C142B Lecture Notes
Campbell / Callis
Chapter 4: The Major Classes of
Chemical Reactions
4.1 The Role of Water as a Solvent
4.2 Precipitation Reactions and Acid-Base Reactions
4.3 Oxidation - Reduction (Redox) Reactions
4.4 Counting Reactants and Products in Precipitation,
Acid-Base, and Redox Processes
4.5 Reversible Reactions: An Introduction to Chemical
Equilibrium
The Role of Water as a Solvent: The solubility of
Ionic Compounds
Electrical conductivity - The flow of electrical current in a solution is a
measure of the solubility of ionic compounds or a
measurement of the presence of ions in solution.
Electrolyte - A substance that conducts a current when dissolved in
water. Soluble ionic compound dissociate completely and
may conduct a large current, and are called Strong
Electrolytes.
NaCl(s) + H2O(l)
Na+(aq) + Cl -(aq)
When sodium chloride dissolves into water the ions become solvated,
and are surrounded by water molecules. These ions are called “aqueous”
and are free to move through out the solution, and are conducting
electricity, or helping electrons to move through out the solution
Fig 4.1 (P135)
Electrical Conductivity of Ionic Solutions
Fig. 4.2
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
Problem: How many moles of each ion are in each of the following:
a)
b)
c)
d)
e)
4.0 moles of sodium carbonate dissolved in water
46.5 g of rubidium fluoride dissolved in water
5.14 x 1021 formula units of iron (III) chloride dissolved in water
75.0 mL of 0.56M scandium bromide dissolved in water
7.8 moles of ammonium sulfate dissolved in water
a) Na2CO3(s)
H2O
moles of Na+ = 4.0 moles Na2CO3 x
2 Na+(aq) + CO3-2(aq)
2 mol Na+
1 mol Na2CO3
= 8.0 moles Na+ and 4.0 moles of CO3-2 are present
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - II
b)
RbF(s)
H2O
Rb+ (aq) + F - (aq)
moles of RbF =
c)
FeCl3(s)
H2 O
Fe+3 (aq) + 3 Cl - (aq)
moles of FeCl3 = 9.32 x 1021 formula units x
moles of Cl - =
moles Fe+3 =
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - II
b)
RbF(s)
H2O
Rb+ (aq) + F - (aq)
moles of RbF = 46.5 g RbF x
1 mol RbF
104.47 g RbF
= 0.445 moles RbF
thus, 0.445 mol Rb+ and 0.445 mol F - are present
H2 O
c) FeCl3(s)
Fe+3 (aq) + 3 Cl - (aq)
moles of FeCl3 = 9.32 x 1021 formula units
x
= 0.0155 mol FeCl3
moles of
Cl -
1 mol FeCl3
6.022 x 1023 formula units FeCl3
3
mol
Cl
= 0.0155 mol FeCl3 x
= 0.0465 mol Cl 1 mol FeCl3
and 0.0155 mol Fe+3 are also present.
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - III
H2O
d) ScBr3 (s)
Sc+3(aq) + 3 Br -(aq)
Converting from volume to moles:
Moles of ScBr3 = 75.0 mL x 13L x
10 mL
Moles of Br - =
e) (NH4)2SO4 (s)
H2O
2 NH4+ (aq) + SO4- 2 (aq)
+
2
mol
NH
4
Moles of NH4 = 7.8 moles (NH4)2SO4 x
= 15.6 mol NH4+
1 mol(NH4)2SO4
+
and 7.8 mol SO4- 2 are also present.
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - III
H2O
d) ScBr3 (s)
Sc+3(aq) + 3 Br -(aq)
Converting from volume to moles:
Moles of ScBr3 = 75.0 mL x 13L x0.56 mol ScBr3 = 0.042 mol ScBr3
10 mL
1L
3
mol
Br
Moles of Br = 0.042 mol ScBr3 x
= 0.126 mol Br 1 mol ScBr3
0.042 mol Sc+3 are also present
H2O
e) (NH4)2SO4 (s)
2 NH4+ (aq) + SO4- 2 (aq)
+
2
mol
NH
4
Moles of NH4 = 7.8 moles (NH4)2SO4 x
= 15.6 mol NH4+
1 mol(NH4)2SO4
+
and 7.8 mol SO4- 2 are also present.
Fig. 4.3
The Solubility of Ionic Compounds in Water
The solubility of Ionic Compounds in water depends upon the relative
strengths of the electrostatic forces between ions in the ionic compound
and the attractive forces between the ions and water molecules in the
solvent. There is a tremendous range in the solubility of ionic
compounds in water! The solubility of so called “insoluble” compounds
may be several orders of magnitude less than ones that are called
“soluble” in water, for example:
Solubility of NaCl in water at 20oC = 365 g/L
Solubility of MgCl2 in water at 20oC = 542.5 g/L
Solubility of AlCl3 in water at 20oC = 699 g/L
Solubility of PbCl2 in water at 20oC = 9.9 g/L
Solubility of AgCl in water at 20oC = 0.009 g/L
Solubility of CuCl in water at 20oC = 0.0062 g/L
The Solubility of Covalent Compounds in Water
The covalent compounds that are very soluble in water are the ones
with -OH group in them and are called “Polar” and can have strong
polar (electrostatic)interactions with water. Examples are compound
such as table sugar, sucrose (C12H22O11); beverage alcohol, ethanol
(C2H5-OH); and ethylene glycol (C2H6O2) in antifreeze.
H
Methanol = Methyl Alcohol
H C O H
H
Other covalent compounds that do not contain a polar center, or the
-OH group are considered “Non-Polar” , and have little or no
interactions with water molecules. Examples are the hydrocarbons in
Gasoline and Oil. This leads to the obvious problems in Oil spills, where
the oil will not mix with the water and forms a layer on the surface!
Octane = C8H18
and / or
Benzene = C6H6
Acids - A group of Covalent molecules which loose
Hydrogen ions to water molecules in solution
When gaseous hydrogen Iodide dissolves in water, the attraction of the
oxygen atom of the water molecule for the hydrogen atom in HI is
greater that the attraction of the of the Iodide ion for the hydrogen atom,
and it is lost to the water molecule to form an Hydronium ion and an
Iodide ion in solution. We can write the Hydrogen atom in solution as
either H+(aq) or as H3O+(aq) they mean the same thing in solution. The
presence of a Hydrogen atom that is easily lost in solution is an “Acid”
and is called an “acidic” solution. The water (H2O) could also be written
above the arrow indicating that the solvent was water in which the HI
was dissolved.
HI(g) + H2O(l)
H+ (aq) + I - (aq)
HI(g) + H2O(l)
HI(g)
H2O(l)
H3O+(aq) + I -(aq)
H+(aq) + I -(aq)
Fig. 4.4
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of sulfuric acid will
loose two protons to yield two Hydronium ions, and one sulfate ion.
What is the molarity of the sulfate and Hydronium ions in a solution
prepared by dissolving 155g of pure H2SO4 into sufficient
water to produce 2.30 Liters of sulfuric acid solution?
Plan: Determine the number of moles of sulfuric acid, divide the moles
by the volume to get the molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the acid molarity.
Solution: Two moles of H+ are released for every mole of acid:
H2SO4 (l) + 2 H2O(l)
2 H3O+(aq) + SO4- 2(aq)
Moles H2SO4 = 155 g H2SO4 x 1 mole H2SO4 = 1.58 moles H2SO4
98.09 g H2SO4
Molarity of SO4- 2 =
Molarity of H3O+ (or simply H+) =
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of sulfuric acid will
loose two protons to yield two Hydronium ions, and one sulfate ion.
What is the molarity of the sulfate and Hydronium ions in a solution
prepared by dissolving 155g of pure H2SO4 into sufficient
water to produce 2.30 Liters of sulfuric acid solution?
Plan: Determine the number of moles of sulfuric acid, divide the moles
by the volume to get the molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the acid molarity.
Solution: Two moles of H+ are released for every mole of acid:
H2SO4 (l) + 2 H2O(l)
2 H3O+(aq) + SO4- 2(aq)
1 mole H2SO4
Moles H2SO4 = 155 g H2SO4 x
= 1.58 moles H2SO4
98.09 g H2SO4
-2
1.58
mol
SO
2
4
Molarity of SO4 =
= 0.687 Molar in SO4- 2
2.30 L solution
Molarity of H+ = 2 x 0.687 mol H+ / 2.30 liters = 0.597 Molar in H+
Fig. 4.5
Precipitation Reactions: A solid product is formed
When ever two aqueous solutions are mixed, there is the possibility of
forming an insoluble compound. Let us look at some examples to see
how we can predict the result of adding two different solutions together.
Pb(NO3) (aq) + NaI(aq)
Pb+2(aq) + 2 NO3- (aq) + Na+ (aq) + I- (aq)
When we add these two solutions together, the ions can combine in the
way they came into the solution, or they can exchange partners. In this
case we could have Lead Nitrate and Sodium Iodide, or Lead Iodide and
Sodium Nitrate formed, to determine which will happen we must look at
the Solubility Rules (P 141) to determine what could form. Rule 3a in
Table 4.1 indicates that Lead Iodide will be insoluble, so a precipitate
will form! The remainder (sodium sulfate) is soluble (Rule 4a).
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
Precipitation Reactions: Will a Precipitate form?
If we add a solution containing Potassium Chloride to a solution
containing Ammonium Nitrate, will we get a precipitate?
KCl(aq) + NH4NO3(aq)
=
K+ (aq) + Cl-(aq) + NH4+(aq) + NO3-(aq)
By exchanging cations and anions we see that we could have Potassium
Chloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium
Chloride. In looking at the solubility rules, Table 4.1:
If we mix a solution of Sodium Sulfate with a solution of Barium Nitrate,
will we get a precipitate? Table 4.1 shows that:
Therefore we get:
Na2SO4(aq) + Ba(NO3)2(aq)
BaSO4( ) + 2 NaNO3(
)
Precipitation Reactions: Will a Precipitate form?
If we add a solution containing Potassium Chloride to a solution
containing Ammonium Nitrate, will we get a precipitate?
KCl(aq) + NH4NO3(aq)
=
K+ (aq) + Cl-(aq) + NH4+(aq) + NO3-(aq)
By exchanging cations and anions we see that we could have Potassium
Chloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium
Chloride. In looking at the solubility rules, Table 4.1 shows all
possible products as soluble, so there is no net reaction!
KCl(aq) + NH4NO3(aq) = No. Reaction!
If we mix a solution of Sodium Sulfate with a solution of Barium Nitrate,
will we get a precipitate? Table 4.1 shows that Barium Sulfate is
insoluble, therefore we will get a precipitate!
Na2SO4(aq) + Ba(NO3)2(aq)
BaSO4(s) + 2 NaNO3(aq)
Fig. 4.6
Predicting Whether a Precipitation Reaction
Occurs (see Table 4.1); Writing Equations:
a) Calcium Nitrate and Sodium Sulfate solutions are added together.
Molecular Equation
Ca(NO3)2(aq) + Na2SO4(aq)
CaSO4(s) + NaNO3(aq)
Total Ionic Equation
Ca2+(aq) + 2 NO3-(aq) +2 Na+(aq) + SO4-2(aq)
CaSO4(s) + 2Na+(aq) + 2 NO3-(aq)
Net Ionic Equation
Ca2+(aq) + SO4-2(aq)
CaSO4 (s)
Spectator Ions are Na+ and NO3b) Ammonium Sulfate and Magnesium Chloride are added together.
In exchanging ions, no precipitates will be formed, accd. To Table 4.1,
so there will be no reactions occurring! All ions are spectator ions!
Acid - Base Reactions : Neutralization Rxns.
An Acid is a substance that produces H+ (H3O+) ions when dissolved
in water.
A Base is a substance that produces OH - ions when dissolved in water.
Acids and Bases are electrolytes, and their strength is categorized in
terms of their degree of dissociation in water to make hydronium or
hydroxide ions, resp.. Strong acids and bases dissociate completely, and
are strong electrolytes. Weak acids and bases dissociate only to a tiny
extent (<<100%) and are weak electrolytes.
The generalized reaction between an Acid and a Base is:
HX(aq) + MOH(aq)
Acid
+
Base
MX(aq) + H2O(l)
=
Salt(aq)
+ Water
Fig. 4.7
Table 4.2 (P 143)
Acids
Selected Acids and Bases
Bases
Strong (100% to H+)
Hydrochloric, HCl
Hydrobromic, HBr
Hydroiodoic, HI
Nitric acid, HNO3
Sulfuric acid, H2SO4
Perchloric acid, HClO4
Strong (100% to OH-)
Sodium hydroxide, NaOH
Potassium hydroxide, KOH
Calcium hydroxide, Ca(OH)2
Strontium hydroxide, Sr(OH)2
Barium hydroxide, Ba(OH)2
Weak (tiny % to H+)
Hydrofluoric, HF
Phosphoric acid, H3PO4
Acidic acid, CH3COOH
(or HC2H3O2)
Weak (tiny % yield of OH-)
Ammonia, NH3
Writing Balanced Equations for
Neutralization Reactions - I
Problem: Write balanced chemical reactions (molecular, total ionic, and
net ionic) for the following Chemical reactions:
a) Calcium Hydroxide(aq) and HydroIodic acid(aq)
b) Lithium Hydroxide(aq) and Nitric acid(aq)
c) Barium Hydroxide(aq) and Sulfuric acid(aq)
Plan: These are all strong acids and bases, therefore they will make
water and the corresponding salts.
Solution:
a) Ca(OH)2(aq) + 2HI(aq)
CaI2(aq) + 2H2O(l)
Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq)
Ca2+(aq) + 2 I -(aq) + 2 H2O(l)
2 OH -(aq) + 2 H+(aq)
2 H2O(l)
Writing Balanced Equations for
Neutralization Reactions - II
b) LiOH(aq) + HNO3(aq)
c)
Ba(OH)2(aq) + H2SO4(aq)
Writing Balanced Equations for
Neutralization Reactions - II
b) LiOH(aq) + HNO3(aq)
LiNO3(aq) + H2O(l)
Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq)
Li+ (aq) + NO3-(aq) + H2O(l)
OH -(aq) + H+(aq)
c)
Ba(OH)2(aq) + H2SO4(aq)
H2O(l)
BaSO4(s) + 2 H2O(l)
Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42 -(aq)
BaSO4(s) + 2 H2O(l)
Ba2 +(aq) + 2 OH -(aq) + 2 H+(aq) + SO42 -(aq)
BaSO4(s) + 2 H2O(l)
Fig. 4.8
Finding the Concentration of Acid from an
Acid - Base Titration
Volume (L) of base (difference in
buret readings) needed to titrate
M (mol/L) of base
Moles of base needed to titrate
molar ratio
Moles of acid which were titrated
volume (L) of acid
M (mol/L) of original acid soln.
Potassium Hydrogenphthalate KHC8H4O4
O
O
C
C
C
K+
O K+
O
O
O
H
C
H+
O
O
= “KHP” = a common acid used to titrate bases
(M = 204.2 g/mol)
Finding the Concentration of Base from an
Acid - Base Titration - I
Problem: A titration is performed between Sodium Hydroxide and
KHP (204.2 g/mol) to standardize the base solution,
by placing 50.00 mg of solid Potassium Hydrogenphthalate in a flask
with a few drops of an indicator. A buret is filled with the base, and the
initial buret reading is 0.55 mL; at the end of the titration the buret
reading is 33.87 mL. What is the concentration of the base?
Plan: Use the molar mass of KHP to calculate the number
of moles of the acid, from the balanced chemical equation, the reaction
is equal molar, so we know the moles of base, and from the difference
in the buret readings, we can calculate the molarity of the base.
Solution: The rxn is: KHP(aq) + OH-(aq) --> KP-(aq) + H2O or
HKC8H4O4(aq) + OH -(aq)
KC8H4O4-(aq) + H2O(aq)
Finding the Concentration of Base from an
Acid - Base Titration - II
moles KHP = 50.00 mg KHP
Volume of base = Final buret reading - Initial buret reading
=
one mole of H+ : one mole of OH-; therefore ____________ moles of
KHP will titrate _____________ moles of NaOH.
molarity of NaOH =
molarity of base =
moles
=
L
Finding the Concentration of Base from an
Acid - Base Titration - II
moles KHP = 50.00 mg KHP x 1.00 g
204.2 g KHP
1000 mg
1 mol KHP
= 0.00024486 mol KHP
Volume of base = Final buret reading - Initial buret reading
= 33.87 mL - 0.55 mL = 33.32 mL of base
one mole of H+ : one mole of OH-; therefore 0.00024486 moles of
KHP will titrate 0.00024486 moles of NaOH.
0.00024486 moles
molarity of NaOH =
= 0.07348679 moles per liter
0.03332 L
molarity of base = 0.07349 M
Fig. 4.9
Fig. 4.10
Fig. 4.11
Highest and Lowest oxidation numbers of
Chemically reactive main-group Elements
Fig. 4.12
+1
+1
-1
1 H
non-metals
1A 2A
+1 +2
3A 4A 5A 6A 7A
+3 +4-4 +5-3 +6-2 +7-1
Li
B
C
N
O
F
3 Na Mg
Al
Si
P
S
Cl
4
Ga Ge As Se
Br
5 Rb Sr
In
I
6 Cs
Tl Pb
2
Be
K Ca
Sn Sb Te
metalloids
metals
Ba
7 Fr Ra
Bi
Po At
Fig. 4.13
Period IA Common Ox #s: Main Group Elements VIIIA
H
He
1
+1 -1 IIA
IIIA IVA
VA
VIA VIIA
Li
Be
B
C
N
O
F
Ne
2
+4,+2 all from
+1
+2
+3 -1,-4 +5 -3 -1,-2
-1
Na
Mg
Al
Si
P
S -1 Cl
Ar
3
+1
+2
+3
+4,-4 +5,+3 +6,+4 +7,+5
-3
+2,-2 +3,+1
Kr
K
Ca
Ga
Ge
As
Se -1 Br
4
+2
+1
+2 +3, +2 +4,+2 +5,+3 +6,+4 +7,+5
-4
-3
-2
+3,+1
Xe
Rb
Sr
In
Sn
Sb
Te -1 I
5
+1
+2 +3,+2 +4,+2, +5,+3 +6,+4 +7,+5 +6,+4
+1
-4
-3
-2
+3,+1 +2
Rn
Cs
Ba
Tl
Pb
Bi
Po -1 At
6
+6,+4 +7,+5
+1
+2
+2 +3,+1 +4,+2
+3
+2,-2 +3,+1
Transition Metals
Possible Oxidation States
VIIIB
IIIB IVB VB VIB VIIB
IB IIB
Sc
Ti
V
Cr +2Mn Fe
Co
Ni
Cu Zn
+3 +4,+3 +5,+4 +6,+3 +7,+6 +3,+2 +3,+2 +2 +2,+1 +2
+2
+3+2 +2
+4,+3
Y
Zr
Nb Mo Tc Ru Rh Pd
Ag
+5,+4 +6,+5 +7,+5 +8,+5
+3 +4,+3
+4,+3 +4,+2 +1
+2
+4,+3 +4
+4,+3
Cd
+2
La
Hf
Ta
W
Re +2Os Ir
Pt
Au Hg
+3 +4,+3 +5,+4 +6,+5 +7,+5 +8,+6 +4,+3 +4,+2 +3,+1 +2,+1
+3
+4
+4
+4,+3 +1
Determining the Oxidation Number of an
Element in a Compound
Problem: Determine the oxidation number (Ox. No.) of each element
in the following compounds.
a) Iron III Chloride b) Nitrogen Dioxide
c) Sulfuric acid
Plan: We apply the rules in Table 4.3, always making sure that the
Ox. No. values in a compound add up to zero, and in a
polyatomic ion, to the ion’s charge.
Solution:
a) FeCl3 This compound is composed of monoatomic ions. The
Ox. No. of Cl- is -1, for a total of -3. Therefore the Fe must be +3.
b) NO2
c) H2SO4
Determining the Oxidation Number of an
Element in a Compound
Problem: Determine the oxidation number (Ox. No.) of each element
in the following compounds.
a) Iron III Chloride b) Nitrogen Dioxide
c) Sulfuric acid
Plan: We apply the rules in Table 4.3, always making sure that the
Ox. No. values in a compound add up to zero, and in a
polyatomic ion, to the ion’s charge.
Solution:
a) FeCl3 This compound is composed of monoatomic ions. The
Ox. No. of Cl- is -1, for a total of -3. Therefore the Fe must be +3.
b) NO2 The Ox. No. of oxygen is -2 for a total of -4. Since the
Ox. No. in a compound must add up to zero, the Ox. No. of N is +4.
c) H2SO4 The Ox. No. of H is +1, so the SO42- group must sum to -2.
The Ox. No. of each O is -2 for a total of -8. So the Sulfur atom is +6.
Recognizing Oxidizing and Reducing Agents - I
Problem: Identify the oxidizing and reducing agent in each of the Rx:
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 (O.N.) to each atom (or ion)
based on the rules in Table 4.3. A reactant is the reducing agent if it
contains an atom that is oxidized (O.N. increased in the reaction). A
reactant is the oxidizing agent if it contains an atom that is reduced
( O.N. 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 - II
b) Assigning oxidation numbers:
S8 (s) + 12 O2 (g)
8 SO3 (g)
____ is the reducing agent and ____ is the oxidizing agent.
c) Assigning oxidation numbers:
NiO(s) + CO(g)
Ni(s) + CO2 (g)
___ is the reducing agent and ____ is the oxidizing agent.
Recognizing Oxidizing and Reducing Agents - II
b) Assigning oxidation numbers:
0
0
S8 (s) + 12 O2 (g)
-2
S [0]
S[+6]
S is Oxidized
8 SO3 (g)
O[0]
O[-2]
O is Reduced
+6
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)
-2
+4
+ CO2 (g)
CO is the reducing agent and NiO is the oxidizing agent
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) Multiply one or both of these numbers by appropriate factors to
make the electrons lost equal the electrons gained, and use the
factors as balancing coefficients.
Step 5) Complete the balancing by inspection, adding states of matter.
REDOX Balancing using Ox. No. Method - I
+2 e-
0
___
2 H2 (g) +___ O2 (g)
0
- 1 e-
electrons lost must = electrons gained
therefore multiply
Hydrogen reaction by 2!
and we are balanced!
-2
___
2 H2O(g)
+1
REDOX Balancing Using Ox. No. Method - II
+2
-1e-
Fe+2(aq) + MnO4-(aq) + H3O+(aq)
+3
Fe+3(aq) + Mn+2(aq) + H2O(aq)
+5 e+2
+7
Multiply Fe+2 & Fe+3 by ___ to correct for the electrons gained by the
Manganese.
__Fe+2(aq)+ MnO4-(aq) + H3O+(aq)
__ Fe+3(aq) + Mn+2(aq) + H2O(aq)
Next balance the # O and H atoms:
5 Fe+2(aq) + MnO4-(aq) +__ H3O+(aq)
5 Fe+3(aq) + Mn+2(aq) +__ H2O(aq)
REDOX Balancing Using Ox. No. Method - II
+2
-1e-
Fe+2(aq) + MnO4-(aq) + H3O+(aq)
+3
Fe+3(aq) + Mn+2(aq) + H2O(aq)
+5 e+2
+7
Multiply Fe+2 & Fe+3 by five to correct for the electrons gained by the
Manganese.
5 Fe+2(aq) + MnO4-(aq) + H3O+(aq)
5 Fe+3(aq) + Mn+2(aq) + H2O(aq)
Next balance the # O and H atoms: Make 4 water molecules from the
oxygens of the MnO4-, which requires 8 protons from the acid (H3O+
ions). These 8 H3O+ ions will also make 8 more water molecules, for
a total of 4+8 = 12 water molecules.
5 Fe+2(aq) + MnO4-(aq) +8 H3O+(aq)
5 Fe+3(aq) + Mn+2(aq) +12 H2O(aq)