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Redox Reactions
Zumdahl 6th Ed. Section Z4.10-Z4.12
4.10 Oxidation-Reduction Reactions
4.11 Balancing Oxidation-Reduction Equations
4.12 Oxidation-Reduction Titrations
Revisit the process for making a salt. A metal and a halogen
(a nonmetal) react. This is an example of a Redox reaction.
•Understand Oxidation Numbers, Ox.Nb.s
•Need Ox.Nb.s to balance redox reactions
•Balance Redox reactions using half cell method.
Problems: 4.55-59
Problems to be used on the exam:
4.60-4.64. (See 4.61c) Be sure to identify spectator ions. 4.6566, 4.69, 4.74, 4.76, 4.79, 4.83
Figure 4.19: Reaction of solid sodium and
gaseous chlorine to form solid sodium chloride:
an oxidation – reduction reaction.
Types of Redox Reactions (Rxns):
Burn a metal, a metal oxide (related to a base)
Burn a non-metal, a non-metal oxide (an acid)
Ox.Nb. Oxidation Numbers
Oxidation - A substance gives up electrons to another
substance. Mg + ½ O2 → Mg2+ + O2- Mg is oxidized
(Oxygen is the oxidizing agent; it accepts electrons)
Reduction - A substance accept electrons from another
substance. Mg + ½ O2 → Mg2+ + O2- O is reduced (Notice
O’s Ox.Nb. (or charge in this case) is reduced; Mg is the
reducing agent causing the reduction of Oxygen.)
•
•
In a chemical reaction, the total number of electrons and the number
of charges are conserved. It is convenient to assign fictitious
charges to the atoms in a molecule and call them “oxidation
states” or “oxidation numbers” (Ox.Nb. Or just O.N.).
Oxidation numbers are chosen so that (a) charges are conserved,
and (b) in ionic compounds the sum of oxidation numbers on the
atoms coincides with the charge on the ion.
Assigning Ox.Nb.s: Use Periodic Table
Rules for giving Ox.Nb.s (Table 4.3 in Text)
Ox Nb; Oxidation Numbers
Determine the oxidation number (Ox.Nb.) of each
element in the following compounds:
1: Iron (III) Chloride
2: Nitrogen Dioxide
3: Sulfuric acid
Strategy: We apply the rules in Table 4.3, always
making sure that the Ox.Nb. values in a compound add up
to zero, and in a polyatomic ion, to the ion’s charge.
Solution:
1: FeCl3 This compound is composed of monoatomic
ions. The Ox.Nb. of Cl- is -1, for a total of -3. Therefore
the ON of Fe is +3, The Roman Numeral, III is the
giveaway here.
Examples of Ox.Nb.s
example: NO2
The Ox.Nb. of oxygen is -2 for a total of -4. Since the
Ox.Nb.s in a compound must add up to zero, the ON of N
is +4.
example: H2SO4
The Ox.Nb. of H is +1, so the SO42- group must sum to 2. The ON of each O is -2 for a total of -8. Therefore S
has the ON +6.
example: OF2 and O2F2 vs. H2O and H2O2
Examples (cont.):
When hydrogen reacts with a metal, it acts as a nonmetal.
BaH2 (Barium Hydride)
H = -1
Ba = +2
NH4NO3
H = +1
O = -2
N = -3
N = +5
Oxidation Reduction
Summary:
• An element’s oxidation
number goes up, it is
oxidized (so it is a
reducing agent)
• An element’s oxidation
number goes down, it is
reduced (so it an
oxidizing agent)
• If you think this is
confusing try:
http://www.youtube.com
/watch?v=LS75NtlH3gI
More Examples of Ox.Nb.s
• What is the oxidation number of sulfur in
Sodium Sulfate, Na2SO4?
• What is the oxidation number of iodine in
the I3- ion?
• What is the oxidation number of oxygen in
O3, ozone?
• OF2 , and O2F2, H2O, H2O2, K2O, K2O2, HgS,
Hg2S.
To Balance RedOx Rxns
• Need Oxidation numbers of species before
and after reaction
• Need the change in OxNb.s to keep the
reaction electrically neutral, so there is no
charge imbalance at the end of the
reaction.
Balance Reaction by inspection (hard)
___ Fe ( s ) + ___ Pt 2+ ( aq ) → ___ Pt ( s ) + ___ Fe3+ ( aq )
• What’s wrong with setting all coefficients to one?
• Look at Full reaction:
___ Fe ( s ) + ___ Pt 2+ ( aq ) + ___ Cl − ( aq )
→ ___ Pt ( s ) + ___ Fe3+ ( aq ) ___ Cl − ( aq )
• Need to make sure charge is not created (or lost).
Reintroduce (fictitious?) spectator ions to balance
charge. So that each side is neutral.
aFe ( s ) + bPt 2+ ( aq ) + 2bCl − ( aq )
→ bPt ( s ) + aFe3+ ( aq ) + 3aCl − ( aq )
• Chloride balance tells us
2b = 3a or a = 2
Balancing REDOX Equations:
The oxidation number method
Step 1) Assign oxidation numbers to all elements in the equation.
Step 2) From the changes in oxidation numbers, identify the oxidized
and reduced species.
Step 3) Compute the number of electrons lost in the oxidation and
gained in the reduction from the oxidation number changes.
Draw tie-lines between these atoms to show electron changes.
Step 4) Choose coefficients for these species to make the electrons lost
equal the electrons gained (or total increase in OxNb = total
decrease in OxNb)
Step 5) Complete the balancing by inspection.
Problem: Calculate the mass of metallic Iron that must be
added to 500.0 liters of a solution containing
0.00040M of Pt2+(aq) ions in solution to reclaim all Pt
via: __ Fe(s) + __ Pt2+(aq) → __ Fe3+(aq) + __ Pt(s)
Solution:
V x M = # moles Pt2+
500.0L x 0.00040 mol Pt2+/ L = 0.20 mol Pt2+
Balance the equation to determine how many
moles of Fe are needed for every mole of Pt.
REDOX Balancing using Ox. No. Method - I
The number of electrons must balance (just like the number of
atoms must balance).
Accepts 2 e- per Pt
+2
0
Fe ( s ) + Pt
2+
( aq ) → Fe ( aq ) + Pt ( s )
3+
Gives 3 e- per Fe
0
+3
electrons lost must = electrons gained
Because the electrons are transferred from one (molecular) species to another
Therefore need 3 Pt atoms for every 2 Fe!
2 Fe ( s ) + 3Pt
2+
( aq ) → 2 Fe ( aq ) + 3Pt ( s )
3+
Balanced and both sides have same (+6) charge. No change in charge.
Problem: Calculate the mass of metallic Iron that must be
added to 500.0 liters of a solution containing
0.00040M of Pt2+(aq) ions in solution to reclaim all Pt
via: 2 Fe(s) + 3 Pt2+(aq) → 2 Fe3+(aq) + 3 Pt(s)
Solution:
moles
⋅ 500liters − Pt 2+ = ( − ) 0.20moles − Pt 2+
liter
Fe
2
SC
Fe
( ) = 0.20moles − Pt 2+
rxn = 0.40 moles − Fe
X = 0.20moles − Pt 2+ *
2+
3
Pt
SC ( Pt 2+ )
3
rxn
X = 0.133mol − Fe * M W ( Fe ) = 0.133mol − Fe ⋅ 55.85 g
= − 7.4 g − Fe
mole ( )
X = M *V = .00040
(
(
)
)
Recognizing Oxidizing and Reducing Agents
Problem: Identify the oxidizing and reducing agent in each of the Rxn:
a) Zn(s) + 2 HCl(aq)
ZnCl2 (aq) + H2 (g)
b) S8 (s) + 12 O2 (g)
8 SO3 (g)
c) NiO(s) + CO(g)
Ni(s) + CO2 (g)
Plan: First we assign an oxidation number (Ox.Nb.) to each atom
(or ion) based on the rules in Table 4.3. The reactant is the reducing
agent if it contains an atom that is oxidized (Ox.Nb. increased in the
reaction). The reactant is the oxidizing agent if it contains an atom that
is reduced ( Ox.Nb. decreased).
Solution:
a) Assigning oxidation numbers:
-1
+1
0
Zn(s) + 2 HCl(aq)
-1
0
+2
ZnCl2 (aq) + H2 (g)
HCl is the oxidizing agent, and Zn is the reducing agent!
Recognizing Oxidizing and Reducing Agents
b) Assigning oxidation numbers:
0
0
S8 (s) + 12 O2 (g)
+6
S [0]
-2
8 SO3 (g)
S is Oxidized
O[0]
O[-2]
O is Reduced
S8 is oxidized and O2 is reduced
S8 is the reducing agent and O2 is the oxidizing agent
c) Assigning oxidation numbers:
C[+2]
C[+4]
C is oxidized
-2
+2
NiO(s)
-2
+2
+ CO(g)
Ni[+2]
Ni[0]
Ni is Reduced
0
Ni(s)
S[+6]
+4 -2
+ CO2 (g)
CO is the reducing agent and NiO is the oxidizing agent
No O changes its O.N. but it is the agent for the change.
REDOX Balancing using Ox. No. Method - I
The number of electrons must balance (just like the number of
atoms must balance).
Accepts +2 e- per O -2
0
___ H2 (g) +___ O2 (g)
0
___ H2O(g)
Donates 1 e- per H +1
electrons lost must = electrons gained
Because the electrons are transferred from one molecular species to another
Therefore need 4 H atoms for the 2 O atoms!
REDOX Balancing using Ox. No. Method - I
2x +2 e-
0
___
2 H2 (g) +_1_ O2 (g)
0
4x -1 e-
-2
___
2 H2O(g)
+1
electrons lost must = electrons gained
Therefore need 4 H atoms for the 2 O atoms!
The half-cell method (next) makes this electron loss and gain
more concrete and is a bit easier to do and more robust.
Balance the thermite reaction:
__ Fe2O3(s) + __ Al(s) → __ Al2O3(s) + __ Fe(s)
Assign Ox.Nb. to each element
Reactants
O: -2 Al:0
Fe:+3
Products
O:-2 Al:+3
Fe:0
Identify the elements that are oxidized and reduced
Fe3+ →Fe0 : Iron is reduced (OxNb decreases by 3)
Al0 →Al3+ : Aluminum is oxidized (OxNb increases by 3)
Make Ox.Nb. total increase = Ox.Nb. total loss
Fe2O3 + 2Al → Al2O3 + 2Fe
Thermite reaction
_ Fe2O3(s) + __ Al(s) → _ Al2O3(s) + _ Fe(l)
+3
0 →
+3
- 3e- per Al
+3e- per Fe
0
Thermite reaction
1 Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(l)
+3
0 →
+3
2x(- 3e- per Al)
2x(+3e- per Fe)
BALANCED!
0
Half (Cell) Reaction Balancing Method
• Break a reaction into two “half-reactions”, one for
each element type involved in the redox.
• Balance each half reaction separately
– Balance O (with water), H with protons, charge
with e
• When done cross multiply and add to cancel out
electrons.
• This uses the ideas of redox but you don’t need to
know the oxidation numbers specifically. Just need
to identify the species being reduced, and the
species being oxidized.
• These reacitons are not fictitious; they are batteries.
Example Half Cell Balance
Cu(s) +HNO3(aq) → Cu2+(aq) + NO(g)
Identify half-reactions, one is ox, other is red
Cu(s) → Cu2+(aq)
Not needed; but nice
0 → +2
to know
HNO3(aq) → NO(g)
+5 → +2
Balance all atoms that are neither H nor O
OK as is
Balance O by adding H2O to side deficient in O
Cu(s) → Cu2+(aq)
HNO3(aq) → NO(g) + 2H2O
Balance H by adding H+ to side deficient in H
Cu(s) → Cu2+(aq)
3H+ +HNO3(aq) → NO(g) + 2H2O
Balance charge by adding e- to side that has + charge
Cu(s) → Cu2+(aq) + 2e3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O
Multiply each eq by factors so electrons cancel out
3x(Cu(s) → Cu2+(aq) + 2e-)
2x(3H+ + HNO3(aq) + 3e- → NO(g) + 2H2O)
Add equations
3Cu(s)+ 6H+ + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
The reaction is balanced in an acidic environment
because there are protons around.
If the problem says, balance in a basic environment, then
first balance in acidic solution (as you just did).
Now add hydroxides to both sides to cancel out the
protons and generate waters. Cancel out excess waters if
necessary.
You are done. Eg: Add OH- and cancel extra water:
3Cu ( s ) + 6 H + ( aq ) + 2 HNO3 ( aq ) → 3Cu 2+ ( aq ) + 2 NO ( g ) + 4 H 2O ( aq )
3Cu ( s ) + 6 H + ( aq ) + 6OH − ( aq ) + 2 HNO3 ( aq )
→ 3Cu 2+ ( aq ) + 2 NO ( g ) + 4 H 2O ( aq ) + 6OH − ( aq )
3Cu ( s ) + 6 H 2O ( aq ) + 2 HNO3 ( aq )
→ 3Cu 2+ ( aq ) + 2 NO ( g ) + 4 H 2O ( aq ) + 6OH − ( aq )
3Cu ( s ) + 2 H 2O ( aq ) + 2 HNO3 ( aq )
→ 3Cu 2+ ( aq ) + 2 NO ( g ) + 6OH − ( aq )
The following redox equation is balanced in acidic
solution. Balance it for a basic solution.
Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3- (aq)
+3H2O(l)
Add OH- equal to number of H+ on both sides
6OH-(aq) +Cr2O72-(aq)+2NO(g)+6H+(aq)→2Cr3+(aq)+2NO3-+3H2O(l)
+6OH-(aq)
Combine OH- and H+ to form water to max extent possible
6H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3-+3H2O(l) +6OH- (aq)
Cancel H2O on both sides to max extent possible
3H2O(l)+Cr2O72-(aq)+2NO(g)→2Cr3+(aq)+2NO3- +6OH- (aq)
RedOx Balancing by Half-Reaction Method
Fe 2+ ( aq ) + MnO4− ( aq ) → Fe3+ ( aq ) + Mn 2+ ( aq )
in acid
Identify Oxidation and Reduction Half Reactions
Fe 2+ ( aq ) → Fe3+ ( aq ) + 1e −
MnO4− ( aq ) + 5e − → Mn 2+ ( aq )
Oxidation Half Rxn
Reduction Half Rxn
Add water as needed to balance O, then add H+ to balance H, then e-s.
8 H + + MnO4− ( aq ) + 5e − → Mn 2+ ( aq ) + 4 H 2O
Sum the two half-reactions (multiplied to balance/cancel electrons):
5 x { Fe 2+ ( aq ) → Fe3+ ( aq ) + 1e − }
1x {8 H + + MnO4− ( aq ) + 5e − → Mn 2+ ( aq ) + 4 H 2O}
8H + + MnO4− ( aq ) + 5 Fe 2+ ( aq ) → 5 Fe3+ ( aq ) + Mn 2+ ( aq ) + 4 H 2O
Redox Balance Problems
• Balance these by any of the three
methods (Half rxn method is most
foolproof).
__ Pb ( s ) + __ PbO2 ( s ) + __ H 2 SO4 ( aq ) → __ PbSO4 ( s )
Fe 2+ ( aq ) + MnO4− ( aq ) → Fe3+ ( aq ) + Mn 2+ ( aq )
in acid
Fe ( CN )6 ( aq ) + Ce 4+ ( aq )
4−
→ Ce ( OH )3 ( s ) + Fe ( OH )3 ( s ) + CO32− ( aq ) + NO3− ( aq )
IO ( aq ) + I
−
3
−
( aq ) → I ( aq )
−
3
in acid
End of Lecture
• Many More Worked Examples in
EXTRA
• Redox reaction balancing and
determining number of moles of product
based on moles of reactants.
• See RedOx titration notes in EXTRA.
Typical stoichiometry problems using a
redox reaction to determine the
stoichiometry of the reaction.