1 suppose we had a 0.3a current producing 15ml of hydrogen gas in 6 min and 30 seconds. how much...
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• 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!!!
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