11

• Suppose we had a 0.3A current producing 15mL of hydrogen gas in 6 min and 30 seconds.

• How much charge did we have per one mole of electrons?

• We need to find the temp, atmospheric pressure and calculate moles… But since the gas was collected over water… What else do we need?

• Lets turn to A-24 for the chart.

Lets look at the Lets look at the hydrolysis of water…hydrolysis of water…

22

Don’t Hesitate… Let’s Don’t Hesitate… Let’s Calculate! Calculate!

33

How close are we to the “correct answer?”How close are we to the “correct answer?”

Current = charge passing

timeCurrent =

charge passingtime

I (amps) = coulombsseconds

I (amps) = coulombsseconds

Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

= = 96,500 C/mol e- 96,500 C/mol e- = = 1 Faraday1 Faraday

Charge on 1 mol e -

= 1.60 x 10-19 Ce -

6.02 x 1023

e -mol

44

Well that is swell, but Well that is swell, but what is the use of this?what is the use of this?

Consider electrolysis of aqueous silver ion.Consider electrolysis of aqueous silver ion.

AgAg++ (aq) + e- ---> Ag(s) (aq) + e- ---> Ag(s)

1 mol e-1 mol e- ---> 1 mol Ag---> 1 mol Ag

If we could measure the moles of e-, we If we could measure the moles of e-, we could know the quantity of Ag formed.could know the quantity of Ag formed.

But how to measure moles of e-?But how to measure moles of e-?

Current = charge passing

timeCurrent =

charge passingtime

I (amps) = coulombsseconds

I (amps) = coulombsseconds

55Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 (aq) solution for 15.0 min. What mass of Ag metal is deposited?min. What mass of Ag metal is deposited?

SolutionSolution

(a)(a) Calc. chargeCalc. charge

Charge (C) = current (A) x time (t)Charge (C) = current (A) x time (t)

= (1.5 amps)(15.0 min)(60 s/min) = 1350 C= (1.5 amps)(15.0 min)(60 s/min) = 1350 C

I (amps) = coulombsseconds

I (amps) = coulombsseconds

66Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

SolutionSolution

(a)(a) Charge = 1350 CCharge = 1350 C

(b)(b) Calculate moles of e- usedCalculate moles of e- used

I (amps) = coulombsseconds

I (amps) = coulombsseconds

1350 C • 1 mol e -96, 500 C

0.0140 mol e -1350 C • 1 mol e -96, 500 C

0.0140 mol e -

0.0140 mol e - • 1 mol Ag1 mol e -

0.0140 mol Ag or 1.51 g Ag0.0140 mol e - • 1 mol Ag1 mol e -

0.0140 mol Ag or 1.51 g Ag

1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min. What (aq) solution for 15.0 min. What mass of Ag metal is deposited?mass of Ag metal is deposited?

(c)(c) Calc. quantity of AgCalc. quantity of Ag

77Your turn to try! Your turn to try!

The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, how If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?long will the battery last?

SolutionSolution

a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb

b)b) Calculate moles of e-Calculate moles of e-

2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -

c)c) Calculate chargeCalculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C4.38 mol e- • 96,500 C/mol e- = 423,000 C

88Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(s) + HSOPb(s) + HSO44--(aq) ---> PbSO(aq) ---> PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, how If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?long will the battery last?

SolutionSolution

a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb

b)b) Mol of e- = 4.38 molMol of e- = 4.38 mol

c)c) Charge = 423,000 CCharge = 423,000 C

Time (s) = Charge (C)

I (amps)Time (s) =

Charge (C)I (amps)

Time (s) = 423, 000 C1.50 amp

= 282,000 sTime (s) = 423, 000 C1.50 amp

= 282,000 s About 78 hoursAbout 78 hours

d)d) Calculate timeCalculate time

99

Michael FaradayMichael Faraday1791-18671791-1867

Originated the terms anode, Originated the terms anode, cathode, anion, cation, cathode, anion, cation, electrode.electrode.

Discoverer of Discoverer of • electrolysiselectrolysis• magnetic props. of mattermagnetic props. of matter• electromagnetic inductionelectromagnetic induction• benzene and other organic benzene and other organic

chemicalschemicalsWas a popular lecturer…. Was a popular lecturer….

Like your teacher!!!!Like your teacher!!!!

I would I would love to love to hear an hear an

oxidation oxidation

Haiku!Haiku!

I would I would love to love to hear an hear an

oxidation oxidation

Haiku!Haiku!

Where did this Faraday Where did this Faraday stuff come from stuff come from

anywaysanyways

1010

Oxidation Haiku!Oxidation Haiku!

• Lost an electron

• But now feeling positive

• Oxidized is cool!

Got any Got any more more

Haiku’s?Haiku’s?

Got any Got any more more

Haiku’s?Haiku’s?

1111

Reduction Haiku!!!

• Gained some electrons

• Gave me a negative mood!

• Now I can say Ger!

I really I really got a got a

coulomb coulomb out of out of

those!those!

I really I really got a got a

coulomb coulomb out of out of

those!those!

1212

YES! EYES! Eoo is related to ∆G is related to ∆Goo, the free , the free energy change for the reaction.energy change for the reaction.

∆∆GGoo = - n F E = - n F Eoo where F = Faraday constant where F = Faraday constant

= 96,500 J/V•mol (C/mol)= 96,500 J/V•mol (C/mol)

and n is the number of moles of and n is the number of moles of electrons transferredelectrons transferred

Michael FaradayMichael Faraday1791-18671791-1867

I know you are I know you are craving to know the craving to know the

answer to one answer to one question… are Equestion… are Eoo and ∆Gand ∆Go o related???related???

Hey! Hey! What What

Gibbs? Gibbs? Why am I Why am I stuck in stuck in

the the

middle?middle?

Hey! Hey! What What

Gibbs? Gibbs? Why am I Why am I stuck in stuck in

the the

middle?middle?

1313

EEoo and ∆G and ∆Goo

∆∆GGoo = - n F E = - n F Eoo

For a For a product-favoredproduct-favored reaction reaction

Reactants ----> ProductsReactants ----> Products

∆∆GGo o < 0 and so E < 0 and so Eo o > 0 > 0

EEoo is positive is positive

For a For a reactant-favoredreactant-favored reaction reaction

Reactants <---- ProductsReactants <---- Products

∆∆GGo o > 0 and so E > 0 and so Eo o < 0 < 0

EEoo is negative is negative

Hey Kids! Hey Kids! Have a Have a Great Great

Day.. Not Day.. Not just a just a

Faraday!Faraday!

Hey Kids! Hey Kids! Have a Have a Great Great

Day.. Not Day.. Not just a just a

Faraday!Faraday!

1414E at Nonstandard E at Nonstandard

ConditionsConditions

• The The NERNST EQUATIONNERNST EQUATION• E = potential under nonstandard conditionsE = potential under nonstandard conditions

• n = no. of electrons exchangedn = no. of electrons exchanged

• ln = “natural log”ln = “natural log”

• If [P] and [R] = 1 mol/L, then E = E˚If [P] and [R] = 1 mol/L, then E = E˚

• If [R] > [P], then E is ______________ than E˚If [R] > [P], then E is ______________ than E˚

• If [R] < [P], then E is ______________ than E˚If [R] < [P], then E is ______________ than E˚

0 0.0257 V [Products] E - ln

n [Reactants]E 0 0.0257 V [Products]

E - ln n [Reactants]

E

more positivemore positivemore positivemore positive

less positiveless positiveless positiveless positive

At 25At 25ooCCAt 25At 25ooCCWalther Nernst, the Walther Nernst, the

famous German famous German physical chemist, physical chemist,

developed an electric developed an electric lamp, known as the lamp, known as the

"Nernst lamp", which "Nernst lamp", which he sold for a very large he sold for a very large

sum of money. A sum of money. A colleague of his, not colleague of his, not without spite asked without spite asked

him whether his next him whether his next project will be making project will be making

diamonds. diamonds.

Walther Nernst, the Walther Nernst, the famous German famous German physical chemist, physical chemist,

developed an electric developed an electric lamp, known as the lamp, known as the

"Nernst lamp", which "Nernst lamp", which he sold for a very large he sold for a very large

sum of money. A sum of money. A colleague of his, not colleague of his, not without spite asked without spite asked

him whether his next him whether his next project will be making project will be making

diamonds. diamonds.

1515

0 0.0592 V E - log Q

nE 0 0.0592 V

E - log Qn

E

0 E -(RT/nF) ln QE 0 E -(RT/nF) ln QE

Other forms of Other forms of the Nernst the Nernst EquationEquation

At 25At 25ooCCAt 25At 25ooCC

Nernst Nernst answered, answered, "No, I can "No, I can

afford to buy afford to buy them now, so them now, so I don't need I don't need

to make to make

them".them".

Nernst Nernst answered, answered, "No, I can "No, I can

afford to buy afford to buy them now, so them now, so I don't need I don't need

to make to make

them".them".

1616

• The cell potential changes as the concentrations change. As reactants are converted to products, the value of Enet must decline from initially positive value to zero

• A potential of zero means that no net reaction occurs. The cell is at equilibrium.

• Therefore:

• E= 0 = Eo – (0.0592 V/n) log K

• log K = nEo/0.0592 (at 25C)

• pg 980 example 20.10

E and the Equilibrium E and the Equilibrium ConstantConstant

I am I am getting getting bored, bored,

make this make this your last your last

example!example!

I am I am getting getting bored, bored,

make this make this your last your last

example!example!

1717In the following reaction….

Fe(s) + Cd2+(aq) Fe2+

(aq) + Cd (s) Eonet = +0.04 V

a.) What is the value of the equilibrium constant?

b.) What are the equilibrium concentrations of Fe2+ and Cd2+ ions if each began with a concentration of 1.0M?

a.) log K = (2.00) (0.04 V)/ 0.05920.0592 = 1.35

K = 22.45

b.) K = 22.45= [Fe2+]/ [Cd2+] = (1.0 +X)/(1.0 –X)

X= 0.9147

Therefore Fe2+= 1.9 M and Cd2+ = 0.10 M

a.) log K = (2.00) (0.04 V)/ 0.05920.0592 = 1.35

K = 22.45

b.) K = 22.45= [Fe2+]/ [Cd2+] = (1.0 +X)/(1.0 –X)

X= 0.9147

Therefore Fe2+= 1.9 M and Cd2+ = 0.10 M

Wow! What an Wow! What an exciting way to exciting way to

end your lectures end your lectures for the year!!! for the year!!!

Wow! What an Wow! What an exciting way to exciting way to

end your lectures end your lectures for the year!!! for the year!!!