Ch17 z5e electrochem
Transcript of Ch17 z5e electrochem
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Electrochemistry pp
Applications of Redox
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17.1 Galvanic CellsOxidation reduction reactions involve a
transfer of electrons.OIL- RIGOxidation Involves LossReduction Involves GainLEO-GER Lose Electrons OxidationGain Electrons Reduction
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ApplicationsMoving electrons = electric current. 8H++MnO4
-+ 5Fe+2 Mn+2 + 5Fe+3 +4H2O
It helps to break the reactions into half reactions.
8H++MnO4-+5e- Mn+2 +4H2O
5(Fe+2 Fe+3 + e-) In the same mixture this happens without
doing useful work, but if separate . . .
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H+
MnO4-
Fe+2
Connected this way the reaction starts.Stops immediately because charge builds
up.
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H+
MnO4-
Fe+2
Galvanic Cell
Salt Bridge allows current to flow
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H+
MnO4-
Fe+2
Galvanic Cell
Electrons flow in the wire, ions flow through the salt bridge
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H+
MnO4-
Fe+2
e-
Electricity travels in a complete circuit Instead of a salt bridge . . .
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H+
MnO4-
Fe+2
Porous Disk
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Reducing Agent
Oxidizing Agent
e-
e-
e- e-
e-
e-
Oxidation
at AnodeReduction at Cathode
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Cell PotentialOxidizing agent pulls the electron.Reducing agent pushes the electron. The push or pull (“driving force”) is called
the cell potential EcellAlso called the electromotive force (emf). Unit is the volt (V). = 1 joule of work per coulomb of charge
transferred (1 V = 1 J/C).Measured with a voltmeter.
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17.2 Standard Reduction PotentialsThe reaction in a galvanic cell is a redox
reaction.So, break it down into two half-reactions.Assign a potential to each.Sum the half-cell potentials to get the
overall cell potential. “Active anodes” - the more active metal is
always the anode (good for multiple choice questions).
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Zn+2 SO4-
2
1 M HCl
Anode
0.76
1 M ZnSO4
H+
Cl-
H2 in
Cathode
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1 M HCl
H+
Cl-
H2 in
Standard Hydrogen ElectrodeThis is the reference
all other oxidations are compared to.
Eº = 0 º indicates standard
states of 25ºC, 1 atm, 1 M solutions.
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Z5e 841 Figure 17.5: Zn/H Galvanic Cell. Notice the electron flow also.
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Cell Potential Zn(s) + Cu+2 (aq) Zn+2(aq) + Cu(s)
The total cell potential is the sum of the potential at each electrode.
Eº cell = EºZn Zn+2 + Eº Cu+2 Cu
We can look up reduction potentials in a table (see, p. 796).
For Br, s/b 1.07 (not 1.09) need for online HW!!) Since one of the 1/2 reactions is oxidation its
table value must be reversed, so change its sign.
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Z5e 842 Fig 17.6 Zn/Cu Galvanic Cell
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Cell Potential pp
Determine the cell potential for a galvanic cell based on the redox reaction . . . Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)
steps followFe+3(aq) + e- Fe+2(aq) Eº = 0.77 V
Cu+2(aq)+2e- Cu(s) Eº = 0.34 VSince one of these must be oxidation,
one of them needs to be reversed.Which one?
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Cell Potential pp
Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)
This is spontaneous only if Eºcell = (+), so
reverse the copper half-reaction.
Fe+3(aq) + e- Fe+2(aq) Eº = 0.77 V
Cu(s) Cu+2(aq)+2e- Eº = -0.34 V
Must balance the e-s, so multiply the Fe 1/2-
reaction by 2, BUT do not multiply Eº. Why?
Cell potential is an intensive property (doesn’t depend on number of times the reaction occurs).
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Cell Potential pp
Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq) 2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V Cu(s) Cu+2(aq)+2e- Eº = -0.34 V Cu(s) + 2Fe+3(aq) Cu+2(aq) + 2Fe+2(aq)
Eº = 0.43 V
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Cell Potential
Be sure to use the correct 1/2-reactions!
For example, table 17.1, p. 796 lists three 1/2-reactions for MnO4
1- Find them.
Answers next slide.
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Cell Potential Three 1/2-reactions for MnO4
1- : MnO4
1- + 4H1+ 3e- MnO2 + 2H2O Eº = 1.68 MnO4
1- + 8H1+ 5e- Mn2+ + 4H2O Eº = 1.51 MnO4
1- + e- MnO42- Eº = 0.56
Pick the reaction that “works” with your overall reaction (look at reactants and products).
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Line Notation pp
solidAqueousAqueoussolidAnode on the leftCathode on the rightSingle line to show different phases.Double line porous disk or salt bridge. If all the substances on one side are
aqueous, a platinum electrode is used.For: Cu(s) + Fe+3(aq) Cu+2(aq) + Fe+2(aq)
Cu(s)Cu+2(aq)Fe+3(aq),Fe+2(aq)Pt(s) Remember to show the electrodes!!
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Complete Galvanic Cell Description (AP Test) pp
The reaction always runs spontaneously in the direction that produced a positive cell potential.
Four things for a complete description:1. Cell Potential and balanced reaction2. Direction of flow3. Designation of anode and cathode4. Nature of all components -- electrodes
& ions (plus inert conductor like Pt if needed). Use line notation.
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Practice pp
Completely describe the galvanic cell based on the following half-reactions under standard conditions.
MnO4- + 8 H+ +5e- Mn+2 + 4H2O
Eº = 1.51 V Fe+2
+2e- Fe(s) Eº = -0.44 V
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Practice - Item 1 pp
Determine cell potential & balanced reaction MnO4
- + 8 H+ +5e- Mn+2 + 4H2O Eº = 1.51 V Fe+2 +2e- Fe(s) Eº = -0.44 V
Since (+) E required, reverse the 2nd reaction Eºcell =1.51 + (+0.44) = 1.95 V Complete, balanced reaction is
2MnO4- + 5Fe(s) + 16 H+ 2Mn+2 + 5Fe2+(aq) + 8H2O(l)
Note: multiplying the half reactions to balance the reaction does NOT multiply the Eº values!!! (intensive property).
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Practice - Item 2 pp
Determine e- flow by inspecting 1/2 rxns & using the direction that gives a (+) Eºcell
MnO4- + 8 H+ +5e- Mn+2 + 4H2O Eº = 1.51 V
Fe(s) Fe+2 +2e- Eº =+0.44 V Eºcell =1.51 + (+0.44) = 1.95 V
So, electrons flow from Fe(s) to MnO4- (aq)
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Practice - Item 3 pp
Designate the anode and cathode MnO4
- + 8 H+ +5e- Mn+2 + 4H2O Eº = 1.51 V Fe(s) Fe+2 +2e- Eº =+0.44 V Eºcell =1.51 - (0.44) = 1.95 VElections flow from Fe(s) to MnO4
- So, oxidation occurs in the compartment
containing Fe(s) -- the anode Reduction occurs in the compartment
containing MnO4- -- Use Pt for the cathode
Note: e- always flow from anode to cathode ”Red cat ate an ox". Red/cat = reduct/cathode
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Practice - Item 4 pp
Describe nature of each electrode & ions present (use line notation)
Complete, balanced reaction is 2MnO4
- + 5Fe(s) + 16 H+ 2Mn+2 + 5Fe2+(aq) + 8H2O(l)
Electrode in Fe/Fe2+ compartment is iron metal
An inert conductor like Pt must be used in MnO4
-1/ Mn+2 compartment (don’t forget).
Line notation is: Fe(s)Fe+2(aq)MnO4-
1(aq),Mn+2(aq)Pt(s)
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pp Figure 17.7: A Schematic of the previous Galvanic Cell
Eº = 1.95 v
Be able to draw this as well as write the line notation for the AP exam.
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17.3 Cell Potential, Work & Gemf = potential (V) = work (J) / Charge(C)E = work done by system / chargeE = -w/q (emf & work have opposite signs)
Use (-)w because it is flowing out from system.Charge is measured in coulombs. -w = qE (where q = the charge)Faraday = 96 485 C/mol e-
q = nF = moles of e- x charge per mole e-
w = -qE = -nFE = G
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Potential, Work and G pp
Gº = -nFE º (at standard conditions) if E º > 0, then Gº < 0 spontaneous if E º < 0, then Gº > 0 nonspontaneous In fact, reverse is spontaneous.
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Potential, Work and GCalculate Gº for the following reaction:
Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq) Fe+2(aq) + 2e-Fe(s) Eº = -0.44 V Cu+2(aq)+2e- Cu(s) Eº = 0.34 V
Gº = -nFE º Answer? -1.5 x 105 J Calculation with units is . . . -(2 mol e-)(96 485 C/mol e-)(0.78 J/C)
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Putting It Together pp
Using Table 17.1, predict if 1 M HNO3 will dissolve gold to form a 1 M Au3+ solution?
What are the half reactions? . . .Gold needs to be oxidized so HNO3 must
be reduced. Look for a half-reaction with HNO3 where NO3
1- is being reduced . . . NO3
1- + 4H1+ + 3e- NO + 2H2O Eº = 0.96 v Au Au3+ + 3e- Eº = -1.50 v
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Putting It Together pp
Using Table 17.1, predict if 1 M HNO3 will dissolve gold to form a 1 M Au3+ solution?
NO31- + 4H1+ + 3e- NO + 2H2O Eº = 0.96 v
Au Au3+ + 3e- Eº = -1.50 v Au + NO3
1- + 4H1+ Au3+ + NO + 2H2O Eºcell = -0.54 v
Since Eºcell = negative, this cannot be spontaneous because . . .
Gº = -nFE º = (-)(3)(96 485)(-0.54) = +156kJ
Since Gº is (+), not spontaneous.
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17.4 Cell Potential and Concentration pp
Notes for online HW1 lb = 453.6 g1 Faraday = 96 485 c/s (not 96 500 c/s)Use 0.0592 in Nernst equation (not
0.0591)
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17.4 Cell Potential and Concentration17.4 Cell Potential and Concentration pppp
Qualitatively - Can predict direction of
change in E from LeChâtelier.
2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s)
Predict if Ecell will be greater or less than
Eºcell of 0.48 v if
[Al+3] = 1.5 M and [Mn+2] = 1.0 M if [Al+3] = 1.0 M and [Mn+2] = 1.5 M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M
Steps . . .
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Cell Potential pp
2Al(s) + 3Mn2Al(s) + 3Mn+2+2(aq)(aq) 2Al2Al+3+3(aq)(aq) + 3Mn(s) + 3Mn(s)
Predict if Predict if EEcellcell will be greater or less than will be greater or less than
EEººcellcell of 0.48 v of 0.48 v if if [Al[Al+3+3] = 1.5 ] = 1.5 MM and and [Mn[Mn+2+2] = 1.0 ] = 1.0 MM
Answer . . .Answer . . . Since a Since a productproduct [ ] [ ] has been raised above has been raised above
1.0 1.0 MM, Le Chatelier predicts a shift , Le Chatelier predicts a shift leftleft and Eand Ecellcell < Eº < Eºcellcell
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Cell Potential pp
2Al(s) + 3Mn2Al(s) + 3Mn+2+2(aq)(aq) 2Al2Al+3+3(aq)(aq) + 3Mn(s) + 3Mn(s)
Predict if Predict if EEcellcell will be greater or less than will be greater or less than
EEººcellcell of 0.48 v of 0.48 v if if [Al[Al+3+3] = 1.0 ] = 1.0 MM and and [Mn[Mn+2+2] = 1.5 ] = 1.5 MM
Answer . . .Answer . . . Since a Since a reactantreactant [ ] [ ] has been raised has been raised
above 1.0 above 1.0 MM, Le Chatelier predicts a shift , Le Chatelier predicts a shift rightright and E and Ecellcell > Eº > Eºcellcell
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Cell Potential pp
2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s)
Predict if Ecell will be greater or less than
Eºcell of 0.48 v if [Al+3] = 1.5 M and [Mn+2] = 1.5 M
Answer . . .Since both [ ]s have been raised above
1.0 M, cannot use Le Chatelier!Must use the Nernst Equation (coming to
your community soon!)
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Le Chatelier, ∆G, & Concentration Cells Figure 17.9
A Concentration Cell That Contains a Sliver Electrode and Aqueous Silver Nitrate in Both Compartments
Since the right compartment has higher [Ag1+] there is a shift right of e-
So, Ag metal is deposited on the right side while [Ag1+] decreases on right side and increases on the left.
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Le Chatelier, ∆G, & Concentration Cells pp
So, Ag metal is deposited on the right side while [Ag1+] decreases on right side and increases on the left.
You will need to recognize this concept to solve for ∆G on the AP exam.
Hint: the electrode with the largest [ ] will always be the cathode (where reduction occurs).
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The Nernst EquationG = Gº +RTln(Q), since G = -nFE . . . -nFE = -nFEº + RTln(Q)
E = Eº - RTln(Q) nF
What is n in Al(s) + Mn2+ Al3+ + Mn(s)? Always have to figure out “n” by balancing.
2Al(s) + 3Mn+2(aq) 2Al+3(aq) + 3Mn(s) Eº = 0.48 V n = 6. Why? . . . n = mole of e-, not mole of compound.
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The Nernst Equation continued G = Gº +RTln(Q)
-nFE = -nFEº + RTln(Q)
E = Eº - RTln(Q) = Eº - 2.303RT log (Q)
nF nF
Since we know R and F, at 25o C, above is aka
E = Eº - 0.0592log(Q)
n Textbook has typo: 0.0591 should be 0.0592 Use 0.0592 for all your calculations!
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The Nernst Equation pp
E = Eº - 0.0592log(Q)
n For concentration cells (i.e., not at 1 M) this
equation must be done separately for each 1/2 cell, then subtract the results (or flip one and add) to get the cell potential.
See the following problem . . .
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We’ll do “a,” “b,” & “c”.We’ll do “a,” “b,” & “c”.
pppp
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Ecell when [Ag1+] on the right = 1.0 M pp
Since [Ag1+] is same on both sides Ecell = Eºcell which is 0 because . . .
Ag1+ + e- Ag Eº = .80v Ag Ag1+ + e- Eº = -.80v
Eºcell = 0
E = Eº - 0.0592log(Q) n
Since log(1) = 0, Ecell = Eºcell
and Eº = 0 from above, so Ecell also = 0.
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Ecell when [Ag1+] on the right = 2.0 M pp
Cathode always has the higher [ ] and e- always flow from the anode to the cathode.
Q = [ ]anode ÷ [ ]cathode
Here, [cathode] is on the right (2.0 M), & in the denominator for Q
E = Eº - 0.0592log(Q)
n
E = 0 - 0.0592log 1.0 = .018 v
1 2.0
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Ecell when [Ag1+] on the right = 0.10 M pp
Cathode always has the higher [ ] and e- always flow from the anode to the cathode.
So, [cathode] is on the left & in the denominator for Q
E = Eº - 0.0592log(Q)
n
E = 0 - 0.0592log 0.1 = .059 v
1 1.0
You will have a test question on this and it is NOT on the pre-test, so . . .
Do p. 832 #53!!
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Ecell when [Ag1+] on the right = 0.10 M pp
Notes for all of these: Eº is always 0 for
concentration cells because the 1/2 reactions cancel.
E = Eº - 0.0592log(Q)
n
E = 0 - 0.0592log anode .
n cathode Memorize this equation!
You will have a test question on this and it is NOT on the pre-test, so . . .
Do p. 832 #53!!
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The Nernst Equation As reactions proceed concentrations of
products increase and reactants decrease. Reaches equilibrium where Q = K and Ecell = 0
Since at equilibrium Ecell = 0 = Eº - RTln(K)
nF
Eº = RTln(K)
nF
nFEº = ln(K) at 25º C aka log(K) = nEº RT 0.0592
Use ln(K) version in online HW even if at 25ºC!!
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Nernst Equation & K pp
Calculate KCalculate Kspsp of silver iodide at 298 K of silver iodide at 298 K AgI(s) AgI(s) Ag Ag++ + I + I-- where Ewhere Eoo
AgI(s) + eAgI(s) + e-- Ag(s) + I Ag(s) + I-- -0.15 v-0.15 vII22(s) + 2e(s) + 2e- - 2I 2I-- +0.54 v+0.54 vAgAg++ + e + e-- Ag(s) Ag(s) +0.80 v+0.80 v
logK = nEº/0.0592logK = nEº/0.0592 1st, get Eº1st, get Eºcellcell for overall reaction. Steps. for overall reaction. Steps. Find overall reaction, then EºFind overall reaction, then Eºcellcell
Only need 1st & 3rd equations to get . . .Only need 1st & 3rd equations to get . . . AgI(s) AgI(s) Ag Ag++ + I + I-- where Ewhere Eoo
cellcell = -0.95 v = -0.95 v
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Nernst Equation & K pp
Calculate Ksp of silver iodide at 298 K AgI(s) Ag+ + I-
where Eocell = -0.95 v
at 25ºC log(K) = nEº = (1)(-0.95) = -16.05 0.0592 0.0592
Ksp = 10-16.05 = 9.0 x 10-17
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17.5 Batteries are Galvanic CellsCar batteries are lead storage batteries.Pb +PbO2 +H2SO4 PbSO4(s) +H2OBe able to recognize the anode &
cathode from the half reactions.
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Figure 17.13One of the Six Cells in Storage Battery a 12-V Lead Storage Battery
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Batteries are Galvanic Cells Dry CellDry Cell - - acidacid battery battery
Zn + 2NHZn + 2NH44++ + 2MnO + 2MnO2 2 Zn Zn+2 +2 + 2NH+ 2NH3 3 + Mn+ Mn22OO33 + H + H22OO
Dry Cell - Dry Cell - alkalinealkaline batterybattery
Zn + 2MnOZn + 2MnO22 ZnO + Mn ZnO + Mn22OO3 3 (in base)(in base)
NiCadNiCad - can be re-charged indefinitely - can be re-charged indefinitelyNiONiO22 + Cd + 2H + Cd + 2H22O O Cd(OH) Cd(OH)2 2 +Ni(OH)+Ni(OH)22
Fuel CellFuel Cell - - reactants are continuously supplied reactants are continuously supplied
CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22O + energy O + energy
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Figure 17.14A Common Dry Cell Battery
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17.6 Corrosion RustingRusting - spontaneous - spontaneous oxidationoxidation.. Most structural metals have Most structural metals have reductionreduction
potentials that are potentials that are lessless positive than O positive than O22 . .
So, they So, they oxidizeoxidize while O while O22 is is reduced.reduced.
FeFe+2+2 +2e +2e-- Fe Fe EEº= º= --0.44 V O0.44 V O22
+ 2H+ 2H22O + 4eO + 4e- - 4OH4OH-- EEº= º= ++0.40 V0.40 V Reverse top half reaction, then add bothReverse top half reaction, then add both Fe + OFe + O2 2 + H+ H22O O FeFe2 2 OO33(rust)(rust) + H+ H++ EºEºcellcell = 0.84 v = 0.84 v
Reaction happens in two places . . . Reaction happens in two places . . .
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Water
Rust
Iron Dissolves - Fe Fe+2
e-
Salt speeds up process by increasing conductivity
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Figure 17.17The Electrochemical
Corrosion of Iron
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Preventing CorrosionCoating - to keep out air and water.Galvanizing - Putting on a zinc coatZinc has a lower reduction potential than
iron, so it is more easily oxidized. So, zinc is a more active metal than iron.Alloying with metals that form oxide coats.Cathodic Protection - Attaching large
pieces of a more active metal like magnesium that get oxidized instead of iron (iron stays reduced).
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Preventing Corrosion Cathodic Protection - Attaching large Cathodic Protection - Attaching large
pieces of an active metal like magnesium pieces of an active metal like magnesium that get oxidized instead of iron.that get oxidized instead of iron.Attach Mg wireAttach Mg wire to iron pipe (and replace to iron pipe (and replace periodically).periodically).
Attach titanium barsAttach titanium bars to ships’ hulls. In to ships’ hulls. In salt water the Ti acts as the anode and is salt water the Ti acts as the anode and is oxidized instead of the steel hull, which oxidized instead of the steel hull, which now acts as the cathode.now acts as the cathode.
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Figure 17.18Cathodic
Protection
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Running a galvanic cell backwards. Put a voltage whose magnitude is bigger than
the potential which reverses the direction of the redox reaction.
Produces a chemical change which would not normally happen because the potential is negative.
Remember: 1 A = 1 C/s and 1 F = 96 485 C Used for electroplating -- depositing the neutral
metal onto the electrode by reducing the metal ions in solution.
17.7 Electrolysis
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1.0 M
Zn+2
e- e-
Anode Cathode
1.10
Zn Cu1.0 M
Cu+2
Galvanic Cell - spontaneous
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1.0 M
Zn+2
e- e-
AnodeCathode
A battery >1.10V
Zn Cu1.0 M
Cu+2
Electrolytic Cell Forces the opposite reaction.
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Figure 17.19(a) A Standard Galvanic Cell
(b) A Standard Electrolytic Cell
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Calculating plating Have to include the charge.Have to include the charge. Measure current Measure current II (in amperes) (in amperes) 1 amp = 1 coulomb of charge per second1 amp = 1 coulomb of charge per second
1 A = 1 C/s1 A = 1 C/s q = q = II x t = the charge x t = the charge q/nF = moles of metalq/nF = moles of metal Mass of plated metalMass of plated metal
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Calculating plating ppHow many minutes must a 5.00 amp
current be applied to produce 10.5 g of Ag from Ag+ Steps follow . . .
Set up a picket fence that includes – Current & time– Quantity of charge (in coulombs)– Moles of electrons– Moles of metal (may be different)– Grams of metal
Arrange the picket fence so the units give you what you’re looking for.
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Calculating plating pp How many minutes must a 5.00 amp
current be applied to produce 10.5 g of Ag from Ag+ Steps follow . . .
The pieces of the picket fence are:– 5.00 amp (rewrite as 5.00
C/s)– 10.5 g Ag– 107.868 g Ag/1mol Ag– 1 mol e-/mol Ag (Ag Ag1+ + 1e-)– 96 485 C/mol e-
– 60 sec/min Your answer? . . .
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Calculating platingCalculating plating pppp
How many minutes must a 5.00 amp current be applied to produce 10.5 g of Ag from Ag+ (amp = C/s)
31.3 minutes31.3 minutes (10.5 g Ag)(1 mol Ag/107.868 g Ag)(1 mol e-/1 mol
Ag)(96 485 C/1 mol e-)(1 s/5.00C)(1 min1 min/60 s)
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Calculating plating pp
An antique automobile bumper is to be chrome plated by dipping into an acidic Cr2O7
2- solution serving as the cathode of an electrolytic cell. MCr = 51.996
Using 10.0 amperes, how long to deposit 1.00 x 102 grams of Cr(s)? Steps . . .
Find overall reaction from 1/2-reactions (to get moles of e-). Steps follow.
The only substances you start with are Cr2O7
2- and H2O. Which is oxidized? Reduced? . . .
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Calculating plating pp
Using 10.0 amperes, how long to deposit Using 10.0 amperes, how long to deposit 1.00 x 101.00 x 1022 grams of Cr(s) by dipping into grams of Cr(s) by dipping into acidic Cracidic Cr22OO77
2-2-(aq) , (aq) , MMCrCr = 51.996? = 51.996?
The only substances you start with are The only substances you start with are CrCr22OO77
2-2- and H and H22O. O. Which is oxidized? Which is oxidized? Which is reduced?Which is reduced? . . . . . .
CrCr22OO772-2- must be reduced must be reduced to get to Cr to get to Cr(s)(s) so so
HH22O must be oxidizedO must be oxidized. Use Table 17.1 p. . Use Table 17.1 p. 796 to find the 1/2-reaction of H796 to find the 1/2-reaction of H22O . . . O . . .
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Calculating plating pp
Use Table 17.1 p. 843 to find the 1/2-Use Table 17.1 p. 843 to find the 1/2-reaction of Hreaction of H22O. Which is it?O. Which is it?HH22OO22 + 2H + 2H1+ 1+ + 2e+ 2e-- 2H 2H220 0 2H2H220 0 O O22 + 4H + 4H1+1+ + 4e + 4e--
OO22 + 2H + 2H220 + 4e0 + 4e- - 4OH4OH1-1- 2H 2H220 0 + 2e+ 2e-- H H22 + 2OH + 2OH1-1-
*****
#2 above is the only one that starts #2 above is the only one that starts with water and is oxidized.with water and is oxidized.
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Calculating plating pp
Using 10.0 amperes, how long to deposit 1.00 x 102 grams of Cr(s) by dipping into acidic Cr2O7
2-(aq) ? , MCr = 51.996
So, the 1/2-reactions needed are . . .6e- + 14H1+ + Cr2O7
2- 2Cr3+ + 7H2O 2H20 O2 + 4H1+ + 4e-
and one more! (Why?)You only got to Cr3+, but you need Cr(s)
Also need Cr3+ + 3e- Cr(s)Now, write the balanced equation . . .
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Calculating plating pp
Using 10.0 amperes, how long to deposit 1.00 x 102 grams of Cr(s) by dipping into acidic Cr2O7
2-(aq) ? , MCr = 51.996
Now, write the balanced equation . . .2H1+ + Cr2O7
2- H2O + 3O2 + 2Cr(s)How many mol e- changed in this reaction
(comprised of three 1/2-reactions)?12 mol e- Now, you can start to do the problem.Your answer in days? . . .
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Calculating plating pp
Using 10.0 amperes, how long to deposit 1.00 x 102 grams of Cr(s) by dipping into acidic Cr2O7
2-(aq)? , MCr = 51.996 with 2H1+
+ Cr2O72- H2O + 3O2 + 2Cr(s) and 12
mol e- moving.Your answer is . . .1.29 days . . . 100.g Cr(s) • 1mol Cr/51.996g Cr(s) • 12 mol e-/2 mol
Cr(s) • 96 486 C/1 mol e- • 1s/10.0C • 1 h/3600 s • 1 d/24 h = 1.29 days
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Calculating plating Electrolysis of a molten salt , MCl, using Electrolysis of a molten salt , MCl, using
3.86 amps for 16.2 min deposits 1.52 g of 3.86 amps for 16.2 min deposits 1.52 g of metal. What is the metal?metal. What is the metal?
Use picket fence to get moles of metal. Use picket fence to get moles of metal. Since g/mol = M, use calculated moles of Since g/mol = M, use calculated moles of metal and 1.52 g to get M. Answer . . . metal and 1.52 g to get M. Answer . . .
Potassium.Potassium. The solution is . . . The solution is . . .
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Calculating plating Electrolysis of a molten salt , MCl, using Electrolysis of a molten salt , MCl, using
3.86 amps for 16.2 min deposits 1.52 g of 3.86 amps for 16.2 min deposits 1.52 g of metal. What is the metal?metal. What is the metal? 3.86 C/s • 16.2 min • 60s/1m • 1mol e-/96 485 C • 1 mol 3.86 C/s • 16.2 min • 60s/1m • 1mol e-/96 485 C • 1 mol MM1+1+/1mol e- = 0.039 = moles of metal./1mol e- = 0.039 = moles of metal.
1.52g/0.039 mol = 39.1 g/mol = potassium1.52g/0.039 mol = 39.1 g/mol = potassium
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Other uses pp
Electrolysis of water.Electrolysis of water. Separating mixtures of ions Separating mixtures of ions A more positive reduction potential A more positive reduction potential
means that reaction proceeds forward. means that reaction proceeds forward. The metal with the most The metal with the most positivepositive
reductionreduction potential is easiest to plate out potential is easiest to plate out of solution (and is the best oxidizer).of solution (and is the best oxidizer).
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Relative Oxidizing Abilities pp An acidic solution has CeAn acidic solution has Ce4+4+, VO, VO22
1+1+, & Fe, & Fe3+3+. Use . Use Table 17.1 p. 796 to predict the order of Table 17.1 p. 796 to predict the order of oxidizingoxidizing ability. Answer . . . ability. Answer . . .
Order of oxidizing ability is the same as the Order of oxidizing ability is the same as the order for order for beingbeing reduced, So. . . reduced, So. . . CeCe4+4+ + e + e-- CeCe3+3+ E = 1.70 vE = 1.70 v VOVO22
1+1+ + 2H + 2H1+1+ + e + e-- VOVO2+2+ + H + H22OO E = 1.00 vE = 1.00 v FeFe3+3+ + e + e-- FeFe2+2+ E = 0.77 vE = 0.77 v
Also, predict which one will be Also, predict which one will be reducedreduced at the at the cathode of an cathode of an electrolyticelectrolytic cell at the cell at the lowestlowest voltage. Answer . . .voltage. Answer . . .
Since Since CeCe4+4+ is the greatest oxidizer it is the most is the greatest oxidizer it is the most easily reduced (needs least voltage).easily reduced (needs least voltage).
Do section 17.8 on your own. Do section 17.8 on your own.