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NAME _____________________________________ NOTES: UNIT 6 (5): REDOX: SPECIES AND
HALF-REACTIONS
I-IX) Oxidation Numbers to Reaction Types
X) How to identify a redox reaction on a worksheet /test ….
A) REVIEW: A REDOX reaction is one in which there is a change in oxidation numbers
when comparing the oxidation numbers of the reactants to the oxidation numbers of the
products
With that understood, most (not all) redox reactions which come the way of the first year chemistry student are
relatively simple and have one thing in common, which allows the student to spot the change in oxidation states.
So, maybe I can help you "cheat" a little bit. Take a look...
B) Only one of the reactions is a redox reaction. Circle the choice representing the redox reaction.
(Hmmmmm!!)
1) Pb(NO3)2 + 2 KI  PbI2 + 2 KNO3
3) CdCO3  CdO + CO2
2) Na2CO3 + 2 HCl  2 NaCl + H2O + CO2
4) 2 AgI + F2  I2 + 2 AgF
1) When trying to determine whether a reaction is a redox, or not, ….
a) the 1st year student, should look for the reaction *with a free element (and that free
element can be on the reactant side, product side …or both!!!)
b) A big exception is:
* 3 O2  2 O3
C) One way to help identify the oxidized and reduced species Fe + CuSO4  FeSO4 + Cu
1) STRATEGY: Assign oxidation numbers to all species and compare
the reactant species to the species it becomes on the product side.
a) The answer must always come from the reactant side, when asked
for: the oxidized or reduced species
reducing agent or oxidizing agent...
+3
+2
+1
0
-1
-2
b) Rationale: the products do not exist, until the reactants undergo the
alterations to their electron clouds (the chemical change), due to electron
-3
transfer … thus, it is to these reactant species that all the action occurs.
Perforce, of that argument, then, the most appropriate response(s) when asked to
identify the oxidized or reduced species (or the agents) … must be a species from the
reactant side of the equation.
c) please note that the name of the reduced form of a nonmetal ends in –ide.
(e.g. Cl-1 = chloride, P-3 = phosphide, O-2 = oxide etc…)
Also, when one says “hydrogen…” it is interpreted as being H0 or H20 , so if you wish
to articulate H+1 as being reduced to H0, you must say “hydrogen plus 1” or something
very close to that effect.
417
2) We can identify the oxidized species because it is the reactant which
becomes * MORE POSITIVE (increases in oxidation #)
The reduced species is the reactant which becomes *MORE NEGATIVE (decreases in
oxidation number
due to a gain of e-
0
a)
0
2 Mg(s)
+ O2(g)

2+ 2-
2 MgO(s)
Mg0 was oxidized because it lost / gained negative charge (e-) and its oxidation
number became more * positive
O20 was reduced because it
lost / gained negative charge (e-) and its oxidation
number became more * negative
 the oxidized and reduced species are from the * reactant
side of the reaction equation!!!
Be sure to include the oxidation state with every answer. Some reactions may have “spectator ions”
Guided Practice: Assign the oxidation numbers to each species and identify the oxidized and reduced species
a) oxidation of aluminum:
0
4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351.4 kJ
0
+3 -2
4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351.4 kJ
reduced species = * O20 or O0
oxidized species = * Al0
because its oxidation state became * more negative
because its oxidation state became * more positive
0
b) Making Ammonia (Haber Process):
reduced species = * N20 or N0
oxidized species = * H20 or H0
0
as a product
as a product
-3 +1
N2(g) + 3H2(g)  2NH3(g)
because its oxidation state became * more negative
because its oxidation state became * more positive
as a product
as a product
418
0
0
+1 -1
c) Making Table Salt: 2 Na(s) + Cl2(g) 
https://www.youtube.com/watch?v=VBReOjo3ri8
2 NaCl(s)
or go to Lonnie’s Lab at: https://www.youtube.com/watch?v=WVonuBjCrNo
https://www.youtube.com/watch?v=oZdQJi-UwYs
reduced species = * Cl20 or Cl0
oxidized species = * Na0
because its oxidation state is more *negative
because its oxidation state is more *positive
+6 -1
d) Purifying Uranium :
reduced species = * U+6
oxidized species = * F-1
on the product side
0
on the product side
0
UF6(s) U(s) + 3 F2(g)
notice there is only 1 reactant … but there
are 2 different species in that reactant.
because its oxidation state is * more negative
because its oxidation state is * more positive
on the product side
on the product side
(note: F6-1 is not correct, because the fluoride ions are not bonded to each other ...each is individually bonded via a covalent bond [surprisingly] to the uranium)
0
e) Production of Laughing Gas:
reduced species: * O20 or O0
+3
f) Purifying Gold:
0
+1 -2
2 N2(g) + 2 O2(g)  2 N2O(g)
oxidized species: * N20 or N0
-2
0
+1 -2
0
Au2S3(s) + 3 H2(g)  3 H2S(g) + 2 Au(s)
reduced species: *Au+3
oxidized species: * H20 or H0
0
g) Purifying Silicon:
+4 -2
0
Mg(s) + SiO2(s)  Si(s) + 2 MgO(s)
reduced species: *Si+4
oxidized species: *Mg0
0
h) Purifying Nickel:
+2 -1
+3 -1
0
2 Al(s) + 3 NiCl2(aq)  2 AlCl3(aq) + 3 Ni(s)
reduced species: *Ni+2
oxidized species: * Al0
+4 -2
i) Making Nitric Acid:
+2 -2
0
+1 -2
+1 +5 -2
4 NO2 + O2 + 2 H2O  4 HNO3(aq)
reduced species: * O20 or O0
oxidized species: * N+4
419
0
+2 +6 -2
0
+2 +6 -2
j) Corrosion of the Statue of Liberty: Fe(s) + CuSO4(aq)  Cu(s) + FeSO4(aq)
reduced species: * Cu+2
oxidized species: * Fe0
+3 -2
l) Iron Ore Reduction:
0
0
+4 -2
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
reduced species: * Fe+3
oxidized species: * C0
N.B. Ore: a general term referring to a metal ion-containing mineral, that may be trapped in a larger mixture known as, a rock.
Iron ore: deemed valuable for its oxidized form of iron: Fe+2 and Fe+3, bonded in a compound with reduced oxygen (oxide).
The reduction of an ore refers to converting the metal CATION back into the metal ATOM by having the ion GAIN electrons back
History Buffs: The Hall Process & later the Bessemer process are often seen as turning points of the Industrial Revolution in the West.
see: http://geo.msu.edu/extra/geogmich/steel_mill.html and a video at: https://www.youtube.com/watch?v=dypdoLm4Rn8
COOL CONCEPT!
(Purely Honors Chemistry)
Disproportionation
Reaction: a type of redox
reaction, in which the same
reactant species is both the
oxidized species and the
reduced species!!!
It’s like shooting yourself
in the foot … you are both
assailant and victim!
+1 -1
m) Challenge: Teeth Whitening:
reduced species: * O-1
+1 -2 +1
n) Challenge:
+1 -2
 2 H2O
2 H2O2
0
+ O2(g)
oxidized species: * O-1
0
2 NaOH + Cl2
reduced species: *Cl20 or Cl0

+1 -1
NaCl
+1 +1 -2
+
+1 -2
NaClO + H2O
oxidized species: *Cl20 or Cl0
Individual Practice: You must assign each oxidation number and then analyzed the change so that you can
identify the oxidized and reduced species of each reaction. You must include the oxidation state of the species
with each answer.
1)
2 O2 (g) +
CH4(g)

CO2(g) + 2 H2O(l)
reduced species: *O20 or O0
2)
Fe(s) + 2 Fe(NO3)3(aq) 
reduced species: *Fe+3 to the Fe+2
oxidized species: * C-4
3 Fe(NO3)2(aq)
oxidized species: *Fe0 to the Fe+2
(interesting n’est pas?)
420
3)
Mg (s) +
H2(SO4)(aq) 
reduced species: * H+1
4)
oxidized species: * Mg0
P4(s) + 6 H2(g) 
4 BCl3 (s) +
reduced species: *P40 or P0
5)
Cl2(g)
+ 2 NaI(s)
2 KClO3(s) 

2 NaCl (s)
2 SO2(g)
+
O2(g) 
+
2 HgO(s) 
2 Hg(l)
reduced species: *Hg+2
I2(s)
3 O2(g)
oxidized species: *O-2
SO3(g)
reduced species: *O20 or O0
8)
+
oxidized species: *I-1
2 KCl(s)
reduced species: *Cl+5
7)
4 BP (s) + 12 HCl(g)
oxidized species: *H20 or H0
reduced species: *Cl20 or Cl0
6)
MgSO4(aq) + H2(g)
oxidized species: *S+4 to the S+6
+
O2(g)
oxidized species: *O-2
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D) The agents: The agents are the opposite of their names. In short, the "agent" is that species whose
presence enables the activity for which it is named. ... HUH????
e.g.) The presence of the oxidizing agent allows oxidation to proceed, hence the
oxidizing agent is the reduced species.
1) Oxidizing Agent: * The reduced species
(a.k.a. an oxidizer or oxidant)
a) a strong oxidizing agent is reduced, easily (readily).
b) a weak oxidizing agent is reduced, but more slowly when compared to a stronger one.
2) Reducing Agent: * The oxidized species
(a.k.a. an antioxidant….Wait a minute … I have heard this term before …. Hey! )
3) Spectator ions: Not always present … but these are ions which do not change in oxidation
state.
TIME FOR A METAPHOR ... MAKE SOME CONNECTIONS!!
To be in agency … is to function upon the behalf of another ….
Think about a sports agent: Does the sports agent play the sport?
Does the travel agent take the arranged trip?
Does the insurance agent purchase the prepared policy?
So, is the oxidizing agent the oxidized species ????
Interpret: The N+5 of NaNO3 is a strong oxidizing agent (oxidizer)
*N+5 is easily / readily reduced (oxidizing agents or oxidizers are reduced species)
Interpret: The Mn+7 of KMnO4 is a strong oxidizing agent (oxidizer)
* Mn+7 is easily / readily reduced
Interpret: Sodium hypochlorite is a stronger oxidizing agent than 3% hydrogen peroxide
* Sodium hypochlorite is more easily reduced than hydrogen peroxide
422
E) A “real world” application: CLOROX bleach / Pool Chlorine
Recall that an oxidizing agent is any substance which causes another substance to * lose
electrons. Bleach and pool chlorine are pretty closely related mixtures.
one or more
The “chlorine” in both mixtures is NOT dichlorine gas (that would poison you upon removal of the cap!).
The “chlorine” is really the Cl+1 in the polyatomic ion of hypochlorite (ClO)-1. Think about the effective nuclear
charge on a nonmetal species, forced into a positive oxidation state! The polyatomic ion is probably
in the mixture, due to the dissolution of a compound like NaClO.
water
NaClO(s) 
Na+1(aq) + (ClO)-1(aq)
Specifically, the decolorizing action of bleaches is due in part to their ability to remove those electrons which are
activated by visible light to produce the various colors. The hypochlorite ion [(ClO)-1], found in many
commercial preparations, is reduced to chloride ion and hydroxide ion forming an alkaline (basic) solution as it
accepts electrons from the colored materials / dyes as shown below.
+1 -2
-1
(ClO)-1 + 2e- + H2O  Cl + 2(OH)-1
In comparison, “color safe bleaches” do not use ClO-1 … Rather, color safe bleaches, such as Oxiclean,
Clorox II, Borateem, or Snowy try to deliver dioxygen to the wash water. The key is the high affinity oxygen
has for electrons and the tendency to become readily reduced. The atoms of dye molecule tend to have a
sufficient hold on their electrons, while the electrons of some minor stains have a weaker hold.
The challenge is to keep the mixture components stable prior to use, but to have them react with wash water and
produce O2(g) on demand. Often, these are stabilized mixtures containing compounds like; sodium percarbonate,
or sodium tetraborate decahydrate (the mineral borax). These compounds decompose into hydrogen peroxide,
which in turn undergoes a disproportionation redox reaction and decomposes into water and oxygen gas.
Na2CO3·1.5H2O2  Na2CO3 + 1.5 H2O + 0.75O2
********************************************
http://bebrainfit.com/lifestyle/drains/zapped-your-brain-on-electromagnetic-fields/
423
PRACTICE
1) Given: F2(g) + 2KCl(aq)  2KF(aq) + Cl2(g)
a) Assign oxidation numbers to each species in the above reaction equation.
b) What species is oxidized? ______
(be sure to record the oxidation number of your answer, even when it is 0)
c) What species is reduced? _______
(be sure to record the oxidation number of your answer, even when it is 0)
d) What species is the oxidizing agent? ______
e) What species is the reducing agent? ______
(be sure to record the oxidation number of your answer, even when it is 0)
(be sure to record the oxidation number of your answer, even when it is 0)
f) How are the species of a redox reaction related to the “agents”?
2) Given: 2 Al(s) + 3 NiCl2(aq)  2 AlCl3(aq) + 3 Ni(s)
a) Assign oxidation numbers to each species in the above reaction equation.
b) What species is oxidized? ______
(be sure to record the oxidation number of your answer, even when it is 0)
c) What species is reduced? _______
(be sure to record the oxidation number of your answer, even when it is 0)
d) What species is the oxidizing agent? ______
e) What species is the reducing agent? ______
(be sure to record the oxidation number of your answer, even when it is 0)
(be sure to record the oxidation number of your answer, even when it is 0)
3) Given: 2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
a) What species is oxidized? ______
Defense: The oxidation state ________________
b) What species is reduced? ______
424
4) Given: Mg(s) + SiO2(s)  Si(s) + 2 MgO(s)
a) What species is the oxidizing agent? ______ Defense: _____________________________
b) What species is the reducing agent? ______
___5) Given: 2 N2(g) + 3 O2(g)  2 N2O3(g)
a)
N20
b) O20
c) N-3
0
6) Given:
Which species is the oxidizing agent?
d) O-2
+4
2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
From a chemical (or redox) point of view, what happens to the three moles of carbon as they react with
the rust? The 3 mol of C0 are _______________ Defense: __________________________________
_______ 7) What is the oxidation number of silicon in the compound H2SiF6(s)?
___8) In which of the following compounds, does oxygen have the most positive oxidation number?
a) NaOH
b) Ag2O
c) NaClO
d) OF2
9) What is meant, in chemical terms, by the phrase: “Fluorine gas (F2(g)) is a strong oxidizing agent.”?
__________________________________________________________________________
425
For question 10 a –d, use your knowledge of chemistry and the following passage, which is an adaptation from
McQuarrie & Rock Descriptive Chemistry 1985 p. 153) Numbers for the lines of print have been provided.
1
3
6
9
Copper is slightly less abundant than nickel and is found in many different ores. An
ore is a general term referring to the rocks that are really mixtures of a number of
valuable minerals (a.k.a. elements or compounds).
Copper generally occurs as various sulfides, although in some ores copper is present in
the form of sulfates, carbonates and other oxygen containing compounds. Deposits of
the free metal are very rare, being found only in Michigan. Most copper-containing
ores have a copper content of less than 1 percent, but some richer ores have up to
4 percent copper.
Copper ores contain other metals and metalloids such as selenium, and tellurium,
which are important by-products when the copper ore is reduced to copper metal.
Some important copper minerals (a.k.a. compounds of copper) are chalcocite, (Cu2S),
chalcopyrite (CuFeS2), and malachite (CuCO3Cu(OH)2), which is a crystalline
mixture of two compounds.
10) a) What is the oxidation number of copper in the mineral (compound) chalcocite (Cu2S) ?
________
b) In lines 5-6 you read: “Deposits of the free metal are very rare…” What does the author mean by the
term “free metal”? _____________________________ What is the oxidation state associated with
the atoms of the free metal copper? _____
c) In line 9 a reference is made to copper ore being reduced to copper metal. Chemically speaking,
what must happen to the metallic ions of an ore, in order to be reduced to the metal? _____________
____________________________________________________________________________________
d) What elements, other than metals and nonmetals may be obtained from refineing copper ores?
e) Calculate the percent composition of copper in malachite. ____________%
426
___11) Given the chemical equation :
KI + Pb(NO3)2  PbI2 + KNO3
you can state that
a) no changes in oxidation states occur
b) lead changes in oxidation state from Pb+2 to Pb0
c) iodide changes in oxidation state from I0
to I-1
d) potassium changes in oxidation state from K+1 to K0
___12) Given the chemical equation:
2 KCl + F2  2 KF + Cl2
The oxidation number of the fluorine atoms in F2 changes from:
a) -1
to -2
c) -1 to 0
b) -1
to +1
d) 0 to -1
____13) Which category is composed of elements that have both positive and negative oxidation states ?
a) the noble gases
b) the transition metals
c) the halogens
d) the alkaline-earth metals
___14) Which of the following reactions is classified as a redox reaction?
a) PCl5 + 4 H2O  H3PO4 + 5 HCl
b) Ca + 2 HNO3  Ca(NO3)2 + H2
c) H2SO4  SO3 + H2O
d) 3BaBr2 +
Al2(S2O3)3  3BaS2O3 + 2AlBr3
___15) Which of the following reactions is classified as a redox reaction?
a) 3 O2  2 O3
b) CaO + SiO2  CaSiO3
c) C2H5OH + 3 O2  2 CO2 + 3 H2O
d) Fe3(PO4)2 + 6 NaOH 
2 Na3PO4 + 3 Fe(OH)2
427
___16) In which of these compounds does Cl have the highest (most positive) oxidation number ?
a) NaClO
c) NaClO3
b) NaClO2
d) NaClO4
____17) What is the oxidation state of nitrogen in the polyatomic ion (NH4)+1?
___18) Given the equation:
4 BCl3 + P4 + 6 H2  4 BP + 12 HCl
a) No redox reaction occurs
c) Phosphorous (P4) is reduced
b) The boron in BCl3 is oxidized
d) Hydrogen is a spectator ion
For 19 – 22 place a checkmark next to the reactions classified as a redox reaction. There may be more
than 1 checked answer
___19)
Mg (s) + H2(SO4)(aq)  MgSO4(aq) + H2(g)
___20)
2 KrF2 (g) + 2 H2O(g)  2 Kr (g) + O2(g) + 4 HF(g)
___21)
4 BCl3 (s) + P4(s) + 6 H2(g)  4 BP (s) + 12 HCl(g)
___22)
2 NH4Cl(s) + Ba(OH)2(aq)  BaCl2(s) + 2 NH3(g) + 2 H2O(l)
A very interesting reaction, n’est pas?
…one of the few reactions with a noble gas
ANSWERS
1) Given: F2(g) + 2KCl(aq)  2KF(aq) + Cl2(g)
a) Assign oxidation numbers to each species
b) What species is oxidized? Cl1c) What species is reduced? F20
d) What species is the oxidizing agent? F20
e) What species is the reducing agent? Cl1f) Answers will vary … think about how you identify an oxidized species …vs. an oxidizing agent … We’ll share answers in class….
2) Given: 2 Al(s) + 3 NiCl2(aq)  2 AlCl3(aq) + 3 Ni(s)
a) Assign oxidation numbers to each species
b) What species is oxidized? Al0
c) What species is reduced? Ni2+
d) What species is the oxidizing agent? Ni2+
e) What species is the reducing agent? Al0
428
3) Given: 2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
a) What species is oxidized? Al0
b) What species is reduced? H1+
4) Given: Mg(s) + SiO2(s)  Si(s) + 2 MgO(s)
a) What species is the oxidizing agent? Si4+
b) What species is the reducing agent? Mg0
5) b) O20
6) From a chemical (or redox) point of view, what happens to the three moles of carbon as they react with
the rust? They are oxidized. Defense: The oxidation number of the C changes from 0 to +4
7) +4
8) d) OF2
9) It is easily / readily reduced
10) a) +1
b)
c)
d)
e)
The pure element
Zero (0)
They gain electrons
metalloids such as Se and Te
Calculate the percent composition of copper in malachite. 128 x 100 = 57.7%
222
11) a) no changes in oxidation states occur
12) d) 0 to -1
13) c) the halogens
14) b) Ca + 2 HNO3  Ca(NO3)2 + H2
15) c)
C2H5OH + 3 O2  2 CO2 + 3 H2O
16) d) NaClO4
17) N = -3
18) c) Phosphorous (P4) is reduced
__19)
Mg (s) + H2(SO4)(aq)  MgSO4(aq) + H2(g)
__20)
2 KrF2 (g) + 2 H2O(g)  2 Kr (g) + O2(g) + 4 HF(g)
__21)
4 BCl3 (s) + P4(s) + 6 H2(g)  4 BP (s) + 12 HCl(g)
Check this out!: For the more sophisticated of you ... and that's everyone !
Take a look at how redox is applied to advanced topics in biochemistry...
Electron Transport System and ATP Production http://highered.mcgrawhill.com/olcweb/cgi/pluginpop.cgi?it=swf::535::535::/sites/dl/free/0072437316/120071/bio11.swf::Electron%20Transport%20System
%20and%20ATP%20Synthesis
Youtube:
http://www.youtube.com/watch?v=kN5MtqAB_Yc
Youtube: http://www.youtube.com/watch?v=xbJ0nbzt5Kw
429
XI) Species of a redox reaction (REVIEW)
oxidized species
reduced species
loses electrons
its oxidation # becomes
more positive, due to the loss
of negative charge
gains electrons
so its oxidation # becomes
more negative, due to the
gain of negative charge
aka: reducing agent
aka: oxidizing agent
spectator ion
no change in oxidation number
A) When a redox reaction is investigated, it can be represented by the separate reduction reaction and
by the separate oxidation reaction. These “portions” are called half-reactions.
1) How to write a reduction half-reaction: * species + e-  new & more negative species
e.g) The reduction of phosphorus to P3-
The reduction of Cl7+ to Cl5+
2) How to write an oxidation half-reaction: * species  new (more positive) species + ee.g) The oxidation of copper to Cu2+
The oxidation of Fe2+ to Fe3+
3) Every oxidation half-reaction must be accompanied by a reduction half-reaction.
a) When the half reactions are balanced separately so that the laws of the conservation
of matter and the conservation of charge are obeyed, the * number of electrons gained
by the reduced species will be equal to the number of electrons lost by the oxidized
species.
430
essential
non essential
essential
non essential
Reduction ½ reaction
Oxidation ½ reaction
exemplar
non-exemplar
exemplar
non-exemplar
4) Complete each reaction by providing the correct number of lost/gained electrons or species
and label each as an oxidation ½ reaction or as a reduction ½ reaction.
Half-Reaction
Type of Half-Reaction
a. Fe0 
Fe+3
b. Cs0 
*Cs+1 + 1e-
c.
+ Se0 
* 2e-
d. Pb0
+
 4e-
e. *3e- + N0 
* 3e-

___________________
Se-2
___________________
+ * Pb+4
___________________
N-3
___________________
f. Mn+6 + * 3e-  Mn+3
g. Cl+5
___________________
2e- + * Cl+7
___________________
___________________
431
Notice, that the blanks are no longer provided. You must now figure out, by analyzing the charges of the
reactants and the products as to what must be "filled in" to complete the half-reaction.
Half-Reaction
Type of Half-Reaction
h.
Si0
 Si-4
__________________
i.
Mn0
 4e-
__________________
j
k
O0
 Ca+2 + 2e-
__________________
 O-2
__________________
Answers : a = oxidation, 3eb = oxidation, Cs+1
c = reduction, 2ed = oxidation, Pb+4 e = reduction, 3e- f=reduction, 3e- g= oxidation, Cl+7
+4
h = reduction, 4e , left side of arrow i= oxid., Mn right side of arrow
j= oxidation, Ca0 left side of arrow k= reduction, 2e- left side of arrow
5) The special difficulties of diatomic elements and writing half-reactions:

a) The reduction of F2 to F1-
* F2 + 2 e- → 2 F-1
stop
Before you go on,
1* Cl2 + 2 e- → 2 Cl-1
analyze the species ... b) The reduction of Cl2 to Cl
With what challenge
c) The reduction of Br2 to Br1must you deal due to
the law of the
conservation of matter
& writing halfd) The reduction of I2 to I1reactions for the
diatomic elements in a
half reaction?
2-
e) The reduction of O2 to O
* Br2 + 2 e- → 2 Br-1
* I2 + 2 e- → 2 I-1
* O2 + 4e- 
2 O-2
f) The reduction of N2 to N3- * N2 + 6 e- → 2 N-3
g) The oxidation of H2 to H1+ * H2 → 2 H1+ + 2eh) The oxidation of O2- to O2
* 2 O-2 → O2 + 4e-
i) The oxidation of Cl1- to Cl2
* 2 Cl-1
j) The reduction of H1+ to H2
*2 H1+ + 2 e- → H2
→ Cl2 + 2e-
432
XII) Balancing Simple Redox Reactions via Half-reactions
 Determine which species were reduced and oxidized (always from the reactant side)
 Write each half reaction
 Balance by mass (balance according to the number of ALL particles: really used for the diatomic elements)
 Balance by charge (obey Law of the Conservation of Charge: if required… multiply all species in one or both
½ rxn(s) by whole number values, in order to get the # electrons in both ½ rxns equal. This produces coefficients)
 Rewrite (recombine) the two half-reactions into1 reaction, using correct coefficients
A) Balance the given reaction :
___Cu2+ + ___Al0  ___Cu0
+ ___ Al3+
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION : ________________________________________________________
RECOMBINED:
* 3 Cu2+ + 2 Al0  3 Cu0 + 2 Al+3
The 2 half reactions are recombined, with the correct coefficients BUT excluding the electrons (for they have been "cancelled out") To recombine, you
could just put the coefficients on the spaces of the original equation.
B) Balance the given reaction :
___H1+
+
___Fe0

___Fe2+
+ ___H20
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION : ________________________________________________________
RECOMBINED : ___________________________________________________________________
C) Balance the given reaction :
___Pb2+ + ___Fe0

___ Pb0
+ ___Fe3+
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION : ________________________________________________________
RECOMBINED : ___________________________________________________________________
D) Balance the given reaction :
___Al0
+ ___Cr+3

___Cr0
+ ___Al3+
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION : ________________________________________________________
RECOMBINED : ___________________________________________________________________
433

E) Balance the given reaction: ___Au+3 + ___ H20
___ H+1
+ ___ Au0
REDUCTION ½ REACTION: ________________________________________________________
OXIDATION ½ REACTION: ________________________________________________________
RECOMBINED: ___________________________________________________________________
F) Balance the given reaction:
Au+3

+ K0
K+1
+ Au0
* Au+3 + 3 e- 
Au0
3(K0  K+1 + 1e-)
Au+3 + 3K0  3K+1
G) Balance the given reaction:
Cr0
+ Cu+  Cu0
+ Au0
+ Cr3+
* 3(Cu+ + 1e-  Cu0 )
Cr0  Cr3+ + 3eCr0 + 3Cu+  3Cu0 + Cr3+
H) Balance the given reaction:
Al+3
+
Ba0

Al0
+ Ba+2
* 2(Al+3 + 3e- 
Al0 )
3(Ba0  2e- + Ba+2 )
2 Al+3
I) Balance:
Ag+
+
Cu0 
Cu+2
*
+
3Ba0

2Al0
multiplying to get 6 e-
+ 3Ba+2
+ Ag0
2(Ag+ + 1 e-  Ag0)
Cu0  Cu+2 + 2e2Ag+
+
Cu0 
Cu+2 + 2 Ag0
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J) Balance:
Al0
+

H1+
Al3+
H20
+
* 3 (2 H1+ 2 e-  H20 )
* 2 (Al0

Al3+ + 3 e- )
+ 6 H1+  2 Al3+ + 3 H20
*2 Al0
K) When correctly balanced using the smallest whole-number ratios, what is the coefficient of H1+?
Sn+4
H20
+

Sn0 + H+1
* Sn+4
a) 1
b) 2
+
2 H20
c) 3

Sn0 + 4 H+1
d) 4
L) When correctly balanced using the smallest whole-number ratios, what is the coefficient of Hg0?
Br20
+
Hg0 
Br-1 + Hg+2
* Br20
a) 1
b) 2
c) 3
+
Hg0 
2 Br-1 + Hg+2
d) 4
M) Determining the moles of electrons in a redox reaction. When asked: How many moles of electrons
are lost / gained, How many moles of electrons are involved / exchanged just multiply the total
number of moles of a species by its oxidation state….
+3 -2
1) In the reaction: 4 Al + 3 O2  2 Al2O3
how many moles of electrons are lost/gained?
Use any species with an oxidation state other than 0. Multiply the moles of that species
by the oxidation number of the species: e.g. using Al+3: There are 4 mol of Al+3 and each
has an oxidation state of +3 … so (4)(3+) = 12 Note: You get the same answer for O-2
This is just an application of (BIG IDEA #1 the Law of the Conservation of Mass, Energy and Charge)
* 6 mol
2) In the reaction: 2 Al + 3 I2  2 AlI3(s) how many moles of electrons are lost by the
aluminum and gained by the iodine?
435
*16 mol
3) Assuming the reaction goes to completion: 8 Cu + S8  8 CuS how many moles of
electrons are gained by the sulfur? (which of course, must be the same number of mols of e- lost by the Cu0 using
the Law of the Conservation of Mass, Energy and Charge)
* 4 mol
*2 mol
4) Assuming the reaction goes to completion: 2 Mg + O2  2 MgO
electrons are lost by magnesium as it is oxidized to Mg2+?
how many moles of
5) Assuming the reaction goes to completion: Fe + CuSO4  Cu + FeSO4 how many moles of
electrons are exchanged between the reduced and oxidized species?
PRACTICE: Base your answers to questions 1 and 2 on the UNbalanced redox reaction below:
Cu(s) + Ag+1(aq) → Cu+2 + Ag(aq)
1) Write the reduction half-reaction. * Ag+1 + 1 e- → Ag OR 2 Ag+1 + 2 e- → 2Ag
2) Balance the redox equation using the smallest whole-number coefficients. * Cu(s) + 2 Ag+1(aq) → Cu+2 + 2 Ag(aq)
3)
4)
6)
7)
5)
436
8)
10)
9)
omit:
11)
12)
14)
13)
Given the equation Mg(s) + Cl2(g)  MgCl2(s) , which half-reaction correctly represents the
reduction that occurs?
(1) Mg(s) + 2e-  Mg+2
(3) Cl2(g) + 2e-  2Cl-1
(2) Mg(s)  Mg+2 + 2e-
(4) 2Cl-1 + 2e-  Cl2(g)
Answers:
1) Ag+1 + 1 e-  Ag0
3) 4
9) 1
4) 1
10) 2
5) 1
11) 2
2) Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag
6) 4
12) 2
7) 1
13) 2
8) 4
14) 3
O
437