1
Electrochemistry pp
Applications of Redox

2
17.1 Galvanic Cells
Oxidation reduction reactions involve a
transfer of electrons.
OIL- RIG
Oxidation Involves Loss
Reduction Involves Gain
LEO-GER
Lose Electrons Oxidation
Gain Electrons Reduction

3
Applications
Moving 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 . . .

4
H+
MnO4
-
Fe+2
Connected this way the reaction starts.
Stops immediately because charge builds
up.

5
H+
MnO4
-
Fe+2
Galvanic Cell
Salt
Bridge
allows
current
to flow

6
H+
MnO4
-
Fe+2
Galvanic Cell
Electrons
flow in
the wire,
ions flow
through
the salt
bridge

7
H+
MnO4
-
Fe+2
e-
Electricity travels in a complete circuit
Instead of a salt bridge . . .

8
H+
MnO4
-
Fe+2
Porous
Disk

9
Reducing
Agent
Oxidizing
Agent
e-
e-
e-
e-
e-
e-
Oxidation
at Anode
Reduction at
Cathode

10
Cell Potential
Oxidizing agent pulls the electron.
Reducing agent pushes the electron.
The push or pull (“driving force”) is called
the cell potential Ecell
Also 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.

11
17.2 Standard Reduction Potentials
The 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).

12
Zn+2
SO4
-
2
1 M HCl
Anode
0.76
1 M ZnSO4
H+
Cl-
H2 in
Cathode

13
1 M HCl
H+
Cl-
H2 in
Standard Hydrogen Electrode
This is the reference
all other oxidations
are compared to.
Eº
= 0
º indicates standard
states of 25ºC,
1 atm, 1 M
solutions.

14
Z5e 841 Figure 17.5: Zn/H Galvanic Cell.
Notice the electron flow also.

15
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.

16
Z5e 842 Fig 17.6 Zn/Cu Galvanic Cell

17
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 follow
Fe+3
(aq)+ e-
→ Fe+2
(aq) Eº = 0.77 V
Cu+2
(aq)+2e-
→ Cu(s) Eº = 0.34 V
Since one of these must be oxidation,
one of them needs to be reversed.
Which one?

18
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).

19
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

20
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.

21
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-
→ MnO4
2-
Eº = 0.56
Pick the reaction that “works” with your overall
reaction (look at reactants and products).

22
Line Notation pp
solidAqueousAqueoussolid
Anode on the leftCathode on the right
Single 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!!

23
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 reaction
2. Direction of flow
3. Designation of anode and cathode
4. Nature of all components -- electrodes
& ions (plus inert conductor like Pt if
needed). Use line notation.

24
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

25
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).

26
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)

27
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 V
Elections 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

28
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)

29
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.

30
17.3 Cell Potential, Work & ∆G
emf = potential (V) = work (J) / Charge(C)
E = work done by system / charge
E = -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

31
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.

32
Potential, Work and ∆G
Calculate ∆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)

33
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

34
Putting It Together pp
Using Table 17.1, predict if 1 M HNO3 will
dissolve gold to form a 1 M Au3+
solution?
NO3
1-
+ 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.

35
17.4 Cell Potential and Concentration pp
Notes for online HW
1 lb = 453.6 g
1 Faraday = 96 485 c/s (not 96 500 c/s)
Use 0.0592 in Nernst equation (not
0.0591)

36
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 . . .

37
Cell Potential pp
2Al(s) + 3Mn2Al(s) + 3Mn+2+2
(aq)(aq) →→ 2Al2Al+3+3
(aq)(aq) + 3Mn(s)+ 3Mn(s)
Predict ifPredict if EEcellcell will be greater or less thanwill be greater or less than
EEººcellcell of 0.48 vof 0.48 v
ifif [Al[Al+3+3
] = 1.5] = 1.5 MM andand [Mn[Mn+2+2
] = 1.0] = 1.0 MM
Answer . . .Answer . . .
Since aSince a productproduct [ ][ ] has been raised abovehas been raised above
1.01.0 MM, Le Chatelier predicts a shift, Le Chatelier predicts a shift leftleft
and Eand Ecellcell < Eº< Eºcellcell

38
Cell Potential pp
2Al(s) + 3Mn2Al(s) + 3Mn+2+2
(aq)(aq) →→ 2Al2Al+3+3
(aq)(aq) + 3Mn(s)+ 3Mn(s)
Predict ifPredict if EEcellcell will be greater or less thanwill be greater or less than
EEººcellcell of 0.48 vof 0.48 v ifif
[Al[Al+3+3
] = 1.0] = 1.0 MM andand [Mn[Mn+2+2
] = 1.5] = 1.5 MM
Answer . . .Answer . . .
Since aSince a reactantreactant [ ][ ] has been raisedhas been raised
above 1.0above 1.0 MM, Le Chatelier predicts a shift, Le Chatelier predicts a shift
rightright and Eand Ecellcell > Eº> Eºcellcell

39
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!)

40
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.

41
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).

42
The Nernst Equation
∆G = ∆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.

43
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!

44
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 . . .

45
We’ll do “a,” “b,” & “c”.We’ll do “a,” “b,” & “c”.
pppp

46
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.

47
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

48
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!!

49
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!!

50
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

51
Nernst Equation & K pp
Calculate KCalculate Kspsp of silver iodide at 298 Kof silver iodide at 298 K
AgI(s)AgI(s) →→ AgAg++
+ I+ I--
where Ewhere Eoo
AgI(s) + eAgI(s) + e--
→→ Ag(s) + IAg(s) + I--
-0.15 v-0.15 v
II22(s) + 2e(s) + 2e--
→→ 2I2I--
+0.54 v+0.54 v AgAg++
+ 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) →→ AgAg++
+ I+ I--
where Ewhere Eoo
cellcell = -0.95 v= -0.95 v

52
Nernst Equation & K pp
Calculate Ksp of silver iodide at 298 K
AgI(s) → Ag+
+ I-
where Eo
cell = -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

53
17.5 Batteries are Galvanic Cells
Car batteries are lead storage batteries.
Pb +PbO2 +H2SO4 →PbSO4(s) +H2O
Be able to recognize the anode &
cathode from the half reactions.

54
Figure 17.13
One of the Six Cells
in Storage Battery a
12-V Lead Storage
Battery

55
Batteries are Galvanic Cells
Dry CellDry Cell -- acidacid batterybattery
Zn + 2NHZn + 2NH44
++
+ 2MnO+ 2MnO22 →→ ZnZn+2+2
+ 2NH+ 2NH33 + Mn+ Mn22OO33 + H+ H22OO
Dry Cell -Dry Cell - alkalinealkaline batterybattery
Zn + 2MnOZn + 2MnO22 →→ ZnO + MnZnO + Mn22OO33 (in base)(in base)
NiCadNiCad - can be re-charged indefinitely- can be re-charged indefinitely
NiONiO22 + Cd + 2H+ Cd + 2H22OO →→ Cd(OH)Cd(OH)22 +Ni(OH)+Ni(OH)22
Fuel CellFuel Cell -- reactants are continuously suppliedreactants are continuously supplied
CHCH44 + 2O+ 2O22 →→ COCO22 + 2H+ 2H22O + energyO + energy

56
Figure
17.14
A
Common
Dry Cell
Battery

57
17.6 Corrosion
RustingRusting - spontaneous- spontaneous oxidationoxidation..
Most structural metals haveMost structural metals have reductionreduction
potentials that arepotentials that are lessless positive than Opositive than O22 ..
So, theySo, they oxidizeoxidize while Owhile O22 isis reduced.reduced.
FeFe+2+2
+2e+2e--
→→ FeFe EEº=º= --0.44 V0.44 V
OO22 + 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 + O22 + H+ H22OO →→ FeFe22 OO33(rust)(rust)+ H+ H++
EºEºcellcell = 0.84 v= 0.84 v
Reaction happens in two places . . .Reaction happens in two places . . .

58
Water
Rust
Iron Dissolves - Fe → Fe+2
e-
Salt speeds up process by increasing
conductivity

59
Figure 17.17
The Electrochemical
Corrosion of Iron

60
Preventing Corrosion
Coating - to keep out air and water.
Galvanizing - Putting on a zinc coat
Zinc 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).

61
Preventing Corrosion
Cathodic Protection - Attaching largeCathodic Protection - Attaching large
pieces of an active metal like magnesiumpieces 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 replaceto iron pipe (and replace
periodically).periodically).
Attach titanium barsAttach titanium bars to ships’ hulls. Into ships’ hulls. In
salt water the Ti acts as the anode and issalt water the Ti acts as the anode and is
oxidized instead of the steel hull, whichoxidized instead of the steel hull, which
now acts as the cathode.now acts as the cathode.

62
Figure
17.18
Cathodic
Protection

63
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

64
1.0 M
Zn+2
e- e-
Anode Cathode
1.10
Zn Cu
1.0 M
Cu+2
Galvanic Cell - spontaneous

65
1.0 M
Zn+2
e- e-
AnodeCathode
A battery
>1.10V
Zn Cu
1.0 M
Cu+2
Electrolytic Cell Forces the opposite
reaction.

66
Figure 17.19
(a) A Standard Galvanic Cell
(b) A Standard
Electrolytic Cell

67
Calculating plating
Have to include the charge.Have to include the charge.
Measure currentMeasure 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 chargex t = the charge
q/nF = moles of metalq/nF = moles of metal
Mass of plated metalMass of plated metal

68
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 . . .
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.

69
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? . . .

70
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)

71
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? . . .

72
Calculating plating pp
Using 10.0 amperes, how long to depositUsing 10.0 amperes, how long to deposit
1.00 x 101.00 x 1022
grams of Cr(s) by dipping intograms of Cr(s) by dipping into
acidic Cracidic Cr22OO77
2-2-
(aq) ,(aq) , MMCrCr = 51.996?= 51.996? TheThe
only substances you start with are Cronly substances you start with are Cr22OO77
2-2-
and Hand H22O.O. Which is oxidized? Which isWhich is oxidized? Which is
reduced?reduced? . . .. . .
CrCr22OO77
2-2-
must be reducedmust be reduced to get to Crto get to Cr(s)(s) soso
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 . . .

73
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--
→→ 2H2H2200
2H2H2200 →→ OO22 + 4H+ 4H1+1+
+ 4e+ 4e--
OO22
+ 2H+ 2H220 + 4e0 + 4e--
→→ 4OH4OH1-1-
2H2H220 +0 +
2e2e--
→→ HH22 + 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.

74
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 . . .

75
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? . . .

76
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+
+
Cr2O7
2-
→ 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

77
Calculating plating
Electrolysis of a molten salt , MCl, usingElectrolysis of a molten salt , MCl, using
3.86 amps for 16.2 min deposits 1.52 g of3.86 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 ofSince 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 . . .

78
Calculating plating
Electrolysis of a molten salt , MCl, usingElectrolysis of a molten salt , MCl, using
3.86 amps for 16.2 min deposits 1.52 g of3.86 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 mol3.86 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

79
Other uses pp
Electrolysis of water.Electrolysis of water.
Separating mixtures of ionsSeparating mixtures of ions
A more positive reduction potentialA more positive reduction potential
means that reaction proceeds forward.means that reaction proceeds forward.
The metal with the mostThe metal with the most positivepositive
reductionreduction potential is easiest to plate outpotential is easiest to plate out
of solution (and is the best oxidizer).of solution (and is the best oxidizer).

80
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 ofTable 17.1 p. 796 to predict the order of
oxidizingoxidizing ability. Answer . . .ability. Answer . . .
Order of oxidizing ability is the same as theOrder of oxidizing ability is the same as the
order fororder 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 beAlso, predict which one will be reducedreduced at theat the
cathode of ancathode of an electrolyticelectrolytic cell at thecell at the lowestlowest
voltage. Answer . . .voltage. Answer . . .
SinceSince CeCe4+4+
is the greatest oxidizer it is the mostis 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.