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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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

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