6.022 x 10 23 particles (formula units/ molecules/atoms) Moles Particles Multiply by 6.022.

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Web viewAtomic Mass. AM = ∑ fraction of isotopes *(mass of isotopes) Avogadro’s Number . 1 mol=6.022* 10 23 particles . Relations. 1 c m 3 =1 mL . 1 m 3 =1 dL

Web viewAtomic Mass. AM = ∑ fraction of isotopes *(mass of isotopes) Avogadro’s Number . 1 mol=6.022* 10 23 particles . Relations. 1 c m 3 =1 mL . 1 m 3 =1 dL

C THREE CALCULATIONS WITH CHEMICAL FORMULAS AND …snorthrup/chem1110/...And contains 6.022 x 1023 oxygen atoms 12.04 x 1023 hydrogen atoms 6. Example: Molar Mass of H 2 = 2 x 1.008

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Formula Mass and the Mole Concept · Element verageA Atomic Mass (amu) Molar Mass (g/mol) Atoms/Mole C 12.01 12.01 6.022 2310 H 1.008 1.008 6.022 2310 O 16.00 16.00 6.022 2310 Na

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Molar Mass - Chemchem.ws/dl-1007/ch02c-molar_mass.pdf · ‣ molar mass grams amu mol atoms (& molecules) molecular scale Avogadro’s molar scale number 1 mol = 6.022 x 1023 Atomic

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What Is a Mole?. One mole = 6.022 x 10 23 items A mole of atoms Or molecules.

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2019 専門IIApplied Chemistry Il Department Applied Chemistry Applicant Number (Problem 3) 0 (two sheets for Problem 3) 1. 6.626 10-34 J s, 6.022 mol-I, 9.109 x Iff31kg, 2.998 x 108

2019 専門IIApplied Chemistry Il Department Applied Chemistry Applicant Number (Problem 3) 0 (two sheets for Problem 3) 1. 6.626 10-34 J s, 6.022 mol-I, 9.109 x Iff31kg, 2.998 x 108

Glossary Glossar - steinhardt.nyu.edu · atomic radius radio atómico atomic weight peso atómico attraction atracción Avogadro's number (6.022 * 10^23) número de Avogadro (6.022*10^23)

Glossary Glossar - steinhardt.nyu.edu · atomic radius radio atómico atomic weight peso atómico attraction atracción Avogadro's number (6.022 * 10^23) número de Avogadro (6.022*10^23)

mgng molecules ftoI¥e x atoms - Open Computing Facility · I mol H atoms H: (11.19g) 1.008 I mol O atoms O: (88.79 g) 16.00 go — 11.10 mol H atoms 5.549 mol O atoms The formula

mgng molecules ftoI¥e x atoms - Open Computing Facility · I mol H atoms H: (11.19g) 1.008 I mol O atoms O: (88.79 g) 16.00 go — 11.10 mol H atoms 5.549 mol O atoms The formula

Topic 15 - Chemical Equilibrium Notes/Web - Topic 15 - Chemical... · Web view0.249 mol N2, 3.21 x 10(2 mol H2, and 6.42 x 10(4 mol NH3 are placed into a 3.50 L reaction vessel at

Topic 15 - Chemical Equilibrium Notes/Web - Topic 15 - Chemical... · Web view0.249 mol N2, 3.21 x 10(2 mol H2, and 6.42 x 10(4 mol NH3 are placed into a 3.50 L reaction vessel at

INTRODUCTION Atomic, molecular and formula weights€¦ ·  · 2010-01-14One mole (1 mol) of sodium ions contains 6.022 × 1023 Na+ ions One mole (1 mol) ... eg 2. How many neon

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Chapter - 1 : Matter In Our Surroundings Science-9_Flowcharts.… · 1 mole of molecules 6.022 ×1023 atoms of C 6.022 ×1023 atoms of H 6.022 ×10 number of that particle 23 6.022

< > Source :- 1 2013 2.pdf · 71341 AL/2013/02/S-11A / All Rights Sri New S llabus w 11 Chemis 11 15 Oog R = 8.314 J mol—I NA = 6.022 x mol-I A 0030ee - B C - (SQ 9 - 14)

< > Source :- 1 2013 2.pdf · 71341 AL/2013/02/S-11A / All Rights Sri New S llabus w 11 Chemis 11 15 Oog R = 8.314 J mol—I NA = 6.022 x mol-I A 0030ee - B C - (SQ 9 - 14)

< > Source :- 1 · 2018. 9. 1. · 30469 AL/2011/02-S..1 Chemistry Two hours * ea 08 * * I EàO 50 (l), (1), (3), (4),.(5) (X) CE5ê)850. 8.314 J K-l 6.022 x 1023 mol-I . @

< > Source :- 1 · 2018. 9. 1. · 30469 AL/2011/02-S..1 Chemistry Two hours * ea 08 * * I EàO 50 (l), (1), (3), (4),.(5) (X) CE5ê)850. 8.314 J K-l 6.022 x 1023 mol-I . @

Full page photo - meshkat-tab.ir khordad 96/shimi.pdf · mol Cly ?gMnOr = x rr/fLCl i mol M nor AYgMnOr mol CI 'V/fgMnq mol MnOr ppm kg Na ikg ppm= ? kg : kg ikg mol mol . / r fmol

Full page photo - meshkat-tab.ir khordad 96/shimi.pdf · mol Cly ?gMnOr = x rr/fLCl i mol M nor AYgMnOr mol CI 'V/fgMnq mol MnOr ppm kg Na ikg ppm= ? kg : kg ikg mol mol . / r fmol

Quick Review: The Mole A very large counting number = 6.022 x 10 23 (6.022 followed by 23 zeros)

Quick Review: The Mole A very large counting number = 6.022 x 10 23 (6.022 followed by 23 zeros)

Assignment #3 KEY - misszukowski.weebly.com #3 KEY April 9, ... 124.OgP4 = 8.29g I mol P4 63.5 g cu ... (based on HCI) = 5.00 x ICF3 mol HCI x -5.00 x I mol HCI mol

Assignment #3 KEY - misszukowski.weebly.com #3 KEY April 9, ... 124.OgP4 = 8.29g I mol P4 63.5 g cu ... (based on HCI) = 5.00 x ICF3 mol HCI x -5.00 x I mol HCI mol

< > Source :- 1 · 71341 AL/2013/02/S-11A / All Rights Sri New S llabus w 11 Chemis 11 15 Oog R = 8.314 J mol—I NA = 6.022 x mol-I A 0030ee - B C - (SQ 9 - 14)

< > Source :- 1 · 71341 AL/2013/02/S-11A / All Rights Sri New S llabus w 11 Chemis 11 15 Oog R = 8.314 J mol—I NA = 6.022 x mol-I A 0030ee - B C - (SQ 9 - 14)

alpastpapers - Connect with chemistry world... · AL/2016/02rr-1 ) u usu016 Occo TWO hours Chemistry I * 08 * * * * 1 50 1Ðæú L16iT6Tnq (x) "6. 6ülTu.4 R = 8.314 J mol - N = 6.022

alpastpapers - Connect with chemistry world... · AL/2016/02rr-1 ) u usu016 Occo TWO hours Chemistry I * 08 * * * * 1 50 1Ðæú L16iT6Tnq (x) "6. 6ülTu.4 R = 8.314 J mol - N = 6.022

An Introduction to Programming Through C++cs101/lectures/Lec2.pdfIndicative example Let us represent Avogadro’s number 6.022 x 1023 • Convert to binary: 1.11111110001010111 x 21001110

An Introduction to Programming Through C++cs101/lectures/Lec2.pdfIndicative example Let us represent Avogadro’s number 6.022 x 1023 • Convert to binary: 1.11111110001010111 x 21001110