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NAME ____________________________________ UNIT 8 (2): REDOX & ELECTROCHEMISTRY
Big Idea 3: Changes in matter involve the rearrangement and/or reorganization
of atoms and/or the transfer of electrons.
Enduring understanding 3.A: Chemical changes are represented Essential knowledge 3.A.1: A chemical change may be represented by a
by a balanced chemical equation that identifies the ratios with
molecular, ionic, or net ionic equation.
which reactants react and products form.
Essential knowledge 3.A.2: Quantitative information can be derived from
stoichiometric calculations that utilize the mole ratios from the balanced
chemical equations. The role of stoichiometry in real-world applications is
important to note, so that it does not seem to be simply an exercise done
only by chemists.
Enduring understanding 3.B: Chemical reactions can be
Essential knowledge 3.B.1: Synthesis reactions are those in which atoms
classified by considering what the reactants are, what the products and/or molecules combine to form a new compound. Decomposition is the
are, or how they change from one into the other. Classes of
reverse of synthesis, a process whereby molecules are decomposed, often by
chemical reactions include synthesis, decomposition, acid-base, the use of heat.
and oxidation-reduction reactions.
Essential knowledge 3.B.2: In a neutralization reaction, protons are
NOTE: Because of their prevalence in the laboratory and in transferred from an acid to a base.
real-world applications, two categories of reactions that are Essential knowledge 3.B.3: In oxidation-reduction (redox) reactions, there
of particular importance are acid-base reactions and
is a net transfer of electrons. The species that loses electrons is oxidized, and
oxidation- reduction reactions.
the species that gains electrons is reduced.
Enduring understanding 3.C: Chemical and physical
transformations may be observed in several ways and typically
involve a change in energy.
Essential knowledge 3.C.1: Production of heat or light, formation of a
gas, and formation of a precipitate and/ or a color change are possible
evidences that a chemical change has occurred.
Essential knowledge 3.C.2: Net changes in energy for a chemical reaction
can be endothermic or exothermic.
Essential knowledge 3.C.3: Electrochemistry shows the interconversion
between chemical and electrical energy in galvanic and electrolytic cells.
Structured Overview for our work on Electrochemistry
review
Electrochemistry
review
Oxidation
and
Reduction
Voltaic
Assignment of
oxidation states
(numbers)
Simple
HalfReaction
Method
Stoichiometry
Balancing a
redox reaction
HalfReaction
Method
in acidic
solutions
Electrolytic
Half
Reaction
Method
in basic
solutions
572
Learning objective 3.1 Students can translate among macroscopic observations of change, chemical equations, and particle views. [See SP
1.5, 7.1; Essential knowledge components of 3.A–3.C]
Learning objective 3.2 The student can translate an observed chemical change into a balanced chemical equation and justify the choice of
equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances. [See SP 1.5, 7.1; Essential knowledge 3.A.1]
Learning objective 3.3 The student is able to use stoichiometric calculations to predict the results of performing a reaction in the
laboratory and/or to analyze deviations from the expected results. [See SP 2.2, 5.1; Essential knowledge 3.A.2]
Learning objective 3.4 The student is able to relate quantities (measured mass of substances, volumes of solutions, or volumes and
pressures of gases) to identify stoichiometric relationships for a reaction, including situations involving limiting reactants and situations in
which the reaction has not gone to completion. [See SP 2.2, 5.1, 6.4; Essential knowledge 3.A.2]
Learning objective 3.5 The student is able to design a plan in order to collect data on the synthesis or decomposition of a compound to
confirm the conservation of matter and the law of definite proportions. [See SP 2.1, 4.2, 6.4; Essential knowledge 3.B.1]
Learning objective 3.6 The student is able to use data from synthesis or decomposition of a compound to confirm the conservation of
matter and the law of definite proportions. [See SP 2.2, 6.1; Essential knowledge 3.B.1]
Learning objective 3.7 The student is able to identify compounds as Brønsted-Lowry acids, bases, and/or conjugate acid-base pairs, using
proton-transfer reactions to justify the identification. [See SP 6.1; Essential knowledge 3.B.2]
Learning objective 3.8 The student is able to identify redox reactions and justify the identification in terms of electron transfer. [See SP
6.1; Essential knowledge 3.B.3]
Learning objective 3.9 The student is able to design and/or interpret the results of an experiment involving a redox titration. [See SP
4.2, 5.1; Essential knowledge 3.B.3]
Learning objective 3.10 The student is able to evaluate the classification of a process as a physical change, chemical change, or
ambiguous change based on both macroscopic observations and the distinction between rearrangement of covalent interactions and
noncovalent interactions. [See SP 1.4, 6.1; Essential knowledge 3.C.1, connects to 5.D.2]
Learning objective 3.11 The student is able to interpret observations regarding macroscopic energy changes associated with a reaction or
process to generate a relevant symbolic and/or graphical representation of the energy changes. [See SP 1.5, 4.4; Essential knowledge
3.C.2]
Learning objective 3.12 The student can make qualitative or quantitative predictions about galvanic or electrolytic reactions based on
half-cell reactions and potentials and/or Faraday’s laws. [See SP 2.2, 2.3, 6.4; Essential knowledge 3.C.3]
Learning objective 3.13 The student can analyze data regarding galvanic or electrolytic cells to identify properties of the underlying
redox reactions. [See SP 5.1; Essential knowledge 3.C.3]
I) Redox Reaction: Any reaction in which there is a change in oxidation number(s) (LEO says GER)
REDuction / OXidation reaction: Electrons are * lost
of * oxidation states (or numbers)
and * gained
resulting in a change
generally for two species. 
any chemical entity: a(n) molecule, atom or ion
A) For every oxidation there must be a reduction. (Law of the Conservation of Charge)
1) Often a redox reaction has a free element as a reactant, which ends up in a compound or
a compound breaking down into (an) element(s). BUT NOT ALWAYS!!!! The surest means
to know is to assign oxidation states to each species ….and to look for a change.
573
B) Oxidation State (a.k.a: * Oxidation Number
): An arbitrarily assigned value which
explains /accounts for/ predicts the number of electrons of a species, involved in making a bond with
a species of a different element. It is NOT a statement of a physical state … just a means of electron
bookkeeping.
1) + or – values for oxidation states apply to species of compounds, or of ions in water.
0 is the oxidation state for pure elements. [for instance: the oxidation state(s) of the oxygen atoms in a
molecule of O2 is 0, since the molecule is produced by species of the same (not different) element(s)]
2) The charge on an ion is only ONE category of oxidation states. Every encounter with an
oxidation state (number), is NOT necessarily an encounter with an ion
a) Oxidation states can be applied to species of a molecule as well. They are used to
describe the number *of shared electrons in a bond between the atoms of
a covalent (molecular) substance.
3) An oxidation number of a species may be * positive
more strongly to another nucleus
or * negative
if its electrons are attracted
if the involved electrons tend
to be gained, relatively speaking.
a) Be aware that classically metals in a compound are assigned a + oxidation state while
nonmetals of a compound may be assigned + or -. Think about this statement and the
existence of nonmetal-nonmetal molecules. In most there must be a + and – species
4) In molecules, the assignment of oxidation state to a species in a compound is closely linked to
the electronegativity values of the bonded species. The electronegativity of a species
reflects the ability of a species to attract bonding electrons, due to the effective nuclear
charge
a) When dealing with inorganic ionic compounds, and binary molecular compounds
(for the most part) the species with the *more positive oxidation # (the species of lower
electronegativity) is written first in a formula
574
C) There only appear be multiple definitions of oxidation and reduction. When analyzed,
they all have a common thread of thought …
What is that common thread?
The Oxidized species of a reaction is
the reactant that:
The Reduced species of a reaction is
the reactant that:
becomes more positive (increases) in
oxidation number
becomes more negative (decreases) in
oxidation number
0
0
+1 -2
16Na + S8 → 8Na2S
completely loses e- (as in an ionic bond)
or experiences a shift of e- away from
the parent nucleus in a covalent bond,
thus becoming a more positive species
0
0
+1 -1
H2 + Cl2 → 2HCl
0
0
+1 -2
16Na + S8 → 8Na2S
completely gains e- (as in an ionic bond)
or experiences a shift of e- towards
itself in a covalent bond, thus
becoming more negative species
0
0
Inorganic /
reaction
chemistry
+1 -1
H2 + Cl2 → 2HCl
the reducing agent (or antioxidant)
the oxidizing agent (or oxidant)
loses hydrogen
gains hydrogen
e.g. alcohol oxidized to an aldehyde
CH3CH2OH → CH3COH
gains oxygen bonds with oxygen
e.g. alcohol oxidized to a carboxylic acid
CH3CH2OH → CH3COOH
The oxidation state of the C in the COOH
group is MORE positive than it was in the
C-O-H group
By increasing the number of highly
electronegative O, more electrons were drawn
away from that C, making it more positive.
loses oxygen
Biological
&/or
Organic
reactions
e.g. Krebs's
cycle
II) Rules for assigning oxidation states
Oxidation states may be whole numbers or fractional values. (remember, they are not necessarily a
statement of physical state … but a bookkeeping process.
For instance, according to Wikipedia, oxygen … which is just about as sloppy as an element can get in
terms of this idea can be:
575
-2 in metal and nonmetal oxides
-1 in peroxides
-½ in superoxides
-⅓ in inorganic ozonides
0 as a pure element
+½ in dioygenyl
+1 in fluorides
+2 in fluorides
e.g.
e.g.
e.g.
e.g.
e.g.
e.g.
e.g.
e.g.
ZnO, CO2
H2O2
KO2
RbO3
O2
O2+ [AsF6]− dioxygenyl hexafluoroarsenate
O2F2
OF2
Additionally, bizarre values for some metals exist. Silver can (for a very short period of time exist
with an assigned oxidation state of +2. The species in this case is used as an oxidizing agent (it is
reduced) to detoxify nerve gas reagents. We will NOT be getting this technical. So, K.I.S.S.S.
Keep It Simple Science Students. Just be aware of the phenomenal variation outside of these walls.
Rule #1: The sum of the oxidation states for a *compound must add up to equal 0
Rule #2: The oxidation number of an element in the elemental state is zero.
Rule #3: In a covalent compound of two elements, the more electronegative element is given a
negative oxidation number and the other element is given a positive oxidation number.
Rule #4: The oxidation number of oxygen in most of its compounds is -2.
Exceptions: in peroxides, the oxidation number of oxygen is -1.
in a binary compound with fluorine the oxidation number may be +1 or +2
Rule #5: The oxidation number of hydrogen in most of its compounds is +1,
Exception: in hydrides, the oxidation number of hydrogen is -1.
Rule #7: In aqueous solution, the charge of a monatomic ion is the oxidation state.
Rule #8: In a polyatomic ion, the sum of the positive and negative oxidation numbers of all atoms in
the formula must equal the charge on the ion.
Rule #9: (Sort of a rule): The oxidation number of a transition metal ion found in a polyatomic ion is a
constant for that polyatomic ion
+1
Proof:
x
-2
Na2Cr2O7
Make your life easier:
-3 +1
CaCr2O7
Al2(Cr2O7)3
The N of (NO3)-1
= * +5
The S of (SO4)-2
= * +6
The P of (PO4)-3
= * +5
x
-2
(NH4)2Cr2O7
576
III) There are a number of different types of redox reactions. The one thing they have in common is
a transfer in electrons (a loss and gain) resulting in a change in oxidation states of species.
A) Redox reactions include but are not limited to:
1) combustion reactions
2) reactions between acids and metals
3) many (but not all) syntheses
4) many (but not all) decomposition reactions
5) All single replacement reactions
6) A host of organic reactions in which O is lost/gained or in which H is lost/gained, which
generally indicates the making or breaking of a double (triple) bond … affecting the
electronic bookkeeping.
PRACTICE 1: Calculate and assign the oxidation number
Answers are on the next page…
Review: 1) Assign "x" to the element you are investigating and known oxidation states to all other species.
2) Multiply every oxidation number by the appropriate subscript to create an equation and set the equation equal to zero (for compounds)
OR equal to the charge for polyatomic ions.
3) Arithmetically solve for "x" by multiplying subscripts by oxidation # and set it all = to 0
a)
S in
Na2SO4
b)
Cl in LiClO4
c)
S in CaS2O3
d)
N in
e)
Mn in KMnO4
f)
Cr in MgCrO4
g)
O in OF2
h)
H in
i)
Cl in LiClO3
KNO3
CaH2
577
j)
Fe in FeCl2
Answers:
a) +6
k)
N in Ca(NO3)2
b) +7
c) +2
d) +5
l)
S in Al2(SO3)3
m)
S in Al2(SO4)3
n)
Cu in CuO
o)
Cr in CrCl3
p)
Cr in Na2Cr2O7
e) +7
f) +6
g) +2
h) -1
i) +5
j) +2
k) +5
l) +4
m) +6
n) +2
q)
o) +3
O in H2O2
p) +6
q) -1 …it’s a
peroxide
PRACTICE 2: Solve for the element in question. The sum, in this case is NOT equal to zero, but to the charge of the
polyatomic ion itself.
a) N in
b) S in (SO4)
e.g.:
-1
(NO3)
x -2
(PO4)-3
x – 8 = -3 (not 0, because it is an ION)
-2
x = +5
c) S in (S2O3)
-2
d) Mn in (MnO4)
-1
ans: a) +5, b) +6 c) +2 d) +7 e) +3
e) C in (C2O4)-2
578
PRACTICE 3: For 1-13 determine the oxidized and reduced species. When asked, identify the reducing agent
(antioxidant) and oxidizing agent (*NB: reducing and oxidizing agents are NOT on your AP exam)
1)

2 O2 (g) +

CH4(g)
CO2(g) + 2 H2O(l)
reduced species: *O20 or O0
oxidized species: * C-4 or CH4
reducing agent: *C-4 or CH4
oxidizing agent: * O2 or O0 or O20
FYI: The terms reducing agent and oxidizing agent are NOT included on the AP Exam
2)
Fe(s) + 2 Fe(NO3)3(aq) 
3 Fe(NO3)2(aq)
reduced species: *Fe+3 to the Fe+2
3)
Mg (s) +
H2(SO4)(aq) 
reduced species: * H+1
4)
5)
4 BCl3 (s) +
MgSO4(aq) + H2(g)
oxidized species: * Mg0
P4(s) + 6 H2(g) 
4 BP (s) + 12 HCl(g)
reduced species: *P40 or P0
oxidized species: *H20 or H0
reducing agent: * H20 or H0
oxidizing agent: * P40 or P0
Cl2(g)
+ 2 NaI(s)

2 NaCl (s)
reduced species: *Cl20 or Cl0
6)
oxidized species: *Fe0 to the Fe+2
2 KClO3(s) 
2 KCl(s)
+
+
I2(s)
oxidized species: *I-1
3 O2(g)
reduced species: *Cl+5
oxidized species: *O-2
reducing agent: * O-2
oxidizing agent: * Cl+5
579
7)
2 SO2(g)
+
O2(g) 
SO3(g)
reduced species: *O20 or O0
8)
9)
2 HgO(s) 
2 Hg(l)
+
O2(g)
reduced species: *Hg+2
oxidized species: *O-2
reducing agent: * O-2
oxidizing agent: * Hg+2
2 H2O2
 2 H2O
+ O2(g)
reduced species: * O-1
10)
oxidized species: *S+4 to the S+6
2 NaOH + Cl2

oxidized species: * O-1
NaCl
+
NaClO + H2O
reduced species: * Cl0 or Cl2
oxidized species: * Cl0 or Cl2
Interesting note for 9 and 10: These are called disproportionation reactions *which is a type of redox
reaction, in which the same reactant species is both the oxidized species and the reduced species!!
11)
12)
Ca3(PO4)2 + 8 C  Ca3P2 + 8 CO
reduced species: *P+5
oxidized species: *C0
reducing agent: * C0
oxidizing agent: * P+5
2 KMnO4 + 5 H2C2O4 + 3 H2SO4  2 MnSO4 + 10 CO2 + 8 H2O + K2SO4
(notice there is no free element … but it is a redox reaction)
reduced species: *Mn+7
oxidized species: *C+3
13) NH4NO3  N2O + 2 H2O
reduced species: *N+5
oxidized species: *N-3
580
14) 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)
15) Interpret: The Mn+7 of KMnO4 is a strong oxidizing agent (oxidizer)
* Mn+7 is easily / readily reduced
16) Interpret: Sodium hypochlorite (NaClO) is a stronger oxidizing agent than a solution that is
3% hydrogen peroxide (H2O2)
* Sodium hypochlorite is more easily reduced than hydrogen peroxide
15) Is the reaction between hydrogen peroxide and ethene a redox reaction? Defend your answer.
H2O2 + C2H4 → C2H4O + H2O
* Yes A redox reaction must have an oxidation and reduction. (A & S), which are shown by changes in
the oxidation states of the reactants, compared to the products (S). The oxygen of hydrogen peroxide is
in a -1 oxidation state. The oxygen atom of the water molecule is in the -2 oxidation state. The
peroxide was reduced … thus there must be an oxidation. The oxidized species is the C of the alkene.
By studying the reaction we can see that the ethane “gained” oxygen, producing C2H4O. When an
organic compound gains oxygen it is said to be oxidized, because the highly electronegative O draws
electrons away from the C to which it is bonded… thus making that C atom more positive in oxidation
state.(P)
581
*3____16) Which category is composed of elements that have both positive and negative oxidation states?
1) the noble gases
2) the transition metals
3) the halogens
4) the alkaline-earth metals
*2____17) Which of the following reactions is classified as a redox reaction?
1)
2)
3)
4)
PCl5 + 4 H2O → H3PO4 + 5 HCl
Ca + 2 HNO3 → Ca(NO3)2 + H2
H2SO4 → SO3 + H2O
3BaBr2 + Al2(S2O3)3 → 3BaS2O3 + 2AlBr3
*3____18) Which of the following reactions is classified as a redox reaction?
1)
2)
3)
4)
3 O2 → 2 O3
CaO + SiO2 → CaSiO3
C2H5OH + 3 O2 → 2 CO2 + 3 H2O
Fe3(PO4)2 + 6 NaOH → 2 Na3PO4 + 3 Fe(OH)2
*4_____19) What is the oxidation state of nitrogen in the polyatomic ion (NH4)+1?
1) +1
2) -1
*3_____20) Given the equation:
3) +3
4) -3
4 BCl3 + P4 + 6 H2  4 BP + 12 HCl
1) No redox reaction occurs
2) The boron in BCl3 is oxidized
3) Phosphorous (P4) is reduced
4) Hydrogen is a spectator ion
For 21-23 place a checkmark next to the reactions classified as a redox reaction. There may be more
than 1 checked answer
* ___ 21)
Mg + H2SO4  MgSO4 + H2
* ___ 22)
2 KrF2 + 2 H2O  2 Kr + O2 + 4 HF
* no___23)
2 NH4Cl + Ba(OH)2  BaCl2 + 2 NH3 + 2 H2O
582
*4_____24) What is the most correct name, using the IUPAC rules for the compound: MnO2
1) manganese (II) oxide
2) manganese (VI) oxide
3) manganese oxide
4) manganese (IV) oxide
And, do you remember these, as Bonus Purgatorio problems from Honors Chemistry???
25) The mineral crocidolite, is a form of asbestos. It has the formula, Na2Fe5(Si4O11)2(OH)2 . The mineral has
ions of Fe+2and Fe+3. How many of the iron ions must be a +2 and how many must be a +3?
*([3] are +2 and [2] are +3) (Descriptive Inorganic Chemistry)
26) Ultramarine is a beautiful blue pigment used in oil-based paints. It has the ability to maintain its depth of
color because it resists oxidation (and thus color change) in bright light. It was more valuable than gold
during the Renaissance!
Known since 7000 B.C, in Afghanistan, ultramarine comes from the mineral lazurite (found in lapis
lazuli, and named for the Persian word for blue). Cufflinks, necklaces from today to those found in the
jewelry and burial masks of the Egyptian Pharaohs are made with lapis lazuli. Some of the “blue”
component of ultramarine of lapis is due to species containing sulfur, as represented here, with the
incomplete formula of: Nax[Al6Si6O24]S2. In this formulation the silicon is in its highest oxidation state,
and the sulfur is present as the disulfide polyatomic ion, (S2)-2. What is the value of “x”?
ans: *(8) (Descriptive Inorganic Chemistry)
583
IV) Balancing Redox (Reduction – Oxidation) Reactions
simple half-reaction method
in acidic solution
in alkaline (basic) soluiton
A) Half-Reaction Method
1) First a review of writing half-reaction:
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.
* 2e-
d. Pb0
+
+ Se0 
 4e-
e. *3e- + N0 
* 3e-
*oxidation
*oxidation
Se-2
*reduction
+ * Pb+4
*oxidation
N-3
*reduction
f. Mn+6 + * 3e-  Mn+3
*reduction
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
h = reduction, 4e-, left side of arrow i= oxid., Mn+4 right side of arrow
j= oxidation, Ca0 left side of arrow k= reduction, 2e- left side of arrow
584
2) 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,
analyze the species ...
b) The reduction of Cl2 to Cl1- * Cl2 + 2 e- → 2 Cl-1
With what challenge
must you deal due to
c) The reduction of Br2 to Br1the law of the
conservation of matter
& writing halfd) The reduction of I2 to I1reactions for the
diatomic elements in a
half reaction?
e) The reduction of O2 to O2-
* 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-
3) 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
PRACTICE
1) Balance the given reaction:
___Cu+2 + ___Al0  ___ Cu0 + ___ Al+3
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.
585
2) Balance the given reaction :
___H+1 + ___Fe0

___Fe+2
REDUCTION ½ REACTION:
*2 H+1 + 2e-  H20
OXIDATION ½ REACTION :
* Fe0  2e- + Fe+2
* 2H+1 + Fe0
RECOMBINED :
3) Balance the given reaction :

___Pb+2 + ___Fe0
Fe+2

*Pb+2 + 2e- → Pb0
OXIDATION ½ REACTION:
* Fe0  Fe+3 + 3e-
* 3 Pb+2 + 2 Fe0
4) Balance the given reaction: ___Au+3
RECOMBINED:
*H20
*2 Au+3
5) Balance the given reaction:
3 Pb0
+ ___ H20 
REDUCTION ½ REACTION: *Au+3
OXIDATION ½ REACTION:

+ H20
___ Pb0 + ___Fe+3
REDUCTION ½ REACTION:
RECOMBINED:
Au+3 + K0 
3K+1
2Fe0  2 Fe+3 + 6e-
which becomes
___ H+1
+ ___ Au0
which becomes
 2 H+1 + 2 e6 H+1
3 Pb+2 + 6e- → 3 Pb0
which becomes
+ 2 Fe+3
+ 3e-  Au0
+ 3 H20 
+ ___H20
2 Au+3
which becomes
3 H20
+ 6e-  2 Au0
 6 H+1 + 6 e-
+ 2 Au0
+ Au0
* Au+3 + 3 e- 
Au0
3(K0  K+1 + 1e-)
Au+3 + 3K0  3K+1 + Au0
586
+ Cu+  Cu0
Cr0
6) Balance the given reaction:
+ Cr3+
* 3(Cu+ + 1e-  Cu0 )
Cr0  Cr3+ + 3eCr0 + 3Cu+  Cu0
Al3+
7) Balance the given reaction:
+
Ba0

Al0
+ Ba2+
* 2(Al+3 + 3e- 
Al0 )
3(Ba0  2e- + Ba+2 )
2 Al+3
8) Balance:
Ag+
+
Cu0 
+
+
H+


2Al0
+ 3Ba+2
2(Ag+ + 1 e-  Ag0)
Cu0  Cu+2 + 2e2Ag+
Al0
3Ba0
multiplying to get 6 e-
Cu2+ + Ag0
*
9) Balance:
+ 3Cr3+
Al3+
+
+
Cu0 
Cu+2 + 2 Ag0
H20
* 3 (2 H1+ 2 e-  H20 )
* 2 (Al0

Al3+ + 3 e- )
*2 Al0
+ 6 H1+  2 Al3+ + 3 H20
10) 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
c) 3
+
2 H20

Sn0 + 4 H+1
d) 4
587
11) 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
+
Hg0 
c) 3
2 Br-1 + Hg+2
d) 4
A BIT MORE COMPLEX … BUT ESSENTIALLY THE SAME TECHNIQUE
DIRECTIONS: Balance by half-reactions. Then apply (substitute in) the coefficients appropriately to the net
reaction, or the “whole” reaction and complete the task by balancing by inspection when needed.
12
13)
1-
* 2 (MnO4)
+
5
H2 S
Red. 1/2 rxn.
*Mn7+ + 5 ,e-
Oxid 1/2 rxn.
* S2-

1+
6 H 
+

S0 + 2 e-
Mn2+
2+
2 Mn
+
5 S +
which becomes
8 H2O
* 2 Mn7+ + 10 e-  2 Mn2+
which becomes * 5 S2-  5 S0 + 10 e-
* 2 (NO3)1- + 2 H+1 + 3 (PO3)3-  2 NO + 3 (PO4)3- + 1 H2O
Red. 1/2 rxn. *N5+ + 3 e-  N2+
*which becomes 2 N5+ + 6 e-  2 N2+
Oxid 1/2 rxn. *P3+  P5+ + 2e-
*which becomes 3 P3+  3 P5+ + 6e-
588
B) Half-Reaction Method in Acidic Solution
1) General Procedure … Read this over carefully … there are key ideas which, when heeded, really help you to succeed
at this process. I have provided you with one of my mnemonics … You won’t find it in the
anywhere (I just made it up), but it helps me to recall the process. Use it as you see fit.
a) Assign oxidation states and identify the oxidized and reduced species
b) Separate the overall reaction in the two halves: the oxidation and reduction … but
don’t include the electrons (yet)
c) Balance each of the halves in the following order:
key ideas
key idea
Mnemonic:
i) Balance elements (Matter) other than H and O
ii) Balance O by adding H2O
iii) Balance H by adding H+ (as in adding acid)
You want MOHE
Matter
Oxygen
Hydrogen
Electrons
d) Balance each half with respect to charge by ADDING
ELECTRONS so that the sum of the charges on BOTH sides
of the equation equal each other. You should add as many e- as necessary.
e) Equate the number of electrons in both half-reactions, by multiplying by the
smallest whole number(s)
f) Recombine the two half-reactions cancelling electrons & other species as is necessary.
2) Balance the redox equation in acidic solution: Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
+2
+7
-2
+3
+2
a) Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq)
oxid
Note: You may find this process easier
taking one taking one ½ rxn at a time
red
b) oxidation:
reduction:
Fe2+(aq) → Fe3+(aq)
MnO4-(aq) → Mn2+(aq)
c) oxidation:
Fe2+(aq) → Fe3+(aq)
reduction:
MnO4-(aq)
2+
→ Mn
(aq)
There are no H or O to balance …so we’ll
move to the next ½ rxn.
+ 4 H2O(ℓ)
Note, 4 H2O molecules were added
to balance the 4 O-2 of the
permanganate ion
now address the change of in the number of H due to the 4 H2O molecules
oxidation:
Fe2+(aq) → Fe3+(aq)
reduction:
8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(ℓ)
589
Fe2+(aq) → Fe3+(aq) + 1 e-
d) oxidation:
Note, 1 e- was added to balance the sum of the
charge on both sides
8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(ℓ)
reduction:
there is a total of +7 charge
+8 + (-1)
there is a total of a +2 charge
balance the charge of the reduction half-reaction by adding 5 e- to the
reactant side, so as to equate the charge on both sides of the reaction
oxidation:
Fe2+(aq) → Fe3+(aq) + 1 e-
reduction:
5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(ℓ)
5 Fe2+(aq) → 5 Fe3+(aq) + 5 e-
e) oxidation:
reduction:
5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(ℓ)
Thus….
oxidation:
5 Fe2+(aq) → 5 Fe3+(aq) + 5 e-
reduction:
5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(ℓ)
f) Recombined:
TRY THIS!
8 H+(aq) + 5 Fe2+(aq) + MnO4-(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(ℓ)
Balance the redox equation in acidic solution: Cu(s) + NO3-(aq) → Cu2+ (aq) + NO2 (g)
Tro p. 863
*0
+5 -2
+2
+4 -2
a) Assign and Identify: Cu(s) + NO3-(aq) → Cu2+(aq) + NO2(g)
* oxid
red
b) Separate out the halves & Balance most matter:
Oxidation: *Cu(s) → Cu2+ (aq)
Reduction: *NO3-(aq) → NO2 (g)
c) Balance O and H
Oxidation: *Cu(s) → Cu2+ (aq)
Reduction: * 2 H+ + NO3-(aq) → NO2 (g) + H2O (ℓ)
d) Balance electrons
Oxidation: *Cu(s) → Cu2+ (aq) + 2eReduction: * 1e- + 2 H+ + NO3-(aq) → NO2 (g) + H2O (ℓ)
e) Balance the 2 half-reactions
Oxidation: *Cu(s) → Cu2+ (aq) + 2eReduction: * 2e- + 4 H+ + 2 NO3-(aq) → 2 NO2 (g) + 2 H2O (ℓ)
f) Recombine: * 4 H+ + Cu(s) + 2 NO3-(aq) → Cu2+ (aq) + 2 NO2 (g) + 2 H2O (ℓ)
590
TRY THIS!
Balance the redox equation in acidic solution: N2H4 (g) + BrO3-(aq) → Br-(aq) + N2 (g)
ans: 3 N2H4(g) + 2BrO3-(aq) → 2 Br-(aq) + 3 N2(g) + 6 H2O(ℓ)
a) Assign and Identify:
oxid. states & species
* -2 +1
+5 -2
-1
0
N2H4 (g) + BrO3 (aq) → Br (aq) + N2 (g)
* oxid
red
b) Separate out the halves & Balance most matter:
Oxidation: *N2H4 (g) → N2 (g)
Reduction: *BrO3-(aq) → Br-(aq)
c) Balance O and H
Oxidation: *N2H4 (g) → N2 (g) + 4 H+ (aq)
Reduction: *6 H+ (aq) + BrO3-(aq) → Br-(aq) + 3 H2O (ℓ)
d) Balance electrons
Oxidation: *N2H4 (g) → N2 (g) + 4 H+ (aq) + 4eReduction: *6 e- + 6 H+ (aq) + BrO3-(aq) → Br-(aq) + 3 H2O (ℓ)
e) Balance the 2 half-reactions
Oxidation: *3N2H4 (g) → 3N2 (g) + 12 H+ (aq) + 12eReduction: *12 e- + 12 H+ (aq) + 2 BrO3-(aq) → 2 Br-(aq) + 6 H2O (ℓ)
f) Recombine: *3 N2H4 (g) + 2BrO3-(aq) → 2 Br-(aq) + 3 N2 (g) + 6 H2O (ℓ)
591
TRY THIS! When the following is balanced correctly, in acidic solution, using the simplest whole number
Brown / LeMay p. 862
coefficients, how many moles of water molecules are there in the balanced equation?
Cr2O72- (aq) + Cl-(aq) → Cr3+ (aq) + Cl2 (g)
ans: choice 4
1) 4 on the reactant side
2) 3 on the product side
a) Assign and Identify:
oxid states & species
* +6 -2
Cr2O72- (aq)
* red
3) 1 on the reactant side
4) 7 on the product side
-1
-
+3
3+
+ Cl (aq) → Cr
(aq)
0
+ Cl2 (g)
oxid
b) Separate out the halves & Balance most matter:
Oxidation: * 2 Cl-(aq) → Cl2 (g)
Reduction: * Cr2O72- (aq) → 2 Cr3+ (aq)
c) Balance O and H
Oxidation: * 2 Cl-(aq) → Cl2 (g)
Reduction: * 14 H+ (aq) + Cr2O72- (aq) → 2 Cr3+ (aq) + 7 H2O (ℓ)
d) Balance electrons
Oxidation: * 2 Cl-(aq) → Cl2 (g) + 2 eReduction: * 6e - +14 H+ (aq) + Cr2O72- (aq) → 2 Cr3+ (aq) + 7 H2O (ℓ)
Beware: 2 x +3 = +6
e) Balance the 2 half-reactions
Oxidation: * 6 Cl-(aq) → 3Cl2(g) + 6 eReduction: * 6e - +14 H+ (aq) + Cr2O72- (aq) → 2 Cr3+ (aq) + 7 H2O (ℓ)
f) Recombine: * 14 H+ (aq) + Cr2O72- (aq) + 6 Cl-(aq) → 2 Cr3+ (aq) + 3 Cl2 (g) + 7 H2O (ℓ)
592
C) Half-Reaction Method in Basic Solution
1) Were a redox reaction to occur in an alkaline solution, it makes sense (hopefully) to balance
using H2O and OH- (instead of H+) … However, there are some special problems with this,
since O and H are found in both water and OH-.
a) Thus, an alternate approach is to first balance the half-reactions as if they were in
an acidic solution ….COUNT the number of H+ in each half-reaction, AND THEN
add the same number of OH- to each side of the half-reaction.
b) This maintains a mass-balance, since we add to both sides. … You are essentially
neutralizing the H+ by adding the OH-, forming water on the side with H+ and OHon the other side. The water molecules can be cancelled, as needed.
2) General Procedure
a) Assign oxidation states and identify the oxidized and reduced species
b) Separate the overall reaction in the two halves: the oxidation and reduction … but
don’t include the electrons (yet)
c) Balance each of the halves as though it were in an acidic solution (MOHE)
i) Balance all elements (Matter) OTHER THAN H and O
ii) Balance O by adding H2O
iii) Balance H by adding H+ (as in adding acid)
d) Balance each half with respect to charge by ADDING ELECTRONS so that
the sum of the charges on BOTH sides of the equation equal each other. You
should add as many e- as necessary.
New key idea
e) Neutralize the added H+ with OH-. Add the same number of OH- to both sides of the
equations … AND CANCEL OUT WATER MOLECULES produced.
f) Equate the number of electrons in both half-reactions, by multiplying by the
smallest whole number(s)
g) Recombine the two half-reactions cancelling electrons & other species as is necessary
e.g.) Balance this equation for a redox reaction that takes place in a basic solution
CN-(aq) + MnO4-(aq) → CNO-(aq) + MnO2(s)
(Brown & LeMay p. 864)
Oxidation: CN-(aq) → CNO-(aq)
Reduction: MnO4-(aq) → MnO2(s)
Question: I didn’t assign oxidation states … how did I know the oxidation half-reaction?
* I noticed oxygen being bonded to the cyanide species … and that is one of the
definitions of oxidation. I can assume, there must have been a decrease in the
oxidation state of C
593
c)
Oxidation: H2O (ℓ) + CN-(aq) → CNO-(aq) + 2 H+ (aq)
Reduction: 4 H+ (aq) + MnO4-(aq) → MnO2(s) + 2 H2O (ℓ)
d)
Begin as though
it were in an
acidic solution
Oxidation: H2O (ℓ) + CN-(aq) → CNO-(aq) + 2 H+ (aq) + 2eReduction: 3e- + 4 H+ (aq) + MnO4-(aq) → MnO2(s) + 2 H2O (ℓ)
NOW! Account for
the added H+ by
adding in OH- to
both sides of the
e)
equation(s) to
keep matter
balanced (and to
neutralize the acid)
makes water
Oxidation: 2 OH-(aq) + H2O (ℓ) + CN-(aq) → CNO-(aq) + 2 H+ (aq) + 2 OH-(aq) + 2eReduction: 3e- + 4 OH-(aq) + 4 H+ (aq)+ MnO4-(aq) → MnO2(s)+ 2 H2O (ℓ) +4 OH-(aq)
makes water
note the H+ and OH- were
combined to make water
Rewrite & Cancel
water …Notice we
cancel water between
reactants and products
…NOT between halfreactions
Rewrite one more
time, showing that
water is produced
f)
1
Oxidation: 2 OH-(aq) + H2O (ℓ) + CN-(aq) → CNO-(aq) + 2 H2O (ℓ) + 2e-
2
Reduction: 3e- + 4 H2O (ℓ) + MnO4-(aq) → MnO2(s) + 2 H2O (ℓ) + 4 OH-(aq)
Oxidation: 2 OH-(aq) + CN-(aq) → CNO-(aq) + H2O (ℓ) + 2eReduction: 3e- + 2 H2O (ℓ) + MnO4-(aq) → MnO2(s) + 4 OH-(aq)
Oxidation: 6 OH-(aq) + 3 CN-(aq) → 3 CNO-(aq) + 3 H2O (ℓ) + 6eReduction: 6e- + 4 H2O (ℓ) + 2 MnO4-(aq) → 2 MnO2(s) + 8 OH-(aq)
g) Recombine: 3 CN-(aq) + H2O (ℓ) + 2 MnO4-(aq) → 3 CNO-(aq) + 2 MnO2(s) + 2 OH-(aq)
Question: Why, when re-combined, are there only 1 water and 2 OH- ?
*When you study the two balanced half-reactions, you can see that the 6 reactant
OH- from the oxidation, are canceled by the 8 product OH- from the reduction,
leaving us with 2 OH-. This reasoning is similar for the waters.
594
TRY THIS!
Brown & LeMay p 901
Assuming an alkaline solution, when balanced in the simplest whole-number ratio,
how many moles of water molecules are indicated?
20.25 f)
MnO4- (aq) + Br-(aq) → MnO2 (s) + BrO3- (aq)
1) 1 mol on the reactant side
2) 2 mol on the product side
*a)
3) 4 mol on the product side
4) 5 mol on the reactant side
ans: choice 1
Oxidation: Br-(aq) → BrO3-(aq)
Reduction: MnO4-(aq) → MnO2(s)
b)
Oxidation: 3 H2O (ℓ) + Br-(aq) → BrO3-(aq) + 6 H+ (aq)
Reduction: 4 H+ (aq) + MnO4-(aq) → MnO2(s) + 2 H2O (ℓ)
c)
Oxidation: 3 H2O (ℓ) + Br-(aq) → BrO3-(aq) + 6 H+ (aq) + 6 eReduction: 3 e- + 4 H+ (aq) + MnO4-(aq) → MnO2(s) + 2 H2O (ℓ)
d)
Oxidation: 6 OH-(aq) + 3 H2O (ℓ) + Br-(aq) → BrO3-(aq) + 6 H+ (aq) + 6 e- + 6 OH-(aq)
Reduction: 3 e- + 4 OH-(aq) + 4 H+ (aq) + MnO4-(aq) → MnO2(s) + 2 H2O (ℓ) + 4 OH-(aq)
Rewrite1
Oxidation: 6 OH-(aq) + 3 H2O (ℓ) + Br-(aq) → BrO3-(aq) + 6 H2O (ℓ) + 6 eReduction: 3 e- + 4 H2O (ℓ) + MnO4-(aq) → MnO2(s) + 2 H2O (ℓ) + 4 OH-(aq)
Rewrite2
Oxidation: 6 OH-(aq) + Br-(aq) → BrO3-(aq) + 3 H2O (ℓ) + 6 eReduction: 3 e- + 2 H2O (ℓ) + MnO4-(aq) → MnO2(s) + 4 OH-(aq)
e)
Oxidation: 6 OH-(aq) + Br-(aq) → BrO3-(aq) + 3 H2O (ℓ) + 6 eReduction: 6 e- + 4 H2O (ℓ) + 2 MnO4-(aq) → 2 MnO2(s) + 8 OH-(aq)
f) Recombine: 2 MnO4- (aq) + Br-(aq) + H2O(ℓ) → 2 MnO2 (s) + 3 BrO3- (aq) + 2 OH-(aq)
595
TRY THIS!
Assuming an alkaline solution, when balanced in the simplest whole-number ratio,
how many moles of water molecules are indicated?
Duncan et.al. p 173
NO2- + Al → NH3 + AlO2- (aq)
1) 1 mol on the reactant side
2) 2 mol on the product side
*a)
3) 4 mol on the product side
4) 5 mol on the reactant side
ans: choice 1
Oxidation: Al → AlO2Reduction: NO2- → NH3
b)
Oxidation: 2 H2O + Al → AlO2- + 4 H+
Reduction: 7 H+ + NO2- → NH3 + 2 H2O
c)
Oxidation: 2 H2O + Al → AlO2- + 4 H+ + 3eReduction: 6e- + 7 H+ + NO2- → NH3 + 2 H2O
d)
Oxidation: 4 H2O + 2 Al → 2 AlO2- + 8 H+ + 6 eReduction: 6e- + 7 H+ + NO2- → NH3 + 2 H2O
… just for kicks, I also balanced out the e- …trying to show that the process is flexible…
e)
Oxidation: 8 OH- + 4 H2O + 2 Al → 2 AlO2- + 8 H+ + 8 OH- + 6 eReduction: 6e- + 7 H+ + 7 OH- + NO2- → NH3 + 2 H2O + 7 OH…at this point I added the OH- ions and in the next section, combined to make water…
Rewrite1
Oxidation: 8 OH- + 4 H2O + 2 Al → 2 AlO2- + 8 H2O + 6 eReduction: 6e- + 7 H2O + NO2- → NH3 + 2 H2O + 7 OHRewrite2
Oxidation: 8 OH- + 2 Al → 2 AlO2- + 4 H2O + 6 eReduction: 6e- + 5 H2O + NO2- → NH3 + 7 OHe) Recombine: OH- + 2 Al + H2O + NO2- → 2 AlO2- + NH3
or rather: OH- + H2O + 2 Al + NO2- → 2 AlO2- + NH3
596
TIME FOR A LITTLE TLC (Think, Learn, Construct)
…answers are on the next page….
(Duncan et. al. p 177)
Blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma with an acidic
potassium dichromate solution. The UNbalanced equation for the reaction is:
Cr2O72- (aq) + C2H5OH (aq) → Cr3+ (aq) + CO2 (g)
a) Identify which species is oxidized and which is reduced
_______________= oxidized species
_________________= reduced species
b) Balance the equation, using the smallest whole number coefficients
c) How many moles of electrons are transferred, according to the balanced equation? _____
d) Identify at least 1 piece of visible evidence that a chemical reaction has occurred. _______________
___________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
597
Answers to TLC:
+6
+2
+6
a) C is oxidized and Cr is reduced
-2
Cr2O72- (aq)
x +1 -2 +1
+3
+4 -2
3+
+ C2H5OH (aq) → Cr
(aq)
+ CO2 (g)
I like this question… It involves a polyatomic ion and a small organic compound … but more than that, I like the
requirement to calculate the oxidation state of the C atoms in ethanol … On average the C atoms are +2. Some of
us might want to analyze the structure of the molecule – but since all of the carbon ends up in a +4 oxidation state
of the carbon dioxide … it is perfectly cool to use the average….remember … this is just bookkeeping…
We can see that the chromium of the dichromate polyatomic ion goes from +6 to +3 (reduction) and that the C
each at a +2 go to a +4 (oxidation)
b) Oxidation: C2H5OH → CO2 (g)
Reduction: Cr2O72- → Cr3+
at this point it may just be easier to write C2H5OH as:
C 2 H6 O
Balance all matter, except for O and H
Oxidation: C2H6O → 2 CO2
Reduction: Cr2O72- → 2Cr3+
Balance the O, using water
Oxidation: 3 H2O + C2H6O → 2 CO2
Reduction: Cr2O72- → 2Cr3+ + 7 H2O
Balance the H, by using H+
Oxidation: 3 H2O + C2H6O → 2 CO2 + 12 H+
Reduction: 14 H+ + Cr2O72- → 2Cr3+ + 7 H2O
These 12 H+ balance the 6 H+ from water and the 6 H+ from ethanol
Balance the charge by adding electrons as necessary
Oxidation: 3 H2O + C2H6O → 2 CO2 + 12 H+ + 12 eReduction: 6 e- + 14 H+ + Cr2O72- → 2Cr3+ + 7 H2O
Equalize the number of electrons lost and gained
Oxidation: 3 H2O + C2H6O → 2 CO2 + 12 H+ + 12 eReduction: 12 e- + 28 H+ + 2 Cr2O72- → 4Cr3+ + 14 H2O
Cancel as necessary
Oxidation: 3 H2O + C2H6O → 2 CO2 + 12 H+ + 12 eReduction: 12 e- + 28 H+ + 2 Cr2O72- → 4Cr3+ + 14 H2O
16
11
+
2Recombine: 16 H + 2 Cr2O7 + C2H6O → 4Cr3+ + 2 CO2 + 11 H2O
c) 12 moles of electrons are transferred
d) I can think of 2 possible visible changes … one is quite obvious … The development of a new gas (CO2(g)) is a
clear signal of a chemical reaction. Also (and far less obvious for a student working on a “paper problem) there
would be a bold color change!!! Recall, many of the period 4 transition ions are colored. Cr+6 is a Period 4 transition
metal ion, hence it probably has a color in solution, (in fact, it is a rich orange) … As it is reduced to the Cr+3 there is
(at least) a loss of orange color &/or a change to a different color (in fact, blue-violet) … because the Cr+3 is (again) a
Period 4 transition metal ion … but of a different oxidation state than the +6 … so it has a different color! Recall that
many transition metal cations have distinct colors, when hydrated, due to the interactions of energy and the quantum
configuration of the e-
598