Chemistry 12
Unit 5 – Oxidation Reduction
Chemistry 12 – Unit 5
Oxidation – Reduction
Introduction
-Demonstration of oxidation – reduction reactions
Definitions: (species means atom, ion or molecule)
Oxidation – a species undergoing oxidation loses electrons
(charge becomes more positive)
Reduction – a species undergoing reduction gains electrons
(charge becomes more negative)
Oxidizing agent – The species being reduced
(gains electrons, causes the other one to be oxidized)
Reducing agent – The species being oxidized
(loses electrons, causes the other one to be reduced)
-
2e
E.g.)
Cu2+ (aq) + Zn (s)
Oxidizing
agent
Cu(s) + Zn2+(aq)
LEO says GER
Losing
Electrons is
Oxidization
Gaining
Electrons is
Reduction
OAR
The
Oxidizing
Agent is
Reduced
To carry it too far…
When LEO the Lion says GER you grab your OAR and
Row Away Outa’ there!
(Reducing Agent is Oxidized)
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
Redox – Short for Oxidation – Reduction
Redox identification
Charge on neutral atom or molecule = 0
Oxidation – Charge gets more + (loses electrons)
Reduction – Charge gets more – (gains electrons)
Reduction (charge decreases)
2+
E.g.) Pb
(aq) +
0
Mg
(s)
0
Pb
(s) +
2+
Mg
(aq)
Oxidation (Charge increases)
Question
In the reaction:
2+
3+
2Fe + Cl2
2Fe + 2Cl
Identify:
a)
b)
c)
d)
e)
f)
g)
h)
The Oxidizing Agent: _____________________
The species being oxidized:________________
The reducing agent:______________________
The species being reduced:________________
The species gaining electrons:______________
The species losing electrons:_______________
The product of oxidation___________________
The product of reduction___________________
Do Ex. 1 (a-e) pp. 192 SW
Half-Reactions
-Redox reactions can be broken up into oxidation & reduction half reactions.
2+
e.g.) Redox rx: Pb
2+
The Pb
+ Zn(s)
(aq)
2+
Pb(s) + Zn
(aq)
(loses/gains) __________ 2 electrons.
2+
Reduction Half-rx: Pb
(aq)
-
+ 2e
Pb(s)
Electrons on the LEFT side (or GER)
Means REDUCTION
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
Write the oxidation half reaction for the following redox rx.
2+
Pb
(aq)
2+
+ Zn(s)
Pb(s) + Zn
(aq)
Ox half rx: ___________________________
(In oxidation reactions, e ‘s are ____ and are
found on the ____ side.) (LEO)
-
-
Note: Half-rx’s always have e ‘s, redox (oxidation-reduction) reactions never show e ‘s!
Given the redox reaction:
2+
F2(g) + Sn
(aq)
-
4+
2F (aq) + Sn
(aq)
Write the oxidation half-rx:__________________________
Write the reduction half-rx:__________________________
Do ex. 2 a-c on p. 192 SW
Oxidation numbers
-Real or apparent charge on an atom in a molecule or ion
In SW. p. 193 -the charge that an atom would possess if the species
containing the atom was made up of ions (even if it’s not!)
Rules to find oxidation number of an atom
1) In elemental form:
(Single atoms of monatomic elements) or (diatomic molecules of diatomic elements)
Oxidation number of atoms = 0
Eg) Mn, Cr, N2, F2, Sn, O2, etc.
The oxidation # of each atom = 0
2) In monatomic ions: oxidation # = charge
Eg)
3+
-oxidation # of Cr = +3
2-
-oxidation # of S = -2
In Cr
S
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
3) In ionic compounds
a) the oxidation # of Alkali Metals is always +1
eg) NaCl
K2CrO4
Ox # of Na & K = +1
b) the oxidation # of Halogens when at the end (right side) of the formula
is always –1
eg) CaCl2 AlBr3
KF
Ox # of Cl, Br and F = -1
Note: Halogens are not always –1! (Only when it is written last in formula.)
4) In molecules or polyatomic ions:
a) Ox. # of oxygen is almost always –2
e.g.) KOH CrO4
2-
Li3PO4
Ox # of
O is –2
b) An exception is Peroxides in which ox. # of O = -1
Hydrogen Peroxide: H2O2
Ox # of O’s = -1
Alkali Peroxides: Na2O2
(Remember, “O” in O2 has an Ox. # of ______)
5) In molecules or ions:
a) Hydrogen is almost always +1
e.g.)
HNO3
H2SO4 HPO4
2-
Every “H” has an ox # of +1
b) An exception is metallic hydrides, which have an ox # of -1
e.g.) NaH
CaH2 (In each one of these Ox. # of H = -1)
(What is the ox # of “H” in NH3? _______)
(And remember ox # of “H” in H2 = ______)
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
Finding oxidation numbers of each atom in a molecule or PAI
In a neutral molecule the total charge = 0
e.g.) NH3
Total charge = 0 (no charge)
In a polyatomic ion – the total ionic charge is written on the top right
e.g.) CrO4
2-
Total ionic charge (TIC) = -2
Oxidation numbers of all atoms add up to total ionic charge (TIC)
e.g.) Find the oxidation # of Cr in CrO4
2-
(Let x = ox # of one Cr atom)
CrO42X + 4 [# of “O”atoms] (-2 [charge of oxygen]) = -2 [total ionic charge]
X – 8 = -2
X = -2 + 8
X = +6 So ox # of Cr here = +6
e.g.) Find ox # of Cl in HClO4
HClO4
+1 + x + 4 (-2) = 0
1+x–8=0
x–7=0
x = +7
e.g.) Find Ox # of Cr in Cr2O7
2-
Cr2O722x + 7(-2) = -2
2x – 14 = -2
2x = +12
x = +6
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
e.g.) Find ox # of P in Li3PO4
Li3 P O4
3(+1) + x + 4 (-2) = 0
3+x–8=0
x–5=0
x = +5
Find Ox # of the underlined element in each of the following:
a) NaH2PO4 _____
b) Na2O2 ______ c) KH ______
Find the ox # of Fe in Fe3O4
Find the ox # of As in As3O5
Read p. 193-194 of SW. Do Exercise 3 on p. 194 of SW.
Changes in oxidation numbers
When an atom’s oxidation # is increased, it is oxidized.
2+
e.g.)
Half-rx: Fe
3+
Fe
-
+ e
More complex:
3+
-When Mn
3+
Mn
-
changes to MnO4 , is Mn oxidized or reduced?
MnO4
-
- What is the ox # of Mn before & after the reaction? Before ___
After ___
- The ox # of Mn is (de/in)____creased.
- In this process, Mn is (oxidized/reduced)__________________
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
Reduction – When an atom’s oxidation # is decreased, it is reduced.
e.g.) Cu(NO3)2
Ox # of
Cu = +2
Cu(s)
Ox # decreases (reduction)
Ox # of
Cu = 0
Redox ID using oxidation #’s
Given a more complex equation – identify atoms which do not change ox #’s
(often “O” or “H” but not always!)
e.g.)
3SO2 + 3H2O + ClO3
-
2-
+
-
3SO4 + 6H + Cl
There are no O2
molecules or
peroxides, so “O” in all
these has an ox # = -2
3SO2 + 3H2O + ClO3
-
2-
+
-
+
-
3SO4 + 6H + Cl
H is (+1) in both
of these so it
doesn’t change
Again:
3SO2 + 3H2O + ClO3
-
2-
3SO4 + 6H + Cl
-
The only atoms left are “S” and “Cl”. Find the Ox #’s of S and Cl in
species that contain them. (Ox # of 1 atom in each case)
3SO2
3SO4
2-
Coefficients are
just for balancing.
SO2
Ox # of S is +4
SO4
2-
Ox # of S is +6
Ox # of S increases so S is being oxidized
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
Note:
R.A.O., the reducing agent is oxidized
The species SO2 is acting as the reducing agent.
The element S is being oxidized so S is losing electrons.
Look at the species with Cl:
ClO3
-
-
Cl
Ox. # of
Cl = +5
Ox. # of
Cl = -1
Decrease in ox # so Cl is being reduced
Therefore, the species acting as the oxidizing agent is __________.
(They may also ask for the atom acting as the oxidizing agent
– this would be Cl in ClO3 )
Eg. –given the reaction:
2-
-
2CrO4 + 3HCHO + 2H2O
2Cr(OH)3 + 3HCOO + OH
-
Find: a) The species being oxidized
c) The reducing agent
d) The species being reduced
e) The oxidizing agent
f) The species losing electrons
g) The species gaining electrons
Notes:
− For hydrocarbons it’s best to rewrite them as simple molecular formulas.
− All O’s are in molecules or ions, no O2 & no peroxides so O remains unchanged
as -2
− All H’s are in molecules or ions, no H2 or metallic hydrides so H remains unchanged
as +1
− The atoms to check for changes are C and Cr.
Oxidation
2CrO4
+6
So…
2-
0
+ 3CH2O + 2H2O
Reduction
+2
2Cr(OH)3 + 3HCO2 + OH
+3
a) the species being oxidized is (CH2O) HCHO (inc. in ox #)
b) the reducing agent is (CH2O) HCHO (RAO)
2c) The species being reduced is CrO4 (decrease in ox #)
2d) The oxidizing agent is CrO4 (OAR)
e) The species losing e ‘s is (CH2O) HCHO (LEO)
2f) The species gaining e ‘s is CrO4 (GER)
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
Given the redox reaction:
-
2-
2MnO4 + 3C2O4 + 4H2O
Find:
a)
b)
c)
d)
e)
f)
2MnO2 + 6CO2 + 8OH
-
The species being reduced: _____________.
The species undergoing oxidation: _____________.
The oxidizing agent: ________________.
The reducing agent: ________________.
The species gaining electrons: ______________.
The species losing electrons: ______________.
Given the balanced redox reaction:
3SO2 + 4NO + 2H2O
3S + 4HNO3
Find:
a)
b)
c)
d)
e)
f)
g)
h)
The oxidizing agent: _____________.
The reducing agent: _____________.
The species being reduced: ___________.
The species being oxidized: ___________.
The species losing electrons: ____________.
The species gaining electrons: ____________.
The product of oxidation: ____________.
The product of reduction: ____________.
Given the following:
6Br2 + 12KOH
Find:
a)
b)
c)
d)
e)
f)
10KBr + 2KBrO3 + 6H2O
The oxidizing agent: ______________.
The reducing agent: ______________.
The species undergoing oxidation: _____________.
The species being reduced: ________________.
The product of oxidation: _______________.
The product of reduction: _______________.
Using oxidation numbers to identify half-reactions
They don’t have to be balanced
e.g.) If NO2
-
-
NO3 is an example of (oxidation or reduction?) _______________.
(“O” does not change it’s ox # (no O2 or peroxides)) so find ox # of N on both sides.
NO2
-
Ox # = +3
NO3
-
Ox # = +5
O.N. = +2
Unit 5-Oxidation-Reduction NOTES
Since ox # increases,
this is an oxidation
Page 9
Chemistry 12
Unit 5 – Oxidation Reduction
e.g.) H2O2
Ox # = -1
This time “O” is the element
changing ox #
H 2O
Ox # = -2
Ox # if “O” decreases so this is a
reduction
O.N. = -1
Find the ∆ O.N. of the element in which it changes and identify each as an
oxidation or reduction
a) C2H5OH
CH3COOH ________________________________
b) Fe2O3
Fe3O4 ______________________________________
c) H3PO4
P4
______________________________________
(P4 is the elemental form of phosphorus)
d) CH3COOH
CH3COH ________________________________
NOTE: When asked if a given reaction is a redox or not:
Look for a change from an element
compound or compound
an element
These will always be redox, because in elemental form ox. # = 0 and in compounds
usually ox. # is not = 0
0
Eg.) Is the reaction: Zn + Cl2
0
+2
ZnCl2 a redox reaction?
-1
Answer: It must be because ∆ON of Zn ( 0
+2 = +2) and ∆ON of Cl (0
-1 = -1)
Do Exercises 4, 5 and 6 on p. 194-195 of SW.
Half-reactions and the reduction table
- Do Experiment 21-A
Unit 5-Oxidation-Reduction NOTES
Page 10
Chemistry 12
Unit 5 – Oxidation Reduction
- Look at “Standard Reduction Table”
Ox agents on left + e-‘s
F2 + 2eAg+ + e-
Stronger ox
agents (More
tendency to be
reduced) (OAR)
(To gain e-‘s)
Reducing agents on right
2FAg (s)
Cu2+ + 2e-
Cu(s)
Zn2+ + 2e-
Zn (s)
Li+ + e-
Stronger reducing
agents
(More tendency to
be oxidized RAO)
(To lose e-‘s)
Li (s)
-So F2 is a stronger ox agent than Ag+, etc.
-The strongest reducing agent on your chart is: ________.
Help in Hunting
- Solid metals mostly on bottom right (less active ones Ag, Au, farther up on the right side)
- Halogens (e.g. Cl2) and oxyanions e.g. BrO3 , MnO4 , IO3 found near top left
2+
2+
+
2+
- Some metal ions found on both sides e.g. Fe , Sn , Cu , Mn can act as OA’s or RA’s
All the half-rx’s are written as reductions:
e.g.)
-
-
F2 + 2e
+
Ag + e
2F
Ag(s)
- The double arrow implies that oxidation’s can also take place (reverse of reductions)
+
Specifically just
reduction
Specifically
just oxidation
e.g.) reduction of Ag
+
Ag + e
Ag(s)
(Same as table- single arrow)
oxidation of Ag
(Reverse of that on table- single arrow)
+
Ag(s)
Ag + e
Write half-reactions for:
2+
- Reduction of Pb
- Oxidation of Pb
2+
- Reduction of Sn
2+
- Oxidation of Sn
_________________________________
_________________________________
_________________________________
________________________________
Unit 5-Oxidation-Reduction NOTES
Page 11
Chemistry 12
Unit 5 – Oxidation Reduction
2+
_________________________________________________
- Oxidation of Fe
2+
_________________________________________________
- Reduction of Fe
- Oxidation of Fe
_________________________________
- Reduction of acidified MnO4- ___________________________
_________________________________
- Oxidation of H2
2+
+
Which is a stronger oxidizing agent:
Ni or Ag ? _________
2+
3+
Fe or Cr ? ________
Must be on the left
side when treating
these as OA’s
Sn
Which is a stronger reducing agent:
Sn
Must look for these
on the right side
2+
or Sn ? ________
4+
2+
or Fe ? ________
2+
Zn or Ba? _________
Cl or Br ? _________
2+
Fe or Au? ________
Which has a greater tendency to lose electrons, Ni or Zn? _____
3+
3+
Which has a greater tendency to gain electrons, Fe or Cr ? _____
Which solid metal has the least tendency to lose electrons? _____
Which solid metal has the greatest tendency to lose e ‘s? _____
2+
Give the formula for an ion that is a stronger oxidizing agent that Ni , but is weaker
2+
than Pb ?
______
Using the reduction table to predict which reactions are spontaneous
An oxidizing agent will react spontaneously with (oxidize) a reducing agent
below it on the right
-
F2(g) + 2e2FS2O8 + 2e
2SO42Li+ + eLi(s)
+
-
K +e
Rb+ + eCs+ + eLi+ + e-
K(s)
Rb(s)
Cs(s)
Li(s)
Look at your reduction chart!
F2, the strongest OA, oxidize (react spontaneously
with) all species below it on the right side from SO42all the way down to Li(s)
Look at the 4th half rx from the bottom
K+ will oxidize only Rb(s), Cs(s) and Li(s), nothing else
on the chart.
A reducing agent on the right will react spontaneously with (reduce) any
oxidizing agent on the left above it
+
e.g.) Li(s) (bottom right) will reduce all species on the left side except Li .
2SO4 (near top right) will reduce only F2
- An OA on the left will not react spontaneously with a RA on the right above it!
e.g.)
3+
Au
2-
will not oxidize (or react spontaneously with) SO4 .
Unit 5-Oxidation-Reduction NOTES
Page 12
Chemistry 12
Unit 5 – Oxidation Reduction
Some points…
+
1) Be very careful with charges e.g. Li is a totally different thing than Li(s).
2) Things don’t react with species which are only on the same side (these are
impossible – not just non-spontaneous.)
+
th
+
+
E.g.) K (4 from bottom on the left) will not oxidize Rb or Cs
+
Li etc. –because they are on the same side only. (Impossible)
E.g.) Li(s) will not reduce Cs(s), Rb(s), K(s), etc. because they are
all on the same side only.
3) Some elements with multiple oxidization numbers e.g.) Sn, Cu, Mn, Fe have
ions on both sides of the chart!
–Look carefully at your table to find these.
o
Note – Don’t worry what E means yet, I will
just use it to let you locate half-reactions.
Notice:
2+
Fe is on the left (OA) at – 0.45
2+
Fe is on the right (RA) at + 0.77
2+
Sn is on the left (OA) at – 0.14
2+
Sn is on the right (RA) at + 0.15
A word about Cu…
+
Notice: Cu is on the left at + 0.52
+
Cu is on the right at + 0.15
-recall that anything on the left will oxidize a species below it on the right.
o
E
+
Cu + e = Cu
0.52
2+
+
Cu + e = Cu
0.15
+
This Cu will
+
oxidize this Cu
+
+
-Since Cu oxidizes and reduces itself, any water solution of Cu is
+
unstable – it won’t remain Cu very long!
(demo Cu in HNO3)
Notice:
2+
o
Mn is on the left at E = -1.19
2+
o
Mn is on the right at E = +1.22
3+
-
2+
Also notice: Cr + e = Cr - 0.41
3+
Cr + 3e = Cr(s) - 0.74
Unit 5-Oxidation-Reduction NOTES
Page 13
Chemistry 12
Unit 5 – Oxidation Reduction
If a redox reaction is non-spontaneous, then the
reverse reaction will be spontaneous!
2+
2+
e.g.) The reaction Sr + Ca(s)
Ca + Sr(s) is non-spontaneous because
2+
Ca is above Sr on the right side.
2+
2+
But the rx: Ca + Sr(s)
Sr + Ca(s) is spontaneous because Sr(s)
2+
is below Ca on the right side
Use the reduction table to answer the following questions:
a)
b)
c)
d)
e)
f)
Will Br2 oxidize Au(s)?………________________
2+
Will Pb(s) reduce Fe ?……..________________
2+
3+
Will Zn react with Cr ?….._______________
2+
3+
Will Mg react with Cr ?…. ______________
Give the symbol of an ion that will oxidize Mn(s) but not Cr(s)…..____________
2+
Give the formula for a compound which will reduce Co but will not reduce
2+
Fe ………………..______________
2+
2+
Which is a stronger reducing agent, Sn or Fe ?(Hint – you must look for
both on the _____ side)….._____________
+
2+
Which is a stronger oxidizing agent, Cu or Sn ?(Hint – you must look for
both on the ____ side)….._____________
g)
h)
Acidified solutions
+
-Any reactions on the table with H in them are acidified or acid solutions.
o
th
e.g.) Look at these: at E = +1.51 (4 from the top)
-
+
-
MnO4 + 8H + 5e
Called acidified
permanganate
solution
2+
Mn
+ 4H2O
Note: Names of many
ions can be found on
the ion table!
o
Give the E corresponding to each of the following:
o
a) acidified iodate …………………..E ________
o
b) acidified dichromate…………..…E ________
o
c) acidified manganese (IV) oxide…E ________
o
d) acidified bromate…………………E ________
o
e) acidified perchlorate…………..…E ________
o
f) acidified oxygen gas……………..E ________
Unit 5-Oxidation-Reduction NOTES
Page 14
Chemistry 12
Unit 5 – Oxidation Reduction
Nitric, Sulphuric & Phosphoric acids
- These acids are shown in ionized form on the table
- Nitric acid (HNO3) is found in two places on the left side.
-
+
-
NO3 + 4H + 3e
-
+
NO + 2H2O
-
2NO3 + 4H + 2e
N2O4 + 2H2O
o
E = + 0.96 v
o
E = + 0.80 v
Don’t worry about coefficients yet.
They are only used for balancing.
- Sulphuric acid is found at + 0.17 v
2-
+
-
SO4 + 4H + 2e
o
H2SO3 + H2O E = + 0.17 v
Find and write the half-reaction for the reduction of phosphoric acid (H3PO4)
Sulphurous acid (H2SO3)
A note about water
-On the top of the table it says “ionic
concentrations are at 1M”
+
-This includes [H ] = 1M with two
exceptions:
-Neutral water is found on the shaded lines at + 0.82 v and – 0.41v
-Neutral water as a reducing agent is on the right side at + 0.82 v
-Neutral water as an oxidizing agent is on the left side at – 0.41 v
-
(Notice H2O is below this at – 0.83 v but in this solution [OH ] = 1M (so it’s basic,
not neutral)
+
(Again H2O is also found at + 1.23 v but here [H ] = 1M so it’s acidic, not neutral)
Unit 5-Oxidation-Reduction NOTES
Page 15
Chemistry 12
Unit 5 – Oxidation Reduction
Questions
a) Will neutral water oxidize Fe(s)? ________ Cr(s)? _______ Na(s)? _______
3+
+
b) Will neutral water reduce Au ? _____ Ag ? _____
2c) Will acidified permanganate oxidize SO4 ? _____ Br ? ______ Zn? _____
d) Will nitric acid react with Ag(s)? _____ Au(s)?_____ I _____Cl ?______
2+
e) Will nitric acid react with Fe
f) Will nitric acid react with Hg to form N2O4? _____
g) Will nitric acid react with Hg to form NO? ______
h) Can you safely put a gold ring in acidified dichromate solution? _____ What about
acidified bromate solution? _____
i) If Cl2 gas is bubbled into water, will it all remain as Cl2, or will some be converted to Cl ? _____
Finding products of spontaneous reactions
eg)
4+
Given Sn + H2S – find the products
See the table at +0.15v and +0.14v
4+
2+
Sn
+ 0.15v
Sn + 2e
+
S(s) + 2H + 2e
H2S + 0.14v
The higher reaction will be reduction ( ), the lower reaction will proceed to the
left ( ) and be an oxidation.
4+
2+
Sn + 2e
Sn
+
S(s) + 2H + 2e
H2S (reversed! Lower one is reversed-is an oxidation)
2+
+
-So the products are Sn , S, and H
(at this point don’t worry about coefficients yet.)
Questions
a) What are the products of the reaction of acidified hydrogen peroxide (H2O2)
and bromide (Br )? _________________
b) What are the products of the reaction when neutral water reacts with:
Ca(s) ____________
Zn(s) ____________
Br2 ______________
Acidified MnO2 _____________
Fluorine gas _______________
Read SW p. 195-199
Do Ex 7-12 p, 199-200 SW
Unit 5-Oxidation-Reduction NOTES
Page 16
Chemistry 12
Notes on Unit 5 –Oxidation-Reduction
Using data to make your own simple Redox table
Example problem:
2+
2+
2+
2+
1) Four metals A, B, C, & D were tested with separate solutions of A , B , C & D .
Some of the results are summarized in the following table:
Solution
2+
2+
2+
2+
Metal
A
B
C
D
(1)
(2)
A
no reaction
reaction
(4)
no reaction
B
(3)
D
reaction
List the ions in order from the strongest to weakest oxidizing agent.
Using data
2+
2+
1) – Since B does not oxidize A : B must be below A on the table.
Oxidizing
Reducing
agents
agents
2+
A + 2e = A
2+
B + 2e = B
2) – Since C
2+
reacts with A: C
2+
NOTE: For the same
element: The more positive
species is always the
Oxidizing Agent.
Eg.) A2+ A
RA
OA
must be above A:
2+
-
C + 2e = C
2+
A + 2e = A
2+
B + 2e = B
2+
2+
3) – Since A reacts with D: A must be above D on the table. But is D
2+
below B ? We don’t know yet.
2+
above or
2+
D
2+
Here
Or here?
C
2+
A
2+
B
Let’s look at the next information:
4) – D
-
2+
does not react with B
Now we know that D
2+
must be below B on the table
So now we have our complete table:
Oxidizing agents
2+
Reducing agents
-
C + 2e = C
2+
A + 2e = A
2+
B + 2e = B
2+
D + 2e = D
-
Unit 5 Notes
At this point its good to go back and recheck that all the data given is
consistent with your table.
So now we have our answer; The ions in order of strongest to weakest ox
2+
2+
2+
2+
agent is: C , A , B , D
Just in case you’re asked, you can see that the order of reducing agent from
strongest to weakest is D, B, A, C.
17
Chemistry 12
Notes on Unit 5 –Oxidation-Reduction
Another example –
Four non-metal oxidizing agents X2, Y2, Z2 and W 2 are combined with solutions of
ions: X , Y , Z and W .
NOTE: For the same
element: The more positive
The following results were obtained;
species is always the
Oxidizing Agent.
(1) X2 reacts with W and Y only.
Eg.) X2 X(2) Y will reduce W 2
List the reducing agents from strongest to weakest
-
-
(1) X2 will be above W & Y , but below Z
Oxidizing agents
Z2 + 2e
X2 + 2e
OA
Reducing agents
2Z
2X
-
-
W &Y
-
RA
-
Are both below X2,
but we don’t know in
which order yet.
-
(2) Since Y reduces W 2, Y must be lower on the right of W 2.
OA’s
Z2 + 2e
X2 + 2e
W 2 + 2e
Y2 + 2e
RA’s
2Z
2X
2W
2Y
To answer the question:
The reducing agents from strongest to weakest are: Y , W , X , Z
Question:
Four solutions A(NO3)2, B(NO3)2, C(NO3)2, and D(NO3)2 are
added to metals, A, B, C, & D
The following information is found:
(1) The metal A will not react with any of the solutions
(2) C(NO3)2 reacts spontaneously with B
(3) B will not react with D(NO3)2
(a) Make a small reduction table showing reductions of the metallic ions. (Don’t
forget to discard the spectator nitrate ions.
Unit 5 Notes
18
Chemistry 12
Notes on Unit 5 –Oxidation-Reduction
(b) List the oxidizing agents in order of strongest to weakest:
_________________________
(c) List the reducing agent in order of strongest to weakest:
_________________________
(d) Would it be safe to store A(NO3)2 solution in a container made of the metal
D? ______________
Do Exercises 14,15,16 & 18 on p. 200 of SW.
Balancing half-reactions
-Some half-rx’s are on the table, but not all.
-Given if the soln. Is acidic or basic.
Pay attention!
-Think of Major Hydroxide (Major
O
H
- (charge))
Major atoms atoms other than O & H
2-
Acid Soln.
-
E.g.) S2O8
HSO3 (acid soln.)
(1) Balance Major Atoms (S in this case)
2-
-
S2O8
2HSO3
(2) Balance “O” atoms, by adding H2O (to the side with less O’s)
2-
-
S2O8
2HSO3 + 2H2O
+
(3) Balance “H” atoms by adding H (to the side with less H’s)
2-
S2O8 + 6H
+
-
2HSO3 + 2H2O
-
(4) Balance charge by adding e ‘s (to the more + side)
2-
S2O8 + 6H
+
-
2HSO3 + 2H2O
TIC = (+4)
TIC = (-2)
-
The left side needs 6e ‘s
to get a –2 charge
So the final balanced half-rx is:
2-
+
S2O8 + 6H + 6
TIC = (-2)
e-
-
2HSO3 + 2H2O
TIC = (-2)
-Always double-check these!
-Don’t miscopy charges, etc.
Unit 5 Notes
19
Chemistry 12
Notes on Unit 5 –Oxidation-Reduction
-
Try this one:
MnO4
Mn
2+
(acid soln)
In basic solution
-Do the first steps of the balancing just like an acid
E.g.) MnO2
-
MnO4
(basic solution)
Major (Mn already balanced)
Oxygen 2H2O + MnO2
MnO4
+
Hydrogen 2H2O + MnO2
MnO4 + 4H
+
Charge 2H2O + MnO2
MnO4 + 4H + 3e
+
In basic solution: write the reaction H + OH
-
H2O or H2O
+
H + OH
-
+
-In whichever way is needed to cancel out the H ’s
-Add to the half-rx
-
E.g.)
+
2H2O + MnO2
MnO4 + 4H + 3e
+
4H + 4OH
4H2O
-
You must write the
whole water equation
+
We need 4H
on the left to
+
cancel the 4H
on the left side
(Take away 2H2O’s
from both sides)
MnO2 + 4OH
-
-
-
MnO4 + 2H2O + 3e
Double-check everything!
Try this one: Pb
-
HPbO2 (basic soln)
-Reactions without H’s or O’s are done in neutral soln -Do Ex 19 a-m p. 203
Unit 5 Notes
20
Chemistry 12
Notes on Unit 5 –Oxidation-Reduction
Balancing overall redox reactions using the half-reaction (half-cell) method
(1) Break up Rx into 2 half-rx’s.
(2) Balance each one (in acidic or basic as given)
(3) Multiply each half rx by whatever is needed to cancel out e ‘s
(Note: balanced half-rx have e ‘s (on left reduction on right oxidation) Balanced redox
don’t have e ‘s)
+
(4) Add the 2 half-rx’s canceling e ‘s and anything else (usually H2O’s, H ’s or OH ‘s) in
order to simplify.
Example:
U
4+
-
2+
+ MnO4
Mn
Balance each ½ rx
4+
2+
U
UO2
(Major (U) balanced already)
4+
2+
Oxygen
U + 2H2O
UO2
4+
2+
+
Hydrogen
U + 2H2O
UO2 + 4H
4+
2+
+
Charge
U + 2H2O
UO2 + 4H + 2e
+ UO2
2+
(acidic)
-
2+
MnO4
Mn
(Major (Mn) balanced already)
2+
MnO4
Mn + 4H2O
+
2+
MnO4 + 8H
Mn + 4H2O
+
2+
MnO4 + 8H + 5e
Mn + 4H2O
Multiply by 5 to get 10e
4+
2+
(U + 2H2O
UO2
+
(MnO4 + 8H + 5e
5U
4+
+
-
4H + 2e ) 5
2+
Mn + 4H2O) 2
-
+
-
5U
Quick check by finding
TIC’s on both sides
-
Try this one:
4+
Multiply by 2 to get 10e
2+
+ 10H2O + 2MnO4 + 16H + 10e
To simplify:
-
5UO2
+
2+
+ 20H + 2Mn
-
-
+ 8H2O + 10e
-
-Take away 10e from both sides
+
-Take away 16H ’s from both sides
-Take away 8H2O’s from both sides
-
+ 2H2O + 2MnO4
2+
5UO2
+
2+
+ 4H + 2Mn
+20 + 0 – 2
+10 + 4 + 4
TIC = +18
TIC = +18
If you have time check all atoms also if TIC’s are not equal you messed up!
Somewhere! Find it!
-
SO2 + IO3
2-
SO4 + I2 (basic solution)
-See examples p.205-207 in SW
Unit 5 Notes
21
Chemistry 12
Notes on Unit 5 –Oxidation-Reduction
Quick notes
-Some redox equations have just one reactant
- Use this as the reactant in both half-rx’s.
- These are called “self-oxidation-reduction” or Disproportionation reactions.
Eg)
Br2
--
-
Br + BrO3 (basic) (found in some hot tubs)
Half rx’s are:
Br2
Br
--
Br2
-
BrO3
Answer: ___________________________________________________
Do Ex 24 a-w p. 207
The more practice the better! See me if you want more!
Balancing redox equations using the oxidation number method
-This is optional
- As long as one method (not guessing!) works for you that’s fine. (This
method or half-rx method.)
- Read examples p. 271-272 SW
- Do any ex 10 a-n & check with key
Redox titrations
-
same as in other units (solubility/acids-bases)
coefficient ratios for the “mole bridge” are obtained by the
balanced redox equation:
TITRATIONS
STANDARD
SAMPLE
mole bridge
Conc. & Volume
moles
moles
Conc. or
or Mass
Volume
mol = M x L
M = mol/L
or: grams x 1 mol = mol
or L = mol/M
MM g
Unit 5 Notes
22
Chemistry 12
Notes on Unit 5 –Oxidation-Reduction
-
Eg) Acidified hydrogen peroxide (H2O2) is used to titrate a solution of MnO4 ions of
2+
unknown concentration. Two products are O2 gas and Mn .
a) Write the balanced redox equation:
-
b) It takes 6.50 mL of 0.200 M H2O2 to titrate a 25.0 mL sample of MnO4 solution.
Calculate the original [MnO4 ].
Finding a suitable solution titrate a sample
Use redox table:
- If sample is on the left (OA)
Use something below it on the right. (RA)
- If sample is on the right (RA) use something above it on the left (OA)
- Good standards will change colour as they react
-
2+
Acidified MnO4 (purple) = Mn (clear)
23+
Acidified Cr2O7 (orange) = Cr (pale green)
Read p. 210-212 carefully – go over the examples! Do ex 26 & 29 p. 213-214 SW.
Unit 5 Notes
23
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Electrochemical Cells
Demonstration (Cu/Zn Cell)
Cu
1 M Zn(NO3)2
1 M Cu(NO3)2
Definitions
Electrochemical cell – A device which converts chemical energy into electrical energy
Electrode – A conductor (usually a metal) at which a half-cell reaction.
(oxidation or reduction) occurs
Anode - The electrode at which oxidation occurs. (A & O are both vowels) LEOA
(label the anode in the diagram above)
Cathode – The electrode at which reduction occurs. (R & C are both consonants) GERC
(Label the anode in the diagram above)
Half-cell reactions
Anode – Oxidation half-rx
2+
Zn(s) à Zn
-
(aq)
+ 2e
Metal atoms are changed to + ions. Metal dissolves and anode loses mass as the cell
operates.
Cathode – Reduction half-rx
2+
Cu
-
+ 2e à Cu
+ ions are changed to metal atoms. New metal is formed so the cathode gains mass.
24
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Flow of electrons
-Since the anode loses e ‘s, (LEOA) and the cathode gains e ‘s (GERC)
-Electrons flow from the anode toward the cathode in the wire (conducting solid)
V
e2+
Anode – Zn à Zn
eCu
Zn
-
+ 2e
2+
Cathode – Cu
-
+ 2e à Cu
-
e ‘s flow from A à C in the wire
Flow of Ions in the salt bridge
-Salt bridge contains any electrolyte (conducting solution)
K+ NO3
Most common – U-tube
containing a salt such as
KNO3 or NaNO3
Porous barrier
which ions can
pass through
Common things which contain
electrolytes, e.g.) potatoes,
apples, oranges, lemons,
frogs, people
Anode
Cathode
Cu2+ + 2e- à Cu
Zn à Zn2+ + 2e-
If there was no SB + ions are used
up, so there would be an excess of
– ions (anions) at the cathode
If there was no SB + ions
(cations) would build up at the
anode
1 M Zn2+
1 M Cu2+
25
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Cations + flow (migrate) toward the cathode
And Anions – flow toward the anode
In the salt bridge
Show with arrows the
direction of flow of all ions
2+
+
2+
(Zn , NO3 , K , Cu )
Also show the direction of
flow of electrons in the wire
Cu
Cu2+
Zn2+
Identifying the anode and cathode
Look at the reduction table
All half-rx’s are reversible (can go forward or backward)
All are written as reductions (GERC)
Their reverse would be oxidations (LEOA)
The half-rx with the greater potential to be reduced is higher on the table
o
(higher reduction potential E )
So the higher half-rx is the cathode (HIC)
2+
2+
(Notice Cu + 2e à Cu is higher than Zn + 2e à Zn so Cu gets to be the cathode)
Also notice that the Anode reaction Is Reversed (AIR)
2+
(Anode rx: Zn à Zn + 2e )
-
Question. Fill in the following table. Use your reduction table:
Metal/ion
Metal/ion
Cathode (HIC)
Cathode Half-rx
+
Fe/Fe
2+
Pb/Pb
Ag/Ag
Zn/Zn
2+
+
-
Ag + e à Ag
Fe (lower)
Fe à Fe
2+
-
+ 2e
2+
Al/Al
3+
Au/Au
Co/Co
Ag (higher)
Anode (AIR) Half-rx
3+
Ni/Ni
Mg/Mg
2+
Anode
2+
2+
Ag/Ag
+
+
H2/H
2+
Sn/Sn
26
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Summary of Electrochemical Cells (ECC’s) so far…
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
15)
Electrochemical cells convert __________ energy into ___________ energy.
___________ is the electrode where oxidation occurs.
Electrons are ________ at the anode.
_________ is the electrode where reduction occurs.
In the half-rx at the cathode, e ‘s are on the ______ side of the equation.
Electrons flow from the ______ toward the ______ in the __________.
Cations ((+) ions) flow from the ____ beaker toward the ______ beaker through the _______.
Anions ((-) ions flow from the _____ beaker through the __________.
The higher half-rx on the table is the one for the ________ and is not reversed.
The lower half-rx on the table is the one for the ________ and is reversed.
Electrons do not travel through the ______ _______, only through the _______.
Ions (cations & anions) do not travel through the wire but only through the _____________.
The salt bridge can contain any ______________.
The anode will ________ (gains/loses) mass as it is __________(oxidized/reduced).
The cathode will ________ mass as it is _______ (oxidized/reduced) .
Read SW p. 215 - 217 in SW.
Ex 34 a-e & 35 a-e p.217 SW.
Standard reduction potentials and voltages
-
Voltage – The tendency for e ‘s to flow in an electrochemical cell. (Note: a cell may have
a high voltage even if no e ‘s are flowing. It is the tendency (or potential) for e ‘s to flow.
-Can also be defined as the potential energy per coulomb. (Where 1C = the
18 charge carried by 6.25x10 e ) 1 Volt = 1 Joule/Coulomb
Reduction potential of half-cells
-The tendency of a half-cell to be reduced. (take e ‘s)
Voltage only depends on the difference in potentials not the absolute potentials.
The voltage of a cell depends only on the difference
in reduction potentials of the two half-cells.
e.g.)
Mrs. A
Mrs. B
$ before buying calculator
$2000
$50
$ after buying calculator
$1980
$30
Difference
$20
$20
-Both people spent $20 on the calculator.
-Relative potentials of half-cells can only be determined by connecting with other half-cells and
reading the voltage.
E.g.) How good a basketball team is can only be determined by playing with other teams and
looking at points (scores).
27
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-
A “standard” half-cell was arbitrarily chosen to compare other half-cells with.
It was assigned a “reduction potential” of 0.000 v
+
It is: 2H
-
(aq)
+ 2e
o
H2(g) E = 0.000 v
- Since potential depends on gas
+
pressure, temp. and [H ] this half cell
is assigned a value of 0.000 v at
o
“standard state” which is 25 C,
+
101.3 kPa and [H ] =1.0 M
Connecting wire
H2 (g) (at standard
pressure 101.3 kPa)
Bubbles of H2
+
Means
standard
state
o
1.0 M H (HCl) solution at 25 C
Potential
Eo = 0.000v
Platinum electrode
Surface for metal, gas &
solution to contact
-The standard half-cell acts as an anode (LEOA) or cathode (GERC) depending on what it is
+
connected to: For example, when the standard half-cell is connected to the Ag/Ag half-cell.
For this cell, the voltage is
0.80 volts with the electrons
flowing toward the Ag.
Found by
experiment
(GERC)
(LEOA)
Draw the Diagram:
So the cathode is the
Ag/Ag+ half-cell
The anode is the
H2/H+ half-cell
Cathode half-rx
Anode half-rx
+
-
o
Ag + e à Ag E = ?
+
o
H2 à 2H + 2e E = 0.00 volts
Voltage of cell = 0.80 volts
28
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o
+
-From this we can see that the E for the Ag/Ag half-cell must be 0.80 V different than that of
+
the standard half-cell. Since the Ag/Ag is the one which is reduced, we say it has a higher
+
reduction potential than the standard. Therefore the reduction potential of the of Ag/Ag
half-cell is +0.80 V.
Another example:
+
2+
The standard (H2/H ) half-cell is connected to the Ni/Ni half-cell.
+
-Electrons are found to flow away from the nickel toward the H2/H half-cell and the
o
voltage (at 25 C, 101.3 Kpa & 1.0 M solutions) is found to be 0.26 volts.
Give the:
Cathode Half-rx: ___________________
Anode Half-rx: _____________________
o
2+
Determine the E for the Ni/Ni half cell: _________________
Now look at Standard Reduction Table.
Notice: -all half-rx’s are written as reductions
o
-The E is the standard reduction potential for the species on the left side.
Eg) Of the following combinations, find the one which gives the highest voltage? ______
+
2+
a) Ag /Ag with Cu /Cu
2+
+
b) Pb /Pb with Ni /Ni
+
+
c) Ag /Ag with Pb /Pb
3+
2+
d) Au /Au with Ni /Ni
-Which combo gives the lowest voltage? _______
Using the reduction table to find the initial voltage of ECC’s at standard state
1) Find the two metals on the reduction table. Higher one is the cathode. (HIC)
2) Write the cathode half-rx as is on the table (cathode - reduction) Include the reduction
o
potential (E ) beside it.
o
3) Reverse the anode reaction (AIR) (anode à oxidation) Reverse the sign on the E
(If (+) à (-) | If (-) à (+))
-
o
4) Multiply half-rx’s by factors that will make e ‘s cancel. DON’T multiply the E ’s by these
factors.
5) Add up half-rx’s to get overall redox reaction.
o
6) Add up E ’s (as you have written them) to get the initial voltage of the cell.
29
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Example:
A cell is constructed using Nickel metal and 1M nickel (II) nitrate along with Fe metal and
1M Iron (II) nitrate.
o
a) Write the equation for the half-rx at the cathode (with the E )
_____________________________ _____________________
o
b) Write the equation for the half-rx at the anode (with the E )
_____________________________ _____________________
c)
o
Write the balanced equation for the overall reaction (with the E )
_____________________________ _____________________
d) What is the initial cell voltage? __________V
Another example:
A cell is constructed using aluminum metal, 1M Al(NO3)3 and lead metal with 1M Pb(NO3)2.
Use the method in the last example to write the overall redox reaction and find the initial cell
voltage.
Overall redox reaction _______________________________________________
Initial cell voltage: _____________________________volts.
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Example
A student has 3 metals: Ag, Zn and Cu; three solutions: AgNO3 , Zn(NO3)2 , and Cu(NO3)2 ,
all 1M. She also has a salt bridge containing KNO3 (aq) wires and a voltmeter.
a) Which combination of 2 metals and 2 solutions should she choose to get the highest
possible voltage?
Metal: _________ Solution: ____________
Metal: _________ Solution: ____________
b) Draw a diagram of her cell labeling metals, solutions, salt bridge, wires, and voltmeter.
c)
o
Write an equation for the half-rx at the cathode. (with E )
_____________________________________ ________________________
o
d) Write an equation for the half-rx at the anode (with E )
____________________________________________________________ _
o
e) Write a balanced equation for the overall redox reaction in the cell (with E )
____________________________________________________________ _
f)
g)
h)
i)
j)
The initial voltage of this cell is _________volts.
In this cell, e ‘s are flowing toward which metal?__________ In the ________
Positive ions are moving toward the _______ solution in the ______ ______.
Nitrate ions migrate toward the ____________ solution in the ______ _______.
__________ metal is gaining mass
As the cell operates.
__________ metal is losing mass
The student now wants to find the combination of metals and solutions that will give the lowest
voltage.
k)
Which metals and solution should she use?
Metal ___________ Solution _____________
Metal ___________ Solution _____________
l)
Find the overall redox equation for this cell.
_________________________________ _
m) Find the initial cell voltage of this cell __________volts.
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Consider the following cell:
e-
Cr
Metal “X”
1M X(NO3)2
1M Cr(NO3)3
The voltage on the voltmeter is 0.45 volts.
o
a) Write the equation for the half-reaction taking place at the anode. Include the E .
o
______________________________________________ E : _________v
b) Write the equation for the half-reaction taking place at the cathode.
o
______________________________________________ E : _________v
c)
Write the balanced equation for the redox reaction taking place as this cell operates.
o
Include the E .
o
______________________________________________E : __________
2+
d) Determine the reduction potential of the ion X .
o
E : __________v
-
e) Toward which beaker (X(NO3)2) or (Cr(NO3)3) do NO3 ions migrate?
_______________________ _
f)
Name the actual metal “X” ________________________________
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Consider the following cell:
e-
Ag
Cd
AgNO3
Cd(NO3)2
The initial cell voltage is 1.20 Volts
o
a) Write the equation for the half-reaction which takes place at the cathode. Include the E
o
____________________________________________________ E = ____ ___v
b) Write the equation for the half-reaction taking place at the anode:
o
____________________________________________________ E = ____ ___v
c)
o
Write the balanced equation for the overall redox reaction taking place. Include the E .
o
____________________________________________________ E = ____ ___v
o
d) Find the oxidation potential for Cd: E = ____ ___v
2+
o
e)
Find the reduction potential for Cd : E = ____ ___v
f)
Which electrode gains mass as the cell operates? _______
+
g) Toward which beaker (AgNO3 or Cd(NO3)2) do K ions move? _______
h) The silver electrode and AgNO3 solution is replaced by Zn metal and Zn(NO3)2 solution.
What is the cell voltage now? __________Which metal now is the cathode? _________
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Consider the following electrochemical cell:
Cu
Ni
Ni(NO3)2
Cu(NO3)2
o
a) Write the equation for the half-reaction taking place at the nickel electrode. Include the E
o
____________________________________________________ E = ____ ___v
o
b) Write the equation for the half-reaction taking place at the Cu electrode. Include the E .
o
____________________________________________________ E = ____ ___v
c)
Write the balanced equation for the redox reaction taking place.
o
____________________________________________________ E = ____ ___v
d) What is the initial cell voltage? _________________ _
-
e) Show the direction of electron flow on the diagram above with an arrow with an “e “ written
above it.
f) Show the direction of flow of cations in the salt bridge using an arrow with “Cations” written
above it.
34
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Voltages at non-standard conditions
Note: When cells are first constructed, they are not at equilibrium. All the voltages calculated
by the reduction table are initial voltages.
-As the cells operate, the concentrations of the ions change:
eg) For the cell: Cu(NO3)2/Cu//Zn/Zn(NO3)2
2+
o
the cathode ½ reaction is: Cu + 2e à Cu
E = + 0.34 v
2+
o
the anode ½ reaction is: Zn à Zn + 2e
E = + 0.76 v
2+
2+
o
the overall reaction is: Cu + Zn à Cu + Zn
E = + 1.10 v
All electrochemical cells are exothermic (they give off energy)
2+
Initially: Cu
-
+ Zn
2+
Cu + Zn
+ energy
strong tendency to form products
Voltage = 1.10 v
2+
2+
As the cell operates [Cu ] decreases (reactants used up) & [Zn ] increases
(products formed). Both these changes tend to push the reaction to the left
(LeChateliers Principle)
2+
2+
Cu + Zn
Cu + Zn + energy
Voltage < 1.10 v
Eventually, these tendencies will be equal. At this point, the cell has reached
equilibrium. At equilibrium the cell voltage becomes 0.00 v.
Question: A cell is constructed using Cr/Cr(NO3)3 and Fe/Fe(NO3)2 with both solutions at
o
1.0 M and the temperature at 25 C.
a) Determine the initial cell voltage.
Answer: _____________v
b) What is the equilibrium cell voltage?
c)
Answer: _____________v
Write the balanced equation for the overall reaction taking place. Write the word
“energy” on the right side and make the arrow double.
_____________________________________________________________
d) Using the equation in (c), predict what will happen to the cell voltage when the
following changes are made:
3+
i) More Cr(NO3)3 is added to the beaker to increase the [Cr ]
Cell voltage: ______________creases
2+
ii) The [Fe ] ions is increased.
Cell voltage ______________creases
iii) A solution is added to precipitate the Fe
2+
ions
2+
The [Fe ] will _____crease & cell voltage will ______crease
35
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iv) Cr
3+
ions are removed by precipitation. Voltage: ______creases
v) The surface area of the Fe electrode is increased (see “conclusion near middle
of page 223 SW) Voltage: ____________________________________
vi) The salt bridge is removed. Voltage_____________________________
Predicting spontaneity from Eo of a redox reaction
o
If E for any redox (overall) reaction is > 0 (positive) the reaction is Spontaneous.
o
If E is < 0 (negative) the reaction is Non-spontaneous
o
When a reaction is reversed the sign of E changes
Example:
o
a) Find the standard potential (E ) for the following reaction:
2+
4+
2MnO4 + 4H2O + 3Sn à 2MnO2 + 8OH + 3Sn
b) Is this reaction as written (forward rx) spontaneous? _____
o
c) Is the reverse reaction spontaneous? _____ E = _____
Solution:
a) Find the two half-rx’s which add up to give this reaction. Write them so what’s
on the left of the overall rx is on the left of the half-rx. (& what’s on right is
on the right) The half-rx for MnO4 à MnO2 in basic soln. is at + 0.60. To
keep MnO4 on the left, this ½rx is written as it is on the table.
-
-
-
o
MnO4 + 2H2O + 3e à MnO2 + 4OH E = +0.60
2+
4+
The rest of the overall rx involves Sn changing to Sn . The ½ reaction for that must be
o
2+
reversed as well as its E . Since Sn must stay on the left side, the half-rx on the table
o
must be reversed as well as its E .
2+
Sn
4+
à Sn
-
o
+ 2e
E = -0.15 V
-
-Now, add up the 2 ½-rx’s to get the overall (Multiply by factors to balance e ‘s
o
–and add up E s.
-
-
-
o
(MnO4 + 2H2O + 3e à MnO2 + 4OH ) 2
2+
4+
(Sn à Sn 2e ) 3
-
2+
2MnO4 + 4H2O + 3Sn
-
E = + 0.60 v
o
E = -0.15 v
4+
à 2MnO2 + 8OH + 3Sn
o
E = +0.45 V
o
So E for the overall redox reaction = + 0.45 v
o
b) Since E is positive, this reaction is spontaneous as written.
o
c) The E for the reverse reaction would be – 0.45 v so the reverse reaction
is non-spontaneous.
36
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Question
o
3+
+
a) Calculate E for the reaction: 3N2O4 + 2Cr + 6H2O à 6NO3 + 2Cr + 12H
b) Is the forward rx spontaneous? _________ The reverse rx? ___________
Read SW. p. 215-224
Do Ex. 35 p. 217 and Ex. 36 a-d & 37-45 on p. 224-226 of SW
Practical Applications of Electrochemical Cells
-See SW. p. 230 - 233
The Lead-Acid Storage Battery (Automobile battery)
Anode (Pb Plate)
Cathode (PbO2 plate)
Pb
PbO2
Electrolyte H2 SO4 (aq)
-This cell is rechargeable (Reactions can be reversed)
Anode ½ reaction (Discharging or operating)
-
+
Pb(s) + HSO4 (aq) à PbSO4 (s) + H
-
(aq)
+ 2e
Ox # = +2
Ox # = 0
You can tell this is oxidation
because ox # of Pb ___creases
and electrons are ____. (LEOA)
Cathode ½ reaction (Discharging or operating)
-
+
PbO2 (s) + HSO4 (aq) + 3H
Ox # of Pb
= _____
-
(aq)
+ 2e à PbSO4 (s) + 2H2O (l)
Ox # of Pb
= _____
GERC
Write the balanced equation for the overall redox reaction (discharging)
37
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The overall redox reaction: (discharging or operating)
+
Pb(s) + PbO2(s) + 2H
Anode Cathode
-
(aq)
+ 2HSO4 (aq) à 2PbSO4 (s) + 2H2O(l) + electrical energy
From _____ in
the electrolyte
White solid forms on
plates as battery
discharges
Notes: As cell discharges the anode (Pb) and cathode (PbO2) disintegrate and the white solid
(PbSO4) forms on both plates.
+
Originally, [H ] & [HSO4 ] is high. i.e.) [H2 SO4] is high. H2 SO4 is denser than H2O therefore the
density (specific gravity) of the electrolyte is high to start with. As the cell discharges,
+
H2 SO4 (H & HSO4 ) is used up and H2O is formed. Therefore, electrolyte gets less dense as
the battery discharges. Condition of the battery can be determined using a “hydrometer” or
battery tester. (Higher the float, the denser the electrolyte)
Adding electrical energy to this reaction will reverse it (recharging)
Charging Reaction:
+
Electrical energy + 2PbSO4(s) + H2O (l) à Pb(s) + PbO2(s) +2H
-
(aq)
+ 2HSO4 (aq)
Supplied by
alternator or
generator
o
The E for the discharging rx, is +2.04 volts. A typical car battery has _____ of these in
(Series/parallel) ____________ to give a total voltage =__________V.
38
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The Zinc-Carbon battery (LeClanche-Cell, Common Dry cell, or regular carbon battery.
Often called “Heavy Duty”)
Cathode
Carbon
MnO2
Carbon
(s)
Zinc Anode
Rod
Steel Case
NH4 Cl & ZnCl2
paste electrolyte
Paste is acidic
(Causes rust
when they leak)
Cathode ½ reaction:
+
(GERC)
2MnO2 (s) + 2NH4 (aq) +2e à 2MnO(OH) (s) + 2NH3 (aq)
4+
Or simplified:
Mn
-
3+
+ e à Mn
Anode ½ reaction:
2+
(LEOA)
Zn (s) + 4NH3 (aq) à Zn(NH3)4 + 2e
2+
-
Or simplified: Zn(s) à Zn + 2e
-Not rechargeable
-Doesn’t last too long – especially with large currents
-Fairly cheap
The alkaline dry cell
+
MnO2
(cathode)
KOH
(Alkaline
electrolyte)
-Operates under basic conditions
- Delivers much greater current
- Voltage remains constant
- More expensive
- Lasts longer
Powdered Zn
anode
Cathode ½ reaction (GERC)
Anode ½ reaction (LEOA)
-
-
2MnO2 + H2O +2e à Mn2O3 + 2OH
Zn(s) +2OH à ZnO + H2O +H2O +2e
Ox # = ___
Ox # = ___
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Fuel cells
-Fuel cells are continuously fed fuel and they convert the chemical energy in fuel to
electrical energy
-more efficient (70-80%) than burning gas or diesel to run generators (30-40%)
-no pollution – only produces water
can use H2 and O2 or hydrogen rich fuels (e.g. methane CH4) and O2.
-used in space capsules –H2 & O2 in tanks H2O produced used for drinking
+ H2 out
H2 in
O2 in
O2 out
Anode
Cathode
Carbon
electrode
impregnated
with catalysts
KOH (aq)
electrolyte
Anode ½ reaction:
2H2(g) + 4OH (aq) à 4H2O(l) + 4e
Cathode ½ reaction:
O2(g) + 2H2O(l) + 4e à 4OH (aq)
Overall reaction:
2H2(g) + O2(g) à 2H2O(l)
Applied electrochemistry—The Breathalyzer Test
After drinking, breath contains ethanol C2 H5 OH.
o
Acidified dichromate (at E = 1.23 on the reduction table) will oxidize alcohol. The
unbalanced formula equation is:
C2 H5 OH + K2 Cr2O7 + H2 SO4 à CH3COOH + Cr2 (SO4)3 + K2 SO4 + H2O
Ox # of C= ___ Ox # of Cr= ___
K2 Cr2O7 is yellow
(orange at higher
concentrations)
Ox # of C= ___
Ox # of Cr= ___
Cr2(SO4)3 is green
-Exhaled air is mixed with standardized acidified dichromate
-Put in a spectrophotometer set at the wavelength of green light
-More alcohol produces more green Cr2 (SO4)3 (green)
-Machine is calibrated with known concentration samples of alcohol to ensure accuracy
Question: In the reaction above name
a) The oxidizing agent _________
b) The reducing agent _________
c) The product of oxidation ________
d) The product of reduction ________
40
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
Electrolytic Cells (ELC’s)
Electrolysis – uses an external power source to cause a non-spontaneous redox reaction to
occur.
The external cell or
power supply
The external
cell pulls e-s
from the Anode
e-
The external cell
forces e-s onto
the Cathode
e+ -
eANODE
In an ELC, the Anode
is connected to the
POSITIVE (+)
terminal of the
external power supply
+
e-
ee-
CATHODE
In an ELC, the
Cathode is connected
to the NEGATIVE (-)
terminal of the
external power supply
In BOTH Electrochemical Cells (ECC’s) and Electrolytic Cells (ELC’s):
OXIDATION takes place at the ANODE (LEOA)
REDUCTION takes place at the CATHODE (GERC)
There are three main types of Electrolytic Cells:
1. Electrolysis of Molten Salts (no H2 O) with Unreactive (Inert) Electrodes
2. Electrolysis of Aqueous Salts (H2 O solution) with Unreactive (Inert) Electrodes
3. Electrolysis of Aqueous Salts (H2 O solution) with Reactive Electrodes
Type 1 - Electrolysis of Molten Salts (no H2O) with Unreactive (Inert)
Electrodes
All molten (melted) salts consist of mobile ions.
No H2 O
present!
Eg.)
Free to
move
NaI(l) à Na+(l) + I-(l)
NaCl(l) à Na+(l) + Cl-(l)
Chemistry 12—Notes on Electrolytic Cells
Page 41
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
An example of a “Type 1” ELC: Electrolysis of molten NaCl (NaCl(l))
Diagram:
Inert Carbon
Electrodes
ANODE
Oxidation of Clhappens here
(LEOA)
CATHODE
Reduction of
Na+ happens
here
Na+(l) + e- à Na(l)
Na+
2Cl- (l) à Cl2(g) + 2e-
Cl Bubbles of
Cl2 gas.
Molten NaCl
Cathode Half-Reaction (Reduction of the Cation)
Looking near the bottom of the Reduction Table, we see the
half- reaction for the reduction of Na+
Na+ + e- à Na(s)
Eo = -2.71 v
Anode Half-Reaction (Oxidation of the Anion)
To write the oxidation of Cl- , we find Cl- on the Right
Side and REVERSE the half-reaction:
2 Cl- à Cl2(g) + 2e- Eo = -1.36 v (the sign on the Eo is changed since the rx. is reversed.)
To find the overall redox reaction with its Eo , we add up the half reactions as follows:
(Na+ + e- à Na(s) ) 2
Eo = -2.71 v
Reduction at Cathode
2 Cl- à Cl2(g) + 2e-
Eo = -1.36 v
Oxidation at Anode
2 Na+ + 2 Cl- à 2 Na(s) + Cl2(g)
Eo = -4.07 v
The Product at the Cathode is Na(s)
The Product at the Anode is Cl2(g)
The industrial application of this cell is
called a Down’s Cell.
Chemistry 12—Notes on Electrolytic Cells
Overall REDOX reaction
In Electrolytic Cells, the Eo for the overall
redox reaction is ALWAYS NEGATIVE.
They are all non-spontaneous!
Page 42
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
A quick method for writing half- reactions for Type 1 Electrolytic Cells (electrolysis of
molten salts with unreactive electrodes)
1.
2.
3.
4.
5.
6.
7.
Dissociate the salt into it’s ions
Underneath the Cation, write C- (representing the Cathode)
Underneath the Anion, write A+ (representing the Anode)
Draw arrow showing where each io n goes
The half-reactions must start with the ion (the ion must be the reactant)
For the Cathode Half-Reaction, write the reduction of the cation (same as on table)
For the Anode Half- Reaction, write the oxidation of the anion (reversed from table)
Example: For the Electrolysis of Molten Potassium Iodide (KI(l))
KI à K+
I-
+
C-
A+
So K+ will be reduced at the
Cathode and I- will be
oxidized at the Anode
Cathode Half-Reaction:
________________________________________ Eo = ____________v
Anode Half-Reaction:
________________________________________ Eo = ____________v
Overall Redox Reaction:
________________________________________ Eo = ____________v
Product at Cathode ________ Product at Anode ________ Minimum Voltage Needed _______v
Sketch this cell, labeling everything:
Chemistry 12—Notes on Electrolytic Cells
Page 43
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
Type 2 Electrolytic Cells – Electrolysis of Aqueous Salts and Non-reactive Electrodes
Because the salts are aqueous, water is present. Looking at the top and bottom shaded lines on
the reduction table, you can see that water can be oxidized or reduced:
The REVERSE of this reaction:
H2 O à ½ O2 + 2H+ + 2 e- Eo = -0.82 v
is the oxidation of water and could take
place at the ANODE
This half-reaction, the way it’s
written, is the reduction of
water and could take place at
the CATHODE.
To find the Half-Reaction at the CATHODE (reduction)
Due to the “Overpotentia l Effect”, the “Arrow on the Left” tells us whether:
1. The Cation is reduced ? or
2. Water is reduced?
All cations (+ ions) ABOVE this
arrow WILL be reduced at the
Cathode in aqueous solution.
(water present)
eg.) Zn2+ + 2e - à Zn(s)
All cations (+ ions) BELOW this
arrow WILL NOT be reduced at
the Cathode in aqueous solution.
The Water will be reduced
instead.
2 H2 O + 2e - à H2 + 2OH-
So when Cu2+ is in solution, what will be reduced at the cathode, Cu2+ or H2 O? _____________
Write an equation for the half-reaction at the Cathode: _________________________ Eo =_____
So when Mg2+ is in solution, what will be reduced at the cathode, Mg2+ or H2 O? ____________
Write an equation for the half-reaction at the Cathode: _________________________ Eo =_____
Chemistry 12—Notes on Electrolytic Cells
Page 44
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
To find the Half-Reaction at the ANODE (Oxidation)
Look at the “overpotential arrow on the RIGHT side of the table
Any anion AVOVE the arrow WILL NOT be
oxidized in aqueous solution. The water will be
oxidized instead:
2 H2 O à + ½ O2(g) + 2 H+ + 2 e(reverse the reaction at +0.82 v)
Any anion BELOW the arrow WILL be
oxidized in aqueous solution. :
Eg. ) 2 Cl- à Cl2 + 2e2 Br- à Br2 + 2e -
So when SO4 2- is in solution, what will be oxidized at the anode , SO42- or H2 O? _________
Write an equation for the half-reaction at the Anode : ________________________ Eo =___
So when I- is in solution, what will be oxidized at the anode , I- or
H2 O? __________
Write an equation for the half-reaction at the Anode : _______________________ Eo =____
A method for finding the Half- Reactions (Oxidation at the Anode) and (Reduction at the
Cathode) in this type of cell is:
1.
2.
3.
4.
Dissociate the salt into it’s ions
Write “H2 O” between the two ions.
Underneath the Cation, write C- (representing the Cathode)
Underneath the Anion, write A+ (representing the Anode)
Chemistry 12—Notes on Electrolytic Cells
Page 45
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
5. Using the rules about the “Overpotential Arrows” determine what is reduced at the
Cathode (the cation or water) and what is oxidized at the Anode (the anion or water).
6. Complete the half-reactions at the Cathode and the Anode (with their Eo ’s)
7. Add half-reactions to get the overall redox reaction if you are asked to.
Let’s do an example: A major industrial process is the electrolysis of brine (NaCl
Anode
NaCl à Na+ H2 O
Cl-
?
?
?
Cathode
?
C-
+
Na
-
NaCl (aq)
Cathode Half-Reaction:
A+
Now, we must look at the Reduction Table:
For the Cathode, look at the “overpotential
arrow” on the LEFT side. Notice that the Na+
is BELOW the arrow. This means Na+ will
NOT be reduced. The water will be reduced
instead. The half- reaction for the reduction
of water is at –1.41 v
+
H2 O
Cl-
(aq) ):
2 H2 O + 2e- à H2(g) + 2 OH-
Eo = -0.41 v
For the Anode, look at the “overpotential arrow” on the RIGHT side. Notice that the Cl- is
BELOW the arrow. This means that Cl- WILL be oxidized (rather than H2 O). So…
Anode Half- Reaction:
Overall Redox Reaction:
2 Cl-
à
Cl2(g)
+ 2e-
Eo = -1.36 v
2 Cl- + 2 H2 O à Cl2(g) + H2(g) + 2 OH- Eo = -1.77 v
The Products at the Cathode would be H2(g) + 2 OH- (the pH near the cathode would ____crease)
The Product at the Anode would be Cl2 (g)
There would be bubbles observed at both electrodes! Draw them into the diagram above!
The minimum voltage required to carry this reaction out would be 1.77 v (just enough to
overcome the –1.77 v Eo )
Now, try to complete the examples on the next page….
Chemistry 12—Notes on Electrolytic Cells
Page 46
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
For the electrolysis of aqueous CuCl2 using platinum (inert) electrodes. Find:
The half-reaction at the Cathode: __________________________________ Eo = ________
The half-reaction at the Anode:
__________________________________ Eo = ________
The overall redox reaction: _______________________________________ Eo = ________
Product(s) at the Cathode:________________ Product(s) at the Anode _________________
The minimum voltage required: ________________ v
For the electrolysis of Na2 SO4(aq) using carbon (inert) electrodes. Find:
The half-reaction at the Cathode: __________________________________ Eo = ________
The half-reaction at the Anode:
__________________________________ Eo = ________
The overall redox reaction: _______________________________________ Eo = ________
Product(s) at the Cathode:_________________ Product(s) at the Anode _________________
The minimum voltage required: ________________ v
For the electrolysis of CuSO4(aq) using inert electrodes. Find:
The half-reaction at the Cathode: __________________________________ Eo = ________
The half-reaction at the Anode:
__________________________________ Eo = ________
The overall redox reaction: _______________________________________ Eo = ________
Product(s) at the Cathode:_________________ Product(s) at the Anode _________________
The minimum voltage required: ________________ v
Chemistry 12—Notes on Electrolytic Cells
Page 47
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
Type 3 Electrolytic Cells – Electrolysis of Aqueous Salts with Reactive Electrodes
In this type of cell, the electrodes are normal metals, not inert ones like platinum or carbon.
Something that’s important:
The metal that the Cathode is made from is NOT reduced. That’s because metals cannot gain
electrons and become negative metal ions (no such ion as Fe2- !)
The Cathode metal is NOT oxidized. That’s because oxidation does not take place at the
cathode!
To summarize:
THE CATHODE METAL NEVER REACTS!!
The Cathode only supplies the SURFACE for the reduction of the Cation (+ ion) or for
the reduction of water (depending on whether the cation is above or below the left
overpotential arrow.) Here’s an example:
An aqueous solution containing the Cu2+ ion is electrolyzed. The cathode is made of Iron.
Write the equation for the Half- Reaction taking place at the Cathode :
The Iron will not react! (It only provides the surface—so there is no half-rx. involving Fe!)
The Cu2+ ion is ABOVE the overpotential arrow on the left, so Cu2+ will be reduced!
The Half-Reaction at the Cathode would be : Cu2+ + 2e- à Cu(s)
Eo = + 0.34 v
When this happens, the Iron Cathode will become coated with copper, and will turn reddish in
colour. (But remember, the iron itself does NOT react!)
Here’s one for you:
An aqueous solution containing the Mg2+ ion is electrolyzed. The cathode is made of Nickel.
Write the equation for the Half- Reaction taking place at the Cathode :
_______________________________________________________________________
What would you observe at the cathode? ______________________________________
Chemistry 12—Notes on Electrolytic Cells
Page 48
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
In a Type 3 cell, there are 3 possibilities for Oxidation at the ANODE:
1. The anion in the solution is oxidized
2. Water is oxidized
3. The Anode Metal is oxidized
The one with the Highest Oxidation Potential (The LOWEST one on the RIGHT of the table)
will be the one that is Oxidized. ( Treat water as if it were at the right overpotential arrow.)
Water behaves as if it were here.
Let’s do an example:
An aqueous solution containing the Cl- ion is electrolyzed. The anode is made of silver.
Write the equation for the Half- Reaction taking place at the Anode:
water
Cl-
Oxidation potential = -1.36 v
Ag
Oxidation potential = -0.80 v
The one with the highest oxidation potential. (The lowest on the RIGHT side) is Ag (the anode)
So the Half- Reaction at the Anode is the oxidation of Ag:
Chemistry 12—Notes on Electrolytic Cells
Ag(s) à Ag+ + e-
Eo = -0.80v
Page 49
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
Here’s an example question. Given the following cell:
Nickel
Iron
Al2 (SO4 )3 (aq)
Identify the Anode ____________________ the Cathode __________________
Write an equation for the half-rx. at the Anode _______________________ Eo =________
Write an equation for the half-rx. at the Cathode ______________________ Eo =________
Write an equation for the overall redox rx. ___________________________ Eo = _______
Product at Anode _____ Product at Cathode ______ Minimum voltage necessary _________
Electrorefining
In Electrorefining, an impure metal is refined by making it the ANODE. The pure metal is the
CATHODE.
Eg.) An ANODE made of IMPURE COPPER could have something like the following make-up:
§ Mostly copper
§ Some Ag or Au (above Cu on the reduction table)
§ Some Zn (below Cu on the reduction table)
§ Some “dirt” –any non- metal impurities
The CATHODE would be made of PURE COPPER
The SOLUTION is an aqueous solution of a compound containing Cu2+ (eg CuSO4 )
Remember, for Electrorefining:
IMPURE ANODE à PURE CATHODE (PC)
Chemistry 12—Notes on Electrolytic Cells
Page 50
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
Use your Reduction Table to help you explain all the information shown in the diagram:
First, we look at the ANODE:
Going towards the Cathode
e-
+
e-
_
The solution is CuSO4(aq)
So there are lots of Cu 2+
ions there to begin with
eIMPURE ANODE
Cu
Cu
Cu
Cu
Ag
Cu
Cu
Cu
Cu
Cu
Ag
Cu
Cu
Cu
Au
Cu
Cu
Cu
Cu
Cu
Au and Ag will NOT oxidize as
long as Cu is present. As the Cu
oxidizes around it, they fall
down, with the dirt, to become
part of the Anode Mud (sludge)
Cu2+
Cu2+
2+
Cu
Once Zn is gone, Cu2+ (the next highest
Ox. Potential) will start oxidizing:
Cu à Cu2+ + 2e-
Cu
Cu
Cu
Cu
Cu2+
Cu
Cu2+
Zn
Cu
dirt
Cu
Cu
Cu2+
Cu2+
Cu2+
Cu2+
Cu2+
Cu2+
Cu2+
2+
Zn2+
SO4 2-
Cu2+
Cu2+
SO4 2-
SO4 2-
Zn2+
Cu
Cu
SO4 2-
Cu2+
Cu2+
Cu2+
Cu2+
Cu2+
Cu2+
Au
Zn will oxidize first (highest
Ox. Potential), putting Zn2+
ions into the solution.
Zn à Zn2+ + 2e-
The solution contains only Cu 2+
SO4 2 and Zn2+ ions. Metals
above Cu (Au & Ag) have fallen
to the Anode mud (not oxidized
into ions!)
Anode Mud
Chemistry 12—Notes on Electrolytic Cells
Page 51
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
An Excess of e-s are forced onto
the Cathode by the external
power supply. The Cathode thus
becomes Negative.
Now, let’s look at the CATHODE
Coming from the Anode
e-
e-
_
+
e-
Cu2+
Cu2+
2+
Cu
SO4 2-
Pure Copper CATHODE
SO4 2-
Cu2+
Cu2+
Cu2+
Cu
Cu
2+
Cu
SO4 2-
SO4 2-
Zn2+
2+
Only Cu will be reduced at the Cathode. The
SO4 2- ion is negative so will not go to the negative
cathode. The Zn2+ ions have a Lower Reduction
Potential than Cu 2+, so they would not be reduced
until all the Cu 2+ is gone. But that NEVER
happens. If Cu 2+ runs low, the anode is replaced
with another one and more Cu 2+ is produced. So
ions like Zn2+ will stay in the solution without
ever being reduced.
Cu
Cu
Cu
Cu
Cu
2+
Cu2+
2+
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
Cu
The Half-Reaction at the
Cathode is:
Cu2+ + 2e- à Cu
So, in summary. In Electorefining:
§ The Impure Metal is the Anode
§ The Pure Metal is the Cathode
§ The Electrolyte (solution) contains the cation of the metal to be purified
Electroplating
An application of an electrolytic cell is Electroplating. In Electroplating :
•
•
•
The object to be Plated is the Cathode (attached to the Negative Terminal of the Battery)
The Electrolyte must contain the Cation of the Metal to be Plated on the Object
The Best Anode is the Metal to be Plated onto the Object
Chemistry 12—Notes on Electrolytic Cells
Page 52
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
Example: We wish to plate and Iron ring with Copper.
•
•
•
The Iron Ring (made by Dwarfs? ) is made the Cathode (connected to the negative)
The Anode is Copper (to keep supplying Cu2+ ions as it is oxidized)
The Electrolyte is aqueous CuSO4 (supplies Cu2+ to start with. SO4 2- does not react here.)
Here is the diagram of the set- up:
+
-
eFe Ring
CATHODE
Cu
ANODE
At the Anode , Cu is
Oxidized to Cu 2+ (LEOA)
Cu
Cu 2+
Cu
Cu 2+
Cu à Cu2+ + 2eCu 2+
SO4
Cu 2+
Cu 2+
SO4 2-
2-
Notice that for every Cu 2+ ion
that is reduced at the Cathode, a
Cu 2+ ion is produced by the
anode
At the Cathode , Cu 2+
ions are Reduced to
form Cu (s) which sticks
to the surface of the
iron ring and Plates it.
(GERC)
Cu2+ + 2e- à Cu
The Electrolyte is CuSO4(aq)
which provides Cu 2+ ions to
start the reduction at the
Cathode. The SO4 2- ions are
spectators in this case.
Electrowinning
The name given to the reduction of ores to produce metals in industry.
Eg.)
Zn2+ +
Comes from Zinc
ores like ZnS,
ZnCl2 etc.
2e -
à
Zn(s)
Zinc Metal
Chemistry 12—Notes on Electrolytic Cells
Page 53
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
Aluminum Production (see p. 246 in SW)
Looking at the reduction table:
We see that Al3+ is BELOW the left
overpotential arrow. Therefore, Al3+
CANNOT be reduced from an aqueous
(water) solution.
In order to reduce Al3+ to it’s the metal Al, you must electrolyze a MOLTEN salt of
aluminum.
The main ore used to produce Aluminum is called Bauxite – Hydrated Aluminum Oxide
.
or
Al2 O3 3H2 O
.
When this is heated, the water is forced off: Al2 O3 3H2 O + heat à Al2 O3 + 3H2 O
The melting point of Alumina (Al2 O3 ) is much too high to melt it
Called
economically. (mp. = 2072 o C)
“Alumina”
So Alumina (Al2 O3 ) is mixed with Cryolite (Na3 AlF 6 )
The melting point of this mixture is ˜ 1000 o C.
Some of the ions present in this mixture of molten salts are: Al3+, O2-, F- and Na+
Remember, there is
NO water in this
mixture. It is molten.
The Half-Reaction at the Cathode is the Reduction of Al3+ ions to Al metal:
Cathode Half-Reaction: Al3+(l) + 3e- à
Al(l)
(At these high temperatures both Al3+ & Al are liquids)
The Half-Reaction at the Anode is the Oxidation of oxide ions using Carbon Electrodes:
Anode Half- Reaction: C(s) + 2O2- à CO2(g) + 4eSome texts state more simply: 2O2- à O2 + 4e -
Chemistry 12—Notes on Electrolytic Cells
Page 54
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
This process is carried out by Alcan in Kitimat, B.C. It uses 10 million amps of electricity.
Alcan has their own power generating plant at Kemano B.C.
Because of high temperature and great amounts of electrical energy used, production of Al is
expensive! Recycling Al used a lot less energy!
Corrosion
Corrosion is Undesirable Oxidation of a metal (usually Fe)
Oxygen from the air is reduced at the Cathode
The Fe becomes the Anode (is oxidized)
Moisture can provide a solution for these processes to take place in.
O2 from the air dissolves in the water
O2
droplet. The dissolved O2 is reduced at
the Cathode and the iron surface becomes
the Anode. The series of reactions that
take place are as follows:
1. Fe à Fe2+ + 2e- (iron is oxidized to Fe2+)
2. ½ O2 + H2O + 2e- à 2 OH- (O2 picks up e-s from the Fe and is reduced to OH- )
3. Fe2+ + 2 OH- à Fe(OH)2(s) (the Fe2+ from Rx.1 and the OH- from Rx.2 form the ppt. (Fe(OH)2(s))
4. Fe(OH)2
Oxidized by O2 and H2 O
5. Some Fe(OH)3(s)
Fe(OH)3(s)
Reacts with H2 O to produce
.
Fe2 O3 3H2 O(s) (Hydrated Iron(III) Oxide)
.
A mixture of Fe(OH)3(s) and Fe2 O3 3H2 O(s) is called RUST
Preventing Corrosion
ü Keep the Fe (or steel) surface protected from moisture and oxygen – Paint or other Coatings
ü Keep the Fe away from metals which are higher on the Reduction Table.
eg.) When Fe and Cu are in close proximity and the conditions are right—moisture is present and there are
some dissolved salts—a type of Electrochemical Cell can form in which Cu becomes the Cathode and
Fe becomes the Anode (and Fe is oxidized)
ü Use what’s called Cathodic Protection
Chemistry 12—Notes on Electrolytic Cells
Page 55
Chemistry 12 Notes on Unit 5-Redox
Electrolytic Cells
In Cathodic Protection:
A metal lower on the Reduction Table (higher oxidation potential) is attached to or brought near
to the Fe.
Any of these metals would
work to Cathodically
protect Iron.
Metals below Mg would
cathodically protects Fe, BUT
they are highly reactive with
water to form explosive H2(g)!
An example would be Mg
In the presence of O2 or any other oxidizing agent, Mg ( with an oxidation potential = +2.37 v)
Will give up e-s more readily than Fe (oxidation potential = +0.45 v ).
This saves the Fe from being oxidized!
Some common examples of Cathodic Protection:
ü Plates of Mg are attached to the hulls of ships. When Mg is oxidized it is replaced.
ü Galvanized Steel is Steel (mainly Fe) coated or mixed with Zn (oxidation potential = +0.76v )
Zn à Zn2+ + 2e-
a strong coating of ZnO is formed, protecting the metal from
further oxidation.
ü Stainless Steel contains Fe with other metals like Cr etc.
Impressed Current
A voltage is applied with an external power supply to keep the Fe slightly negative. Thus, the
Fe surface has excess e-s so it doesn’t have to oxidize in order to supply e-s to O2 .
The End of Chem 12 Notes! (And you thought we’d never get there!)
Chemistry 12—Notes on Electrolytic Cells
Page 56
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Chemistry 12
Electrochemical and Electrolytic Cells—Summary
In BOTH Cells:
Losing
Electrons
Oxidation at the
Anode
Electrochemical Cells
Higher Half-Rx. on Table Is the Cathode
o
Gaining
Electrons
Reduction at the
Cathode
Electrolytic Cells
Anode is +
Cathode is –
SPONTANEOUS E is Positive +
At the Cathode:
Reduction of the Cation in the beaker.
Eg. Cu2+ + 2e- à Cu
NON-SPONTANEOUS Eo is Negative At the Cathode:
Reduction of Cation in the Electrolyte
Eg. Cu2+ + 2e- à Cu or
Reduction of Water: H2O +2e- à H2 + 2OH-
At the Anode:
Oxidation of the Metal Anode Electrode
Eg. Pb à Pb2+ + 2e-
At the Anode:
Oxidation of the Anion in Solution:
Eg. 2Cl- à Cl2 + 2e- or
Oxidation of Water:
H2O à ½ O2 + 2H+ + 2e- or
Oxidation of the Metal Anode Electrode:
Eg. Cu à Cu2+ + 2e(whichever has the highest oxidation
potential ie. lowest on the right side of table)
Electrons go from A à C in the wire
External power supply pushes e- ‘s on to the
Cathode, making it – and takes e- ‘s from the
Anode making it +
Cations move toward à Cathode
Anions move toward à Anode
Cations(+) attracted to the – Cathode where
they (or water) are reduced.
Anions(-) attracted to the + Anode, where
they, water or the anode is oxidized
In the Salt Bridge
Applications include: Automobile (Pb/Acid)
Battery, Zn/C (Dry) Cells, Alkaline Cells,
Fuel Cells
Applications include: Electrolysis to
decompose compounds into elements,
Downs Cell, Electrorefining, Electroplating,
Electrowinning (eg. Production of Al)