Download Chapter 4-5

General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Review Chapters 4:5 Chemical Reactions
Chem 1050
Chapter 4-5
Dr. Ghumman
Chapter 4:
4-1
4-2
4-3
4-4
4-5
Chem 1050
Chemical Reactions and Chemical Equations
Chemical Equations and Stoichiometry
Chemical Reactions in Solution
Determining the Limiting reagent
Other Practical Matters in Reaction Stoichiometry
Chapter 4-5
Dr. Ghumman
Chemical Reactions and
Chemical Equations
• Chemical reactions- Reactants are converted to products.
– are accompanied by a chemical change.
• Some physical evidence for a chemical reaction
– Color change
– Precipitate formation
– Gas evolution
– Heat absorption or evolution
– A color change
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1
Chemical Reaction
Nitrogen monoxide + oxygen → nitrogen dioxide
Step 1: Write the reaction using chemical symbols.
Step 2: Balance the chemical equation.
2NO +
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O2 → 2 NO2
Chapter 4-5
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Molecular Representation
Chem 1050
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Balancing Equations
• Never introduce extraneous atoms to balance.
NO + O2 → NO2 + O
(incorrect)
• Never change a formula for the purpose of balancing an
equation.
NO + O2 → NO3 (incorrect)
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2
Balancing Equation Strategy
• Balance elements that occur in only one compound on each
side first.
• Balance free elements last.
• Balance unchanged polyatomic ions as groups.
• Fractional coefficients are acceptable and can be cleared at
the end by multiplication.
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Writing and Balancing equation: combustion of
triethylene glycol
• Chemical Equation:
C6H14O4(l) + 15/2 O2 (g) → 6 CO2 (g) + 7 H2O (l)
1. Balance C
2. Balance H
3. Balance O
4. Multiply by two to remove the fractional coefficient
2 C6H14O4 (l) + 15 O2 (g)→ 12 CO2 (g) + 14 H2O(g)
and check all elements.
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Chemical Equations and Stoichiometry
• Stoichiometry includes all the quantitative relationships
involving:
– atomic and formula masses
– chemical formulas.
• Mole ratio is a central conversion factor.
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3
Example 4-6
Additional Conversion Factors in a Stoichiometric
Calculation: Volume, Density, and Percent Composition.
An alloy used in aircraft structures consists of 93.7% Al and
6.3% Cu by mass. The alloy has a density of 2.85 g/cm3. A
0.691 cm3 piece of the alloy reacts with an excess of
HCl(aq). If we assume that all the Al but none of the Cu
reacts with HCl(aq), what is the mass of H2 obtained?
2 Al(s)+ 6 HCl(aq) → 2 AlCl3(aq) + 3 H2 (g)
Plan the strategy:
cm3 alloy → g alloy → g Al → mole Al → mol H2 → g H2
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Example 4-6
We need 5 conversion factors!
Write the Equation and calculate;
× 97.3 g Al ×
mH2 = 0.691 cm3 alloy × 2.85 g alloy
100 g alloy
1 cm3
1 mol Al × 3 mol H2 × 2.016 g H2
2 mol Al
26.98 g Al
1 mol H2 = 0.207 g H2
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Chemical Reactions in Solution
• Close contact between atoms, ions and molecules
necessary for a reaction to occur.
• Solvent
– We will usually use aqueous (aq) solution.
• Solute
– A material dissolved by the solvent
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Molarity
•
Calculating the mass of Solute in a solution of Known
•
Molarity (M) = moles of solute/L of solution (mol.L-1)
We want to prepare exactly 0.2500 L (250 mL) of an
0.250 M K2CrO4 solution in water. What mass of K2CrO4
should we use?
Volume → moles → mass
Plan strategy
Write equation and calculate:
mK2CrO4 = 0.2500 L × 0.250 mol× 194.02 g = 12.1 g
1.00 mol
1.00 L
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Solution Dilution
Mi × Vi
M=
n
V
M i × V i = ni = nf = M f × V f
Mf =
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M i × Vi
Vi
= Mi
Vf
Vf
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Determining Limiting Reagent
•
•
Limiting Reagent L.R):The reactant that is completely
consumed determines the quantities of the products
formed.
Reaction stops when the limiting reagent is consumed.
Determining the Limiting Reactant in a Reaction.
1. Assume each reactant in turn is the limiting reagent and
calculate its capacity to yield a given product
2. The reactant that will give the smallest product yield is the
LR
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Other Practical Matters in
Reaction Stoichiometry
• Theoretical yield- is the expected yield from reactants.
– Calculated maximum obtainable yield of product based
on the100% consumption of the LR.
Actual yield- is the amount of product actually produced.
Percent yield =
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Actual yield
× 100%
Theoretical Yield
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Theoretical, Actual and Percent Yield
• When actual yield = 100 % , the reaction is said to be
quantitative.
• Side reactions reduce the percent yield.
• By-products are formed by side reactions.
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Limiting Reagent
• Example: A 109.5 g sample of zinc is added to 150.0mL
of 2.50 M HNO3 (aq) producing 12.38 g of N2O (g).
Calculate the % yield of N2O.
Solution :4Zn (s) + 10 HNO3 (aq) → 4Zn(NO3)2(aq) + N2O(g) + 5H2O(l)
– Calculate moles of each reactant
– Then calculate the amount of N2O produced by using in turn all the
moles of each reactant.
– The reactant giving the least moles of N2O is the limiting reactant.
– Theoretical yield is based on the # of moles of N2O calculated by
using LR.
– Then calculate the % Yield = Actual Y/Theoretical Y x 100%
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Gravimetric Analysis
Gravimetric analysis: Weighing a precipitate formed in a reaction
accurately to determine the unknown. ( Lab)
Example:A 4.270 g of an unknown metal bromide XBr3 was dissolved in
H2O and treated with excess of AgNO3. If 8.140 g of AgBr were
recovered
a) Calculate the no. of moles of AgBr formed .
b) Calculate no. of moles of XBr3 reacted.
c) Calculate the molar mass of element X and Identify the element X.
Solution:
XBr3 (aq) + 3AgNO3(aq) →3AgBr(s) + X(NO3)3 (aq)
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Solution Stoichiometry
Example: A 20.00 mL aliquot of dilute aqueous H2O2 (d=
1.04 g.mL-1) requires 46.90mL of 0.1451M KMnO4(aq)
for complete reaction. Calculate the mass% of H2O2 in the
solution.
5 H2O2(aq) + 2MnO4-(aq) + 6H+(aq) → 5O2(g) + 2Mn2+(aq) + 8H2O(l)
Solution: need mass of solution and mass of H2O2
KMnO4 is strong electrolyte
(100%)
KMnO4(aq) → K+(aq) + MnO4-(aq)
• mol MnO4- = CV → n H2O2 →Mass of H2O2 →Mass of solution using
density and volume of the solution.
• Mass % of H2O2 = ( g H2O2/ g of sol’n.) x 100%
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Chapter -5
The Nature of Aqueous Solutions
5-2
Precipitation Reactions
5-3
Acid-Base Reactions
5-4
Oxidation-Reduction: Some General Principles
5-5
Balancing Oxidation-Reduction Equations
5-6
Oxidizing and Reducing Agents
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Aqueous reactions
Aqueous reactions can be grouped into three general categories;
a. precipitation, b. acid-base and c. Oxidation reactions
– Reactions are driven from reactants to products by some energetic
force that pushes them along.
1. Precipitation Reactions
• Driving force = removal of material (ppt) from solution.
• e.gPb (NO3)2 (aq) + 2 KI(aq) → 2KNO3 (aq) + PBI2 (S) yellow solid
• Acid- base neutralization reactions - An acid reacts with a base to
produce a salt and water.Driving Force = formation of water
• Oxidation- Reduction (Redox) reaction. One or more electrons are
transferred between reactants.
• Driving force is decrease in electrical potential.
• Mg(s) + I2 (g) → MgI2 (s)
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Electrolytes
Some solutes can dissociate into ions.
• Most electrolytes are ionic compound
• molecular e.g HCl (g) → H+ (aq) + Cl- (aq)
Strong Electrolytes- completely dissociate in water
Weak Electrolytes- partially dissociate in
water (weak conductor of electricity).
• CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
Establish an equilibrium between forward
and backward reactions- dynamic process
• Non electrolytes- do not dissociate (poor conductor)
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Representation of Electrolytes using Chemical
Equations
A strong electrolyte:Good conductor of electricity
MgCl2(s)
→ Mg
2+(aq)
+ 2 Cl-(aq)
A weak electrolyte: fair conductor of electricity
CH3CO2H(aq)
← CH CO
3
2
-(aq)
+ H+(aq)
A non-electrolyte:
– CH3OH(aq)
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Calculating Ion concentrations in a Solution of a
Strong Electolyte.
What are the aluminum and sulfate ion concentrations in
0.0165 M Al2(SO4)3?.
Balanced Chemical Equation:
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
[Al] =
0.0165 mol Al2(SO4)3
x
1L
2 mol Al3+
1 mol Al2(SO4)3
= 0.0330 M Al3+
Sulfate Concentration:
[SO42-] = 0.0165 mol Al2(SO4)3
1L
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×
3 mol SO421 mol Al2(SO4)3
= 0.0495 M SO 24
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Precipitation Reactions
•
•
Soluble ions can combine to form an insoluble compound.
Precipitation occurs.
•
A qualitative test for Cl- ion in tap water
Ag+(aq) + Cl-(aq) → AgCl(s)
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Net Ionic Equation
Overall Precipitation Reaction:
AgNO3(aq) +NaI (aq) → AgI(s) + NaNO3(aq)
Complete ionic equation:
Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) →AgI(s) + Na+(aq) +
NO3-(aq)
Spectator ion
Net ionic equation:
Ag+(aq) + I-(aq) → AgI(s)
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Solubility Rules
•
Compounds that are soluble:
– Alkali metal ion and ammonium ion salts
Li+, Na+, K+, Rb+, Cs+
– Nitrates, perchlorates and acetates
NO3-
ClO4-
NH4+
CH3CO2 -
• Compounds that are mostly soluble:
– Chlorides, bromides and iodides
Cl-, Br-, I• Except those of Pb2+, Ag+, and Hg22+.
– Sulfates
SO42• Except those of Sr2+, Ba2+, Pb2+ and Hg22+.
• Ca(SO4) is slightly soluble.
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Solubility Rules
• Compounds that are insoluble:
– Hydroxides and sulfides
OH-, S2• Except alkali metal and ammonium salts
• Sulfides of alkaline earths are soluble
• Hydroxides of Sr2+ and Ca2+ are slightly soluble.
– Carbonates and phosphates
CO32-, PO43• Except alkali metal and ammonium salts
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Acids
• Svante Arrhenius 1884 Acid-Base theory.
Acids provide H+ in aqueous solution.
• Strong acids:completely ionize in solution
HCl(aq)
→
H+(aq) + Cl-(aq)
• Weak acids: partially ionize in solution
CH3CO2H(aq)
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←
→
H+(aq) + CH3CO2-(aq)
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Bases
• Bases provide OH- in aqueous solution.
• Strong bases:
NaOH(aq)
• Weak bases:
HO
→
Na+(aq) + OH-(aq)
NH3(aq) + H2O(l)
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2
←
→
OH-(aq) + NH4+(aq)
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• Svante Arrhenius 1884 Acid-Base theory.
• Acids have ionizable hydrogen ions.
– CH3CO2H or HC2H3O2
• Bases have OH- combined with a metal ion.
KOH
or are identified by chemical equations
Na2CO3(s) + H2O(l)→ HCO3-(aq) + 2 Na+(aq) + OH-(aq)
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Acid-Base Reactions
• Milk of magnesia
Mg(OH)2
• Reaction with strong acid:
Mg(OH)2(s) + 2 H+(aq) → Mg2+(aq) + 2 H2O(l)
• Reaction with weak acid:
Mg(OH)2(s) + 2 CH3CO2H(aq) → Mg2+(aq) + 2 CH3CO2-(aq)
+ 2 H2O(l)
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Acid-Base Reactions
• Limestone and marble.
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2CO3(aq)
But: H2CO3(aq) → H2O(l) + CO2(g)
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
Note :CO32- acting as a base instead of OH-.
Our definition must be expanded from simple Arrhenius theory (BronstedLowry and Lewis)
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Oxidation number
• Oxidation number- a means of determining whether the
atom is neutral, electron rich or electron deficient.
• May not represent ionic charges.
• Oxidation numbers can be calculated.
• By comparing the oxidation number of a species before
and after a reaction we can tell that whether a species has
gained or lost electrons (reduced or oxidized).
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Rules for Assigning Oxidation numbers (or states)
1.
Oxidation number, (ON) of an atom in a free element is zero
e.g. 0 for H in H2
2. ON of a monatomic ion = charge on ion
e.g. +1 for Na+ , and –2 for sulfur S23. An atom in polyatomic ion or in a molecular compound usually has the
same ON it would have if it were a monatomic ion e.g. (OH-) O = -2
and H = +1
4. ON for Hydrogen (H) is +1 in all compounds
except ionic hydrides (e.g.NaH , -1 for H).
5. ON for oxygen (O) = -2 in all compounds
except H2O2 (O = -1) and other peroxides.
6. ON for halogens is -1,
except in compounds where halogen is bonded to oxygen e.g. Cl2O (Cl is
+1 and O = -2)
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Rules for Assigning Oxidation numbers
•
•
•
•
•
7. Sum of oxidation numbers must be
– a. zero for neutral molecules
– b. equal to ionic charge for ions
– e.g. AlCl3 ON of Al is +3 and -1 for Cl
– NH3 ON = +1 for H , therefore ON = -3 for N
– ClO4 - ON = -2 for O, therefore ON = +7 for Cl
Example:Find out the oxidation no. of S in H2SO4
First assign the oxidation no. to the easy atoms.
H (+1) and O (-2) then 2 (+1) + S? + 4 (-2) = 0
S = 0 -2(+1)-4(-2) = +6
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Practice problems: Assigning oxidation numbers
• Assign oxidation numbers to atoms in the following
compounds or ions.
• Fe in Fe2O3,
•
•
•
•
•
•
Cr in Cr2O7 2- (dichromate ions)
P in H3PO4
C in C2O42HNO2
Hg22+
C in CH3OH
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Oxidation and Reduction
• Oxidation-is Loss of electrons
– O.S of some element increases in the reaction as
electrons are lost.
– Electrons are on the right of the equation.
• Reduction-Gain of electron
– O.S of some element decreases in the reaction.
– Electrons are on the left of the equation.
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Oxidizing and Reducing Agents.
• An oxidizing agent (oxidant ):
– Contains an element whose OS decreases in a redox
reaction (reduced)
– Gains electrons
• A reducing agent (reductant):
– Contains an element whose OS increases in a redox
reaction ( oxidized).
– Loses electrons.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Zn is reducing agent and Cu2+ is oxidizing agent.
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Zinc in Copper Sulfate
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
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Oxidation-Reduction: Some
General Principles
An Oxidation-reduction or redox reaction involves the
transfer of electrons e.g.
Mg (s) + 2 H+(aq) → Mg2+(aq) + H2(g)
• Oxidation and reduction always occur together.
• H+ is reduced to hydrogen gas.
– Each H+ ion has gained one eMg (s) is oxidized to Mg2+ ions.
– Mg atom has lost 2 e• Number of electron lost = number of electrons gained
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Oxidation State Changes
• Production of Iron metal from its ore hematite in blast
furnace.
• Assign oxidation states:
3+ 2-
2+ 2-
0
4+ 2-
Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)
Fe3+ is reduced to metallic iron (oxidizing agent).
CO(g) is oxidized to carbon dioxide (reducing agent).
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Identifying redox reaction
• Which of the following are redox reactions?
• SnO2 (s) +2C (s) →Sn (s) + 2CO(g)
• Na2SO4 (aq) + Pb(ClO4)2 (aq) →PbSO4(s) + 2NaClO4(aq)
• HNO3(aq) + NH3 (aq) →NH4NO3(aq)
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Oxidizing and Reducing Agents.
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The activity series
Redox reactions of metals can be predicted from the activity series
– ranks the elements in order of their reducing ability in aqueous
solution
1. Any element higher in the activity series will react with
the ion of any element lower in the activity series.
2. Position of Hydrogen- indicates which metal will react with aq. acid
H+(aq) to produce H2 gas
3. Most reactive metals- top of the activity series (strong reducing agent)
4. Least reactive metal - bottom of the activity series ( weak reducing
agent)
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The activity series of elements
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The activity series
• Metals at the top of the series react even with water to
produce H2 gas
• Metals in the middle of the series react with aqueous acid
but not with water
• Metals at the bottom of the series react with neither
aqueous acid nor water
• Fe (s) + 2H+ (aq) → Fe2+ (aq) + H2(g)
• Cu(s) + H+ (aq) → No reaction
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Balancing Oxidation-Reduction Equations
Few can be balanced by inspection.
• Systematic approach required.
1. The Half-Reaction Method.
2. Oxidation number method ( you only need to know one
method).
• You should know how to balance both ionic and molecular
equation.
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Half-Reactions
• Represent a reaction by two half-reactions.
Oxidation:
Zn(s) → Zn2+(aq) + 2 e-
Reduction:
Cu2+(aq) + 2 e- → Cu(s)
Overall:
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
• Doesn’t work well for the molecular equation
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Balancing in Acid
• Write the equations for the half-reactions.
– Balance all atoms except H and O.
– Balance oxygen using H2O.
– Balance hydrogen using H+.
– Balance charge using e-.
• Equalize the number of electrons by multiplying with a
suitable coefficient.
• Add the half reactions.
• Check the balance (mass balance and charge balance).
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Balancing in Basic Solution
• OH- appears instead of H+.
• Treat the equation as if it were in acid.
– Then add OH- to each side to neutralize H+.
– Remove H2O appearing on both sides of equation.
• Check the balance MB and CB.
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Points to Remember
• For acidic solutions, the final equation should only have
H+ and H2O in addition to what was originally present.
• For basic solutions, only OH- and H2O should be added.
• For molecular equations, only H2O can be added.
For basic solutions, only OH- and H2O should be added.
Mass balance: the number of atoms of each type must be
same on both sides of the equation.
Charge balance : The sum of the charges must be same on
both sides of the equation.
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Balancing redox Reactions by the Oxidation number
method
Example: Using the oxidation -number method, balance the following
reaction which takes place in acidic solution.
• MnO4 - (aq) + SO2 (aq) → Mn2+ (aq) + HSO4 -(aq)
• Solution :
1. Unbalanced ionic equation
• MnO4- (aq) + SO2 (aq) → Mn2+ (aq) + HSO4-(aq)
2. Balance all atoms other than hydrogen and oxygen. This is same as
unbalanced equation.
3. Assign oxidation numbers to all atoms.
•
•
+7 -2
•
MnO4- (aq) + SO2 (aq) → Mn2+ (aq) + HSO4-(aq)
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+4
+2
+6
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Balancing redox Reactions by the Oxidation number
method
Atoms which have changed oxidation number:
Mn: +7 → +2( gained 5e-) ; S: +4 → +6 (lost 2e-)
reduced
oxidized
5. Net increase in ON of oxidized atoms = 2
Net decrease in ON of reduced atoms = 5
Multiply net increase with 5 and the net decrease by 2 to give equal
number of electrons
• 2MnO4- (aq) + 5SO2 (aq) → 2Mn2+ (aq) + 5HSO4-(aq)
6. Reactant side of the equation has two less O, so add 2H2O.
2MnO4- (aq) + 5SO2 (aq) + 2H2O→
→ 2Mn2+(aq) + 5HSO4-(aq)
7. Reactant side of equation has one less hydrogen, so add one H+
H+ (aq) + 2MnO4- (aq) + 5SO2 (aq) + 2H2O→
→ 2Mn2+(aq) + 5HSO4-(aq)
+1
+(-2) = -1
2(+2) + (-5)= -1
8. Check your answer to make sure all atoms and charges are balanced.
Chem 1050
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Balancing Redox reactions by half reaction method
• Half reaction method of balancing redox reaction focuses
on the transfer of electrons
• Key- Overall reaction can be broken into two parts or half
reactions;
• 1. Oxidation part of the reaction called oxidation half
reaction.
• 2. Reduction part of the reaction called reduction half
reaction.
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Example: Using the half reaction method balance the
following reaction which takes place in acidic medium
• Cr2O72- (aq) + HC2O4- (aq)→ Cr3+ (aq) + CO2 (g)
• Write unbalanced ionic equation:
• 2. Two unbalanced half reactions:
• Cr2O72- (aq) → Cr3+ (aq) Cr is reduced from +6 to+3
• HC2O4- (aq)→ CO2 (g) C is oxidized from +3 to +4.
3. Balance each half reaction for atoms other than H and O.
• Cr2O72- (aq) →2 Cr3+ (aq)
• HC2O4- (aq)→ 2CO2 (g)
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Add water to balance O and H+ to balance H.
• 14 H+ (aq) + Cr2O72- (aq) →2 Cr3+ (aq) + 7H2O (l)
• HC2O4 - (aq)→ 2CO2 (g) + H+(aq)
5. Balance each half reaction for charge:
• 6e- +14 H+ (aq) + Cr2O72- (aq) →2 Cr3+ (aq) + 7H2O (l)
• HC2O4 - (aq)→ 2CO2 (g) + H+(aq)+ 2 e6. Make the electron count the same in both reactions by multiplying a
suitable number.
• 6e- +14 H+ (aq) + Cr2O72- (aq) →2 Cr3+ (aq) + 7H2O (l)
• 3 x [HC2O4 - (aq)→ 2CO2 (g) + H+(aq)+ 2 e-]
• 7. Add the two half reactions together, canceling any thing that appears
on both sides
11H+(aq) + Cr2O72-(aq) + 3HC2O4-(aq) →
2Cr3+(aq)+7H2O(l)+6CO2(g)
8. Check to make sure that all atoms and charges are balanced.
Chem 1050
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Notes on balancing redox reactions
•
•
Disproportionation reaction, in which same element is oxidized and
reduced e.g. in H2O2 reaction, O atom is oxidized and reduced in
products.
It is a common type of redox reaction that can be difficult to balance.
-1
-2
0
2H2O2(aq) →2H2O(l) + O2(g)
0
-1
+5
• 3Br2(aq) + 6 OH-(aq) → 5Br-(aq) + BrO3-(aq) + 3H2O(l)
The best way to balance is by half reaction method e.g.
Oxi.
Br2 → BrO3 Red.
Br2 → BrChem 1050
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Reverse disproportionation reaction
• Reverse disproportionation reaction
HS-(aq) + HSO3-(aq) →S2O32-(aq)
• S2O32-(aq) is product in both half reaction
Oxi. HS-(aq) →S2O32-(aq)
Red. HSO3-(aq) →S2O32-(aq)
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Practice Problems
•
•
Balance the the reactions below under the conditions indicated.
Zn(s) +HNO3 (aq) →Zn(NO3)2 (aq) + NH4NO3(aq)
•
MnO2(s) +HNO2(aq) → Mn2+(aq) + NO3-(aq) (acidic)
•
MnO4 -(aq) +NO2-(aq) → MnO2(s) + NO3 -(aq)
(basic)
•
Cr2O72-(aq) + H2O2(aq) → Cr3+(aq) +O2(aq)
(acidic)
Chem 1050
Chapter 4-5
Dr. Ghumman
Example
Identifying Oxidizing and Reducing Agents.
Hydrogen peroxide, H2O2, is a versatile chemical. Its uses
include bleaching wood pulp and fabrics and substituting
for chlorine in water purification. One reason for its
versatility is that it can be either an oxidizing or a reducing
agent. For the following reactions, identify whether
hydrogen peroxide is an oxidizing or reducing agent.
H2O2(aq) + 2 Fe2+(aq) + 2 H+ → 2 H2O(l) + 2 Fe3+(aq)
Iron is oxidized and peroxide is reduced.
Chem 1050
Chapter 4-5
Dr. Ghumman
21
Stoichiometry of Reactions in Aqueous Solutions:
Titrations
• Titration
– Carefully controlled addition of one solution to another.
• Equivalence Point
– Both reactants have reacted completely.
• Indicators
– Substances which change colour near an equivalence
point.
Chem 1050
Chapter 4-5
Dr. Ghumman
Example 5-10
Standardizing a Solution for Use in Redox Titrations.
A piece of iron wire weighing 0.1568 g is converted to
Fe2+(aq) and requires 26.42 mL of a KMnO4(aq) solution
for its titration. What is the molarity of the KMnO4(aq)?
Solution: Write a balanced equation for the reaction;
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 4 H2O(l) + 5 Fe3+(aq) +Mn2+(aq)
Chem 1050
Chapter 4-5
Dr. Ghumman
Cont’d
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 4 H2O(l) + 5 Fe3+(aq) + Mn2+(aq)
Determine KMnO4 consumed in the reaction:
nKMnO4 = 0.1568 g Fe ×
1 mol Fe 1 mol Fe2+
×
×
55.847 g Fe 1 mol Fe
−
1 mol MnO4 1 mol KMnO4
×
= 5.615×10−4 mol KMnO4
5 mol Fe2+ 1 mol MnO4 −
Determine the concentration:
[ KMnO4 ] =
Chem 1050
5.615 × 10 −4 mol KMnO4
= 0.02140 M KMnO4
0.02624 L
Chapter 4-5
Dr. Ghumman
22
Chem 1050
Chapter 4-5
Dr. Ghumman
23