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2014/08/29
Stoichiometry …
Chapter 3: Chemical Reactions
End of chapter exercises: Ch 3: 3, 11, 16, 17, 20, 23,
27, 37, 43, 47, 51. Ch 20: 1, 5
Chemical Equations
→ chemical formulas are used to represent
reactions and the amounts of substances
involved.
Depict a reaction using a balanced chemical
equation.
i.e. quantities of substances consumed and
produced → stoichiometry
For a 100% conversion:
• If total reactant = 10 g → total product = 10 g.
• or 1000 reactant atoms → 1000 atoms in products.
• P4 (s)
+
6 Cℓ2 (g)
→
4 PCℓ3 (ℓ)
4 P – atoms
6 x 2 = 12 Cℓ atoms
4 P – atoms + 4 x 3 = 12 Cℓ atoms
• Balancing Chemical Equations
• 2 Fe (s) +
3 Cℓ2 (g)
→
•
2 Fe – atoms
3 x 2 Cℓ = 6 Cℓ atoms
2 FeCℓ3 (s)
2 Fe atoms + 2 x 3 = 6 Cℓ atoms
→ ensures the same number of atoms of each element
appears on both sides of the equation.
e.g. reaction of solid phosphorus, P4, with chlorine gas,
Cℓ2, to form liquid phosphorus trichloride
• P4 (s)
+
6 Cℓ2 (g)
→
4 PCℓ3 (ℓ)
• Reactants – left of arrow
• Products – right of arrow
• Physical states of reactants and products can be
indicated: solid (s); liquid (ℓ); gas (g); aqueous
solution (aq).
• Law of conservation of matter: Matter cannot be
created or destroyed → changes are due to a
rearrangement of atoms.
→ should contain the smallest possible whole number
coefficients.
• Formulas for reactants and products must be correct.
• Subscripts in formulas of reactants / products cannot
be changed to balance an equation, i.e. H2O to H2O2 or
CO2 to CO – this would change the identity of the
substance.
→ placing a coefficient in front of formula changes the
amount of substance, not its identity.
H2O: one molecule H2O → 2 x H – atoms and 1 x O – atoms
2 H2O: two molecules H2O → 4 x H – atoms and 2 x O – atoms
H2O2: one molecule H2O2 → 2 x H – atoms and 2 x O – atoms
e.g. Combustion Reactions:
Balance the following:
• Burning of fuel in oxygen to produce CO2, H2O and
energy (exothermic).
e.g. Methane CH4 (g) + O2 (g) → CO2 (g) + H2O (g)
• Start: balance element that occurs in fewest chemical
formulas – i.e. C or H.
→ one atom C in both reactant and product
→ one molecule CH4 contains 2x as many H – atoms as
in H2O
1. Synthesis of urea:
CO2 (g) + NH3 (g) → H2NCONH2 (s) + H2O (ℓ)
2.Reaction to produce superphosphate fertiliser:
Ca3(PO4)2 (s) + H2SO4 (aq) → Ca(H2PO4)2 (aq) + CaSO4 (s)
3. Production of diborane, B2H6:
NaBH4 (s) + H2SO4 (aq) → B2H6 + H2 (g) + Na2SO4 (aq)
4. Decomposition of ammonium dichromate:
• (NH4)2Cr2O7 (s) →
N2 (g) + H2O (ℓ) + Cr2O3 (s)
5. Catalytic oxidation of ammonia:
• NH3 (g) + O2 (g)
→ NO (g)
+
H2O (ℓ)
6. CH3OH (ℓ) + O2 (g) → CO2 (g) + H2O (ℓ)
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Combustion Reactions ….
• If a third element is present – its mass can be
determined by subtracting the masses of C and H
from the original compound mass.
• Q1. Hexachlorophene, a compound made up of
carbon, hydrogen, chlorine and oxygen atoms, is an
ingredient used in germicidal soaps. Combustion of a
1.000 g sample of hexachlorophene yields 1.407 g
CO2, 0.1340 g H2O and 0.5228 g Cl2 gas
• a)What is the empirical formula of hexachlorophene?
• B) If the molar mass of hexachlorophene is 406.9
g.mol-1, what is its molecular formula?
→ determine moles of CO2 = moles of C
→ determine mass of C
→ determine moles of water
→ determine moles of H (moles of H2O x 2)
→determine mass of H
→ determine moles and mass of additional elements
→ subtract mass of C and H (and others) from the
original mass to determine mass of O
→ determine moles of O
→ determine mole ratios → empirical formula
→ compare molar masses → molecular formula
3.2 Introduction to Chemical Equilibrium
• Chemical reactions are reversible → many reactions
lead to incomplete conversion of reactants to
products.
e.g. reaction of nitrogen (N2) and hydrogen (H2) to form
ammonia (NH3).
•
N2 (g) +
3 H2 (g)
→
2 NH3 (g)
• As ammonia is produced, some breaks down to form
N2 and H2 (decomposition).
•
2 NH3 (g) →
N2 (g) +
3 H2 (g)
• At start: Reaction to produce NH3 predominates.
• As reactants are consumed, the rate of production of
NH3 slows down (forward reaction).
Product favoured reactions – reactants are completely /
largely converted into products at equilibrium, e.g.
combustion reactions.
• Reactant favoured – small amount of reactant
converted into product, e.g. ionisation of weak acids
(i.e. acetic acid).
• CH3CO2H (aq) +
H2O (ℓ)
⇔
CH3CO2+
(aq)
+
H3O (aq)
• Very small amount of acetate and hydronium ions
formed.
• Simultaneously, as NH3 concentration increases, rate
of decomposition increases (reverse reaction).
• Eventually, rate of forward reaction = rate of reverse
reaction.
→ system has reached chemical equilibrium.
→ reaction vessel will contain all 3 substances, i.e. N2,
H2 and NH3.
→ both forward and reverse reactions are still
occurring (at the same rates).
• i.e. Dynamic equilibrium …. Use double arrows ⇔
• N2 (g)
+
3 H2 (g)
⇔
2 NH3 (g)
3.4 Chemical Reactions in Aqueous Solution
4 major categories of reactions:
Precipitation
Acid – Base
Gas – Forming
Oxidation – reduction
Most are ion – exchange reactions → ions of
reactants change partners.
•
A+B- +
C+D- →
A+D- +
C+B•
•
•
•
•
•
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ion – exchange reactions ...
e.g. Ag+NO3- (aq) + K+Cℓ- (aq) → AgCℓ (s) +
K+NO3- (aq)
• BaCℓ2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HCℓ (aq)
• Recognising exchanges helps us to predict products
of precipitation, acid – base and gas forming
reactions.
• When ionic solids dissolve – each ion is separated
from the oppositely charged ions that surround it
• Interactions between the water and the ions allow
the solid to dissolve:
• All ionic compounds that are soluble in water are
electrolytes.
• NaCℓ
→ Na+ (aq) + Cℓ- (aq)
• 1 mol
1 mol
1 mol
= 2 moles of
ions
• 100% dissociation – strong electrolyte, i.e. solute
dissociated completely into ions = good conductor of
electricity.
• Ba(NO3)2 →
Ba2+ (aq)
+
2 NO3- (aq)
• NO3-: polyatomic ions stay together as a unit.
Solubility Guidelines - pg 122 Text book
Na+,
K+
+
• Soluble: All salts of
(alkali metals) and NH4
(ammonium)
• Salts of nitrates (NO3-), chlorates (CℓO3-), perchlorate
(CℓO4-), acetate (C2H3O2-)
• Cℓ-, Br-, I- salts except Ag+, Pb2+, Hg22+
• F- salts except group 2 metals and Pb2+
• SO42- salts except Ca2+, Sr2+, Ba2+, Pb2+
Ions are free to move about in the solution –
movement is generally random, i.e. positive and
negative ions dispersed uniformly.
• If electrodes are placed in solution and connected to
a power supply – positive ions (cations) move
towards the negative electrode and negative ions
(anions) move towards the positive electrode
→ Movement no longer random
→ ions are able to conduct charge in a solution.
• Compounds whose aqueous solutions conduct
charge are called electrolytes.
Weak electrolyte – conducts poorly
→
few ions – low concentration
•
– low solubility.
• Non-electrolytes – compounds whose aqueous
solutions do not conduct electricity.
→ particles are molecules not ions, e.g. glucose
(C6H12O6), ethylene glycol (C2H6O2).
• Many compounds dissolve in water but do not ionise
or dissociate. Most molecular compounds stay
together as a whole.
• Some molecular compounds i.e. acids and bases,
react with water and form ions → electrolytes.
Not all ionic substances dissolve completely – some
dissolve only to a small extent, others are insoluble.
• Insoluble: salts of CO32-, PO43-, oxalate (C2O42-),
chromate (CrO42-) except alkali metals & NH4+
• Metal sulphides (S2-), except alkali metals
• Most metal hydroxides and oxides, except alkali
metals, ammonium & Ba(OH)2
e.g. Ca(OH)2 is poorly soluble. From +/- 5 g into 100 mL
water, only 0.17 g dissolves @ 10 oC.
• Ca(OH)2 (s)
⇔
Ca2+ (aq)
+
2 OH- (aq)
• Most of Ca(OH)2 will remain in the solid form.
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3.6 Precipitation Reactions
• Produce water insoluble product
• Reactants generally water-soluble ionic compounds
(dissociate to give relevant cations and anions)
• React to form a solid precipitate by exchanging ions.
• e.g. Ag+NO3- (aq) + K+Cℓ- (aq) →
AgCℓ (s)
+
K+NO3- (aq)
• Decide which ions are formed in solution when
reactants dissolve.
• Determine whether a cation from one reactant will
form an insoluble product with an anion of another
reactant.
Spectator ions still necessary, but could be any soluble
anion or cation.
• Writing and balancing net ionic equations:
e.g. reaction between aqueous solutions of Iron(III)
chloride and potassium hydroxide to give iron(III)
hydroxide and potassium chloride.
→ Write balanced equaKon
e.g. sodium carbonate mixed with copper(II) chloride.
• Na2CO3 →
2 Na+ (aq) +
CO32- (aq)
• CuCℓ2
→
Cu (aq)
+
2 Cℓ- (aq)
+
•
→
CuCO3 (s) + 2 Na (aq) + 2 Cℓ- (aq)
+
• Na and Cℓ present in solution before and after the
reaction – spectator ions (do not participate).
• Simplify:
• Cu2+ (aq) + CO32- (aq) →
CuCO3 (s)
• = Net ionic reaction equation
• Only the ions and nonelectrolytes that take part in
the reaction need to be included in the net ionic
equation.
e.g. a solution of lead nitrate reacts with potassium
chromate:
Pb(NO3)2 (aq) + K2CrO4 (aq) →
Pb2+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + CrO42- (aq) →
Q. A solution of nickel(II) chloride is added to a solution
of ammonium sulphide:
• NiCℓ2 (aq) + (NH4)2S (aq) →
• Decide on the solubility of each compound.
• Identify the ions in solution. All soluble ionic
compounds dissociate = electrolytes.
Q1. A mixture contains solid sodium bromide and solid
sodium phosphate. 5.000 g of this mixture was
dissolved in a quantity of water. A solution of
magnesium nitrate was added, drop-wise, to the first
solution, until no further precipitate was formed. The
precipitate was filtered off, dried and weighed. Mass of
the precipitate was 2.731 g. Determine the % sodium
bromide in the mixture.
ACIDS and BASES
•
•
•
•
•
•
•
•
•
Acids have characteristic properties:
Change the colour of litmus from blue to red.
Neutralise the effect of a base.
Produce bubbles of CO2 gas when added to metal
carbonates:
CaCO3 (s) + 2 HCℓ (aq) → CO2 (g) + H2O (ℓ) + CaCℓ2(aq)
React with many metals to produce hydrogen gas (H2):
Zn (s) + H2SO4 (aq) → H2 (g) + ZnSO4 (aq)
React with metal hydroxides to form a salt and water:
NaOH (aq) + HNO3 (aq) → H2O (ℓ) + NaNO3 (aq)
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Bases:
Change the colour of litmus from red to blue.
Oppose the action of an acid, i.e. neutralises it.
Soluble bases are called alkalis.
Different definitions of ‘acid’ and ‘base’. The 2 most
common:
• Arrhenius
• Brønsted-Lowry
•
•
•
•
•
Strength of acid is related to the extent to which acid
ionises.
• Complete ionisation, e.g. HCℓ, HNO3, H2SO4 → strong
electrolytes, strong acids.
• Many acids, e.g. sulphuric acid, can provide more
than 1 mole of H+ ions per mole of acid = polyprotic.
• e.g. sulphuric acid, H2SO4 – diprotic, ionisation in 2
steps:
• Strong acid:
H2SO4 (ℓ) →
H+ (aq) + HSO4(aq)
100% ionised
• Weaker acid:
HSO4- (aq) →
H+ (aq) + SO42(aq)
< 100% ionised
A base is a substance that, when dissolved in water,
increases the concentration of hydroxide ions , OH-, in
the solution.
• – Hydroxide ion is characteristic of bases.
• Most metal hydroxides are insoluble except alkali
metals, NH4+ and Ba+.
• Water soluble hydroxide compounds, i.e. NaOH and
KOH, are strong bases and strong electrolytes.
• NaOH (s) →
Na+ (aq) + OH- (aq)
100% dissociated
Arrhenius – proposed that acids, bases and salts can
dissolve in water and forms ions → i.e. electrolytes.
– focuses on the formation of hydrogen ions (H+) and
hydroxide ions (OH-) in aqueous solutions.
• An acid is a substance that, when added to water,
increases the concentration of H+ - ions (in the form
of hydronium ions, H3O+ ions) in the solution.
• Hydrogen chloride gas, HCℓ (g), ionises in water to
form a hydronium ion and a chloride ion.
•
HCℓ (g) + H2O (ℓ) → H3O+ (aq) + Cℓ- (aq)
• Completely converted into ions in solution, i.e. HCℓ is
a strong acid and a strong electrolyte.
1st ionisation = complete, therefore strong acid and
strong electrolyte.
2nd ionisation = incomplete
Hydrogen sulphate ion (HSO4-) is only partially ionised in
aqueous solution = weak electrolyte and weak acid.
• Similarly, other acids, i.e. acetic (C2H3O2H), hydrofluoric
(HF), are incompletely ionised in aqueous solution.
CH3COOH (ℓ) + H2O (ℓ) ↔ CH3COO- (aq) + H3O+ (aq)
→ weak electrolyte and weak acids.
Many weak acids exist in solution primarily as molecules
– only a small fraction ionise to produce H+ (aq) ions and
corresponding anion.
• Ammonia (NH3) forms hydroxide ions as a result of
the reaction with water:
• NH3 (aq) + H2O (ℓ) ↔ NH4+ (aq) + OH- (aq) <
100% ionised
• Only a small concentration of ammonium and
hydroxide ions is present in a solution of NH3 – weak
electrolyte and weak base.
• Amines react in a similar way:
│
–N:
│
│
+
H2O →
–N–H
+ OH- (aq)
│
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Brønsted-Lowry theory – viewed acid-base reactions in
terms of the transfer of a proton (H+) from one species
to another.
describes acid – base reactions in terms of equilibria
→ helps to predict whether a reacKon is product
or reactant favoured, based on acid and base
strength.
• Acid is a proton donor.
• Base is a proton acceptor. This definition includes OHbut also allows the inclusion of other types of bases.
• An acid – base reaction is the transfer of a proton to
a base, to form a new acid and new base.
Some species are amphiprotic – function as either an
acid or a base, depending on the reaction.
e.g. H2O: with ammonia, water acts as the acid,
donating a proton.
• With HCℓ, acts as a base, accepting a proton.
Also HCO3- (aq):
• Acid: HCO3- (aq) ↔ CO32- (aq) + H+ (aq)
• Base: HCO3- (aq) + H+ (aq) ↔ H2CO3 (aq)
Written as an equilibrium reaction, with the
equilibrium favouring the weaker acid and base.
e.g. Hydrogen chloride – 100% ionised = strong acid.
Equilibrium favours products.
• HCℓ (aq) + H2O (ℓ) ↔
H3O+ (aq) + Cℓ- (aq)
• Acid
base
acid
base
• Strong electrolyte
• Ammonia – weak base: reactants favoured
• NH3 (aq) + H2O (ℓ) ↔ NH4+ (aq) + OH- (aq)
• Weak base acid
acid
base
•
Weak electrolyte << 100% ionised
Oxides of non-metals and metals.
• Acids – generally have one or more H – atoms in the
molecular formula that can ionise in water to form H+
- ions (H3O+).
• Less obvious compounds also form acidic solutions,
i.e. oxides of non-metals, CO2, SO2, NO2 – react with
water to form H+ - ions.
CO2 (g) + H2O (ℓ) ↔ H2CO3 (aq) ↔ H+ (aq) + HCO3- (aq)
• HCO3- (aq)
↔ CO32- (aq) + H+ (aq)
Oxides that react with water to produce H+ - ions are
acid oxides.
2 NO2 (g) + H2O (ℓ) ↔ HNO3 (aq) + HNO2 (aq)
Nitric acid nitrous acid
Oxides of metals are basic oxides – they form basic
solutions if they dissolve in water.
• CaO (s) +
H2O (ℓ)
→
Ca(OH)2
• Na2O (s) +
H2O (ℓ)
→
2 NaOH (aq)
• Acid or base?
– SeO2
– MgO
– P4O10
Q5.Write a balanced equation for the reaction when
phosphoric acid donates a proton to water.
• Write the net ionic equation showing the dihydrogen
phosphate ion acting as:
• a Brønsted acid in a reaction with water:
• a Brønsted base in a reaction with water:
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Reactions of Acids and Bases.
• Acids and bases in aqueous solution react to produce
a salt and water.
• HCℓ (aq) + NaOH (aq) → H2O (ℓ) + NaCℓ (aq)
• “Salt” – any ionic compound whose cation comes
from a base (Na+ from NaOH) and whose anion
comes from an acid (Cℓ- from HCℓ).
• Both HCℓ and NaOH are strong electrolytes in water,
so complete ionic equation.
Na+ and Cℓ- are on both sides of the equation
(spectator ions)
→ net ionic equaKon:
• H3O+ (aq) + OH- (aq) → 2 H2O (ℓ)
• This is always the net ionic equation when a strong
acid reacts with a strong base.
→ Neutralisation reaction i.e. on completion of
reaction, if the same number of moles of acid and base
are mixed, the solution is neutral.
• Other ions remain unchanged – Na+ (aq) + Cℓ- (aq)
• Evaporate the water – solid salt remains e.g. NaCℓ (s)
Acetic acid and sodium hydroxide:
3.8 Gas Forming Reactions
CH3CO2H (aq) + NaOH (aq) ↔ H2O (ℓ) + NaCH3CO2 (aq)
• Several different chemical reactions lead to gas
formation:
• Metal carbonates / bicarbonates + acid → metal
salt, carbon dioxide and water.
• Na2CO3 (aq) + 2 HCℓ (aq) → CO2 (g) + H2O (ℓ) + 2
NaCℓ (aq)
• NaHCO3 (aq) + HCℓ (aq) → CO2 (g) + H2O (ℓ) +
NaCℓ (aq)
Weak acid, partially ionised
Molecular species predominant
→ in ionic equaKons, aceKc acid wriVen as molecular,
CH3CO2H
• CH3CO2H (aq) + Na+ (aq) + OH- (aq) ↔ H2O (ℓ) +
Na+ (aq) + CH3CO2- (aq)
+
• Only Na (aq) ions are spectator ions.
• Net ionic equation:
• CH3CO2H (aq) + OH- (aq) ↔ H2O (ℓ) + CH3CO2- (aq)
• Metal sulphides + acid → hydrogen sulphide + metal
salt
• Na2S (aq) + 2 HCℓ (aq) → H2S (g) + 2 NaCℓ (aq)
• Most common are those leading to CO2
formation.
Q6. Barium carbonate is used in brick, ceramic and
glass manufacturing. Write a balanced, net ionic
equation, that shows what happens when barium
carbonate is treated with nitric acid:
1. Identify reactants, write formulas:
• Metal sulphites + acid → sulphur dioxide + water +
metal salt
• Na2SO3 (aq) + 2 HCℓ (aq) → SO2 (g) + H2O (ℓ) + 2
NaCℓ (aq)
• Ammonium salt + strong base → ammonia + water +
metal salt
• NH4Cℓ (aq) + NaOH (aq) → NH3 (g) + H2O (ℓ) +
NaCℓ (aq)
2. Recognise this as gas forming reaction between
metal carbonate and acid.
• Products:
• Anion of salt:
• Cation of salt:
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Q7. Ammonium sulphate reacts with sodium hydroxide:
Q8. Rhodocrosite, a red mineral, consists largely of
manganese II carbonate. Write an equation for the
reaction of the mineral with hydrochloric acid. Name
the products.
Q9. Sodium sulphite and acetic acid react.
Q10. Write a balanced, net ionic equation for the
following, and classify the reaction:
– Potassium carbonate reacts with copper II nitrate.
– Perchloric acid reacts with aluminium hydroxide
– Nitric acid reacts with sodium sulphide
• Metal loses its electrons to oxygen, forming an ionic
compound of the metal ion and oxide ion.
•
2 Ca (s) + O2 (g) → 2 CaO (s)
• As Ca is oxidised, oxygen is changed from neutral O2
to two negative O2- ions.
→ when an atom, ion or molecule has become more
negatively charged, i.e. gained electrons, it is
reduced.
•
Gain of electrons = reduction
• Oxidation and reduction have to go together – one
reactant loses electrons, another must gain them.
Can be shown by dividing the general redox
reaction into 2 parts or half reactions.
X + Y → Xn+ + YnX
→ Xn+ + neY + ne→ YnX transfers e- to Y: X is oxidised (reducing
agent)
• Y accepts e- from X: Y is reduced (oxidising
agent).
•
•
•
•
20.1 Oxidation – Reduction Reactions
→ A reacKon involving the transfer of one or more
electrons from one species to another.
• Corrosion of iron (rusting) involves the conversion of
solid metal into a metal compound due to the metal
reacting with substances in the environment.
→ when a metal corrodes it loses electrons and forms a
cation, becoming positively charged.
•
Loss of electrons = oxidation
•
4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)
For Fe (s) or Ca (s) to become oxidised – oxygen must
accept the electrons → oxygen is the oxidising agent.
For oxygen to become reduced – Fe (s) or Ca (s) must
donate the electrons → Fe / Ca is the reducing agent.
• Oxygen is not always involved.
• All redox reactions occur due to the transfer of
electrons between substances.
• In every oxidation – reduction reaction,
• one reactant is reduced (oxidising agent) and
• one reactant is oxidised (reducing agent).
e.g. React copper metal with an aqueous silver salt, i.e.
AgNO3 (aq) → Ag+ (aq) + NO3- (aq)
• 2 Ag+ (aq) + Cu (s) → 2 Ag (s) + Cu2+ (aq)
• Ag+ ions accept electrons from copper – reduced to
Ag (s). Ag+ is the oxidising agent.
• Ag+ (aq) + e→
Ag (s)
• Cu (s) donates electrons to Ag+ (aq) – oxidised to
Cu2+ (aq). Cu is the reducing agent.
• Cu (s)
→
Cu2+ (aq) + 2e-
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Need to keep track of the electrons being gained by the
substance being reduced, or lost by the substance
being oxidised.
• Concept of oxidation numbers or oxidation states is
used as a way of following electrons.
• Oxidation number of an atom in an ion or molecule is
the “charge” that the atom has, or appears to have.
• Oxidation = increase in oxidation number
• Reduction = decrease in oxidation number
Rules/ guidelines for assigning oxidation numbers:
• except (i) in peroxides (e.g. Na2O2 or H2O2) which
contain O22- ion – ON = -1;
• or (ii) superoxides, e.g. KO2 – ON = - ½
• Or (iii) When combined with F (oxygen takes on a
positive ON)
• ON of H is +1 when bonded to non-metals and -1
forming binary compounds with metals (metal ion is
positive), e.g. CaH2
• Algebraic sum of oxidation numbers of all atoms in
neutral compound is zero.
• In polyatomic ion, sum of ON = charge of the ion.
• e.g. HCℓO4
H = +1; O = -2; Cℓ = ?
• PO43- : P = ?
• Determine the ON of sulphur in each of the
following:
• H2S;
S8;
SCℓ2; Na2SO3;
SO42-
Recognising oxidation – reduction reactions:
e.g. 2 Na (s) + Cℓ2 (g) →
2 NaCℓ (s)
• N in NO3- changes from x + 3(-2) = -1, x = +5
• to N in NO2: x + 2(-2) = 0, x = +4
Decrease in ON, therefore reduction (oxidising agent).
•
NO3- (aq) + 1 e- →
NO2 (g)
Na → Na+ + 1e- Cℓ2 + 2e- → 2 Cℓ-
Cu (s) + 2 NO3- (aq) + 4 H3O+ (aq) → Cu2+ (aq) + 2 NO2 (g)
+ 6 H2O (ℓ)
• If an un-combined element is converted into a
compound.
• If a well-known oxidising agent or reducing agent is
involved e.g. O2; Br2; Cℓ2; F2
• Cu (s):
ON = 0 whereas Cu2+ (aq): ON = +2 ;
therefore oxidation.
• Cu (s)
→
Cu2+ (aq) + 2e(reducing agent)
• Atom in elemental form / pure element – ON is zero;
e.g. H atom in H2; P in P4; Cu
• Monatomic ions – ON = charge on the ion; e.g. K+ =
+1; S2- = -2; Mg2+ = +2
• ON of fluorine is -1 in all compounds.
• Cℓ, Br, I: ON = -1 in compounds except when
combined with oxygen or fluorine. Cℓ in NaCℓ = -1;
Cℓ in CℓO- = +1
• ON of oxygen usually -2 in both molecular and ionic
compounds .....
• Oxidation number of an element depends on the
compound in which it occurs:
• P2O5
NaH
Cr2O72SnBr4
N2O
• Which of the following are redox reactions? Identify
the oxidising and reducing agents.
• NaOH (aq) + HNO3 (aq) → NaNO3 (aq) + H2O (ℓ)
• Cu (s) + Cℓ2 (g)
→ CuCℓ2 (aq)
2• 2 S2O3 (aq) + I2 (g) → S4O62- (aq) + 2 I- (aq)
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20.1 Balancing Oxidation – Reduction Equations:
• Redox reactions must be balanced for both mass and
charge.
• Same number of atoms appear in both reactants and
products.
• Sum of electric charges of all species must be the
same on each side of equation arrow, (i.e. gains and
losses in e-s must be balanced – number of electrons
lost in oxidation = number of electrons gained in
reduction).
• Cu (s) + 2 Ag+ (aq)
→
Cu2+ (aq) + 2 Ag (s)
• Separate the redox reactions into 2 half-reactions –
one for oxidation and one for reduction.
• Each half-reaction is balanced for mass and charge.
• For net chemical reaction – add the two halfreactions.
Sn2+(aq) + Fe3+(aq) → Sn4+(aq) + Fe2+(aq)
In acid medium :
• Determine oxidation numbers
• Divide into half reactions
• Ensure number of electrons lost in oxidation =
number of electrons gained in reduction.
• Oxidation – electrons are products
• Reduction – electrons are reactants.
• Reactions involving oxoanions, e.g. SO42-, NO3-, CℓO-,
CrO42-, and organic compounds in aqueous solutions
often require the addition of water (H2O), H+ (aq) in
acidic solution, or OH- (aq) in basic solution, for
balancing.
• Assign oxidation states to see which atoms are
gaining electrons and which are losing electrons.
• Divide the equation into 2 half reactions – one
oxidation and one reduction.
• Balance each half reaction:
• - balance elements other than H and O
• - balance O by adding H2O as needed
• - balance H by adding H+ as needed
• - balance the charge by adding electrons as needed,
(check no. of electrons in each step corresponds to
the change in ON determined in step 1).
In acid medium :
e.g. The reaction between permanganate (MnO4-) and
oxalate (C2O42-) in acidic solution produces CO2 and
Mn2+.
• If necessary, multiply the half reactions by integers so
that number of electrons lost in one half reaction =
number of electrons gained in the other.
• Add the 2 half reactions and simplify by removing
species appearing as both reactants and products.
• Check the number of atoms of each kind on the left
side of the equation is the same as on the right hand
side.
• Check total charge on the left = total charge on the
right.
11. Cr2O72- (aq) + Cℓ- (aq) → Cr3+ (aq) + Cℓ2 (g)
12. Mn2+ (aq) + NaBiO3 (s) → Bi3+ (aq) + MnO4- (aq)
13. Cℓ2 (g) + S2O3 (aq) → Cℓ- (aq) + SO42- (aq)
14. I2 + CℓO- → IO3- + Cℓ15. I2
→
I- + IO3-
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In basic solution:
• Use OH- and H2O
• e.g.
Aℓ (s) + S (s) → Aℓ(OH)3 (s) + HS- (aq)
• ON:
0
0
+3
-2
•
e.g. CN- (aq) + MnO4- (aq) → CNO- (aq)
ON:
-1
+7
+1
+ MnO2 (s)
+4
Q16. Test for nitrates using Devarda’s alloy in strong
alkaline solution:
NO3- (aq) + Aℓ (s) →
NH3 (g) + AℓO2-
Q17. Balance the following in basic medium
Br2 (ℓ) → BrO3- (aq) + Br- (aq)
-
-
Q18. H2O2 + Cℓ2O7 → CℓO2 + O2
Or may balance as per acid solution and then
“neutralise” the H+ using OH• e.g. CN- (aq) + MnO4- (aq) → CNO- (aq) + MnO2 (s)
• ON: -1
CN-
+7
+1
CNO-
H+
+4
e-
•
(aq) + H2O →
+ 2
+ 2
(x3)
• MnO4- + 4 H+ + 3 e- →
MnO2 + 2 H2O
(x2)
• 3 CN- + 3 H2O + 2 MnO4- + 8 H+ → 3 CNO- + 6 H+
+ 2 MnO2 + 4 H2O
• Simplify:
3 CN- + 2 MnO4- + 2 H+ → 3 CNO- + 2 MnO2 + H2O
Neutralise and simplify:
3 CN- + 2 MnO4- + H2O → 3 CNO- + 2 MnO2 + 2 OH-
Ionic and Molecular equations:
e.g. Balance the following reaction using half
reactions:
K2SO3 (aq) + KMnO4 (aq) + H2SO4 (aq) →
K2SO4 (aq) + MnSO4 (aq) + H2O (ℓ)
Q19. N2H4 (g) + KCℓO4 → NO (g) + KCℓ (aq) +
H2O (ℓ) (Basic medium)
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