21/3/2023 Dr.

Wafa Hatim Balla

Lec (7)
Chemical reactions

Chemical reactions
❖ Chemical reaction is a process that leads to the chemical transformation of one
set of chemical substances to another.
Types of chemical reaction:
1. Acid-Base Neutralization:
✓ A neutralization reaction is a reaction between an acid and a base.
✓ Generally, aqueous acid-base reactions produce water and a salt, which is an
ionic compound made up of a cation other than H+ and an anion other than
OH- or O2- : acid + base→ salt + water
✓ For example, when a HCl solution is mixed with a NaOH solution, the
following reaction occurs: HCl(aq) + NaOH(aq) →NaCl(aq) +H2O(l)
2. Precipitation Reactions:
✓ In which, two solutions of soluble salts are mixed resulting in an insoluble
solid (precipitate) forming.
✓ For example, when an aqueous solution of lead(II) nitrate [Pb(NO3)2] is added
to an aqueous solution of potassium iodide (KI), a yellow precipitate of lead
iodide (PbI2) is formed: Pb(NO3)2(aq) + 2KI(aq) →PbI2(s) + 2KNO3(aq)
3. Oxidation-Reduction Reactions:
A redox reaction (reduction and oxidation reaction) is a reaction in which there is a
transfer of electrons. Reduction and oxidation always happen at the same time.

Oxidation Number: The charge of the atom would have in a molecule (or an ionic
compound) if electrons were completely transfer.

➢ Rules for Assigning an Oxidation Number

❖ General Rules
1. For an atom in its elemental form (Na, O2): O.N. = 0
2. For a monatomic ion: O.N. = ion charge
3. The sum of O.N. values for the atoms in a molecule or formula unit of a
compound equals to zero, Or equals to the ion’s charge if it is a polyatomic
ion)
❖ Rules for Specific Atoms or Periodic Table Groups
1. For Group 1A (1) :O. N. = +1 in all compounds
2. For Group 2A (2) :O. N. = +2 in all compounds
3. For hydrogen: O.N. = +1 in combination with nonmetals O.N. = -1 in
combination with metals and boron
4. For fluorine: O.N. = -1 in all compounds
5. For oxygen: O.N.= -1 in peroxides O.N. = -2 in all other compounds (except
with F)
6. For Group 7A (17): O.N. = -1 in combination with metals, nonmetals (except
O), and other halogens lower in the group
Example (1): Determine the oxidation number (O.N.) of each element in these
compounds: a) CaO (s) b) KNO3 (s) d) CaCO3 (s) e) N2 (g)
Solution:

Definitions:
Oxidation: is the addition of oxygen or removal of hydrogen/electron, results in an
increase in Oxidation number. e.g. Cu+ → Cu2+ + e-
Reduction: is the addition of hydrogen/electrons or removal of oxygen, results in a
decrease in Oxidation number. e.g. Fe3+ + e- → Fe2+
The oxidizing agent is the species doing the
oxidizing by takes electrons. The oxidizing
agent is therefore reduced.
The reducing agent is the species doing the
reducing by donates electrons. The reducing
agent is therefore oxidized.
❖ Balancing redox equations:
1. Determine which species is oxidized and which species is reduced,
2. Write half reactions for oxidation and reduction processes
✓ Oxidation reaction will have e-’s on the right side of equation
✓ Reduction reaction will have e-’s on the left side of the equation
3. Balance half reactions including charge balance
4. Multiply each half reaction by a factor so that the total number of e-’s
transferred in each reaction are equal
5. Add the resulting half reactions together to get the overall balanced redox
equation

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Example (2):
Balance the following equation:
𝟑+
𝑭𝟐(𝒈) + 𝑨𝒍(𝒔) →𝑭− (𝒂𝒒) + 𝑨𝒍(𝒂𝒒)
oxid. state 0 0 -1 +3
Fluorine is reduced (0 → -1), Aluminum is oxidized (0 → +3)
Half reactions:
oxidation: 𝑨𝒍(𝒔) → 𝑨𝒍𝟑+ − − −
(𝒂𝒒) + 𝟑𝒆 , reduction: 𝑭𝟐(𝒈) + 𝟐𝒆 → 2𝑭(𝒂𝒒)
Balance e-’s transferred:
Multiply oxidation rxn by 2: 𝟐𝑨𝒍(𝒔) → 𝟐𝑨𝒍𝟑+ (𝒂𝒒) + 𝟔𝒆
−

Multiply reduction rxn by 3: 𝟑𝑭𝟐(𝒈) + 𝟔𝒆− → 6𝑭− (𝒂𝒒)

𝟑+
Add half reaction to get net reaction: 𝟑𝑭𝟐(𝒈) + 𝟐𝑨𝒍(𝒔) →𝟔𝑭−
(𝒂𝒒) + 𝟐𝑨𝒍(𝒂𝒒)

❖ Balancing Oxidation-Reduction Equations in Acidic Solution:

1. Divide the “skeleton” equation into two separate reduction and oxidation “half-
reactions”.
2. For each half-reaction:
a) Balance all elements except hydrogen and oxygen.
b) Balance oxygen by adding water (H2 O(l) )
+
c) Balance hydrogen by adding 𝐻(𝑎𝑞) ion.
d) Balance charge by adding electrons.
3. If required, multiply each balanced half-reaction by the smallest whole number
necessary to equalize the number of electrons in the two half reactions.
4. Sum the two half-reactions to obtain the overall equation. If possible, simplify
the overall equation by cancelling like species.
5. Check that the elements and charges balance on both sides of the equation.

Example (3): Balancing Redox Equations for Reactions Run in Acidic

𝟑+
Conditions: 𝑪𝒓𝟐 𝑶𝟕 𝟐− −
(𝒂𝒒) + 𝑯𝑵𝑶𝟐 (𝒂𝒒) → 𝑪𝒓(𝒂𝒒) + 𝑵𝑶𝟑(𝒂𝒒) (acidic)
oxid. State +6 +3 +3 +5
✓ Step #1: Write the skeletons of the oxidation and reduction half-reactions.
2−
Oxidation: HNO2 → NO− 3 , and Reduction: Cr2 O7 → Cr
3+

Step #2a: Balance all elements other than H and O.

2−
Oxidation: HNO2 → NO− 3 , and Reduction: Cr2 O7 → 2 Cr
3+

Step # 2b: Balance the oxygen atoms by adding H2O molecules on the side of the
arrow where O atoms are needed.
2−
Oxidation: HNO2 + H2 O → NO− 3 , and Reduction: Cr2 O7 → 2 Cr
3+
+7 H2 O

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Step # 2c: Balance the hydrogen atoms by adding H+ ions on the side of the arrow
where H atoms are needed.
Oxidation: HNO2 + H2 O → NO− +
3 + 3H , and
Reduction: Cr2 O2− +
7 + 14H → 2 Cr
3+
+7 H2 O
Step # 2d: Balance the charge by adding electrons, e-.
❖ Oxidation: HNO2 + H2 O → NO− 3 + 3H ,
+
Sum of charges is zero sum of charges is +2
• If we add two electrons to the right side, the sum of the charges on each side
of the equation becomes zero. 𝑯𝑵𝑶𝟐 + 𝑯𝟐 𝑶 → 𝑵𝑶− +
𝟑 + 𝟑𝑯 + 𝟐𝒆
−

❖ Reduction: Cr2 O2− +

7 + 14H → 2 Cr
3+
+7 H2 O
Sum of charges is +12 sum of charges is +6
• If we add six electrons to the left side, the sum of the charges on each side of
the equation becomes +6.
𝟔𝒆− + 𝑪𝒓𝟐 𝑶𝟐− +
𝟕 + 𝟏𝟒𝑯 → 𝟐 𝑪𝒓
𝟑+
+𝟕 𝑯𝟐 𝑶

✓ Step 3#: For the chromium half-reaction to gain six electrons, the nitrogen half-
reaction must lose six electrons. Thus, we multiply the coefficients in the
nitrogen half-reaction by 3.
𝟑𝑯𝑵𝑶𝟐 + 𝟑𝑯𝟐 𝑶 → 𝟑𝑵𝑶− +
𝟑 + 𝟗𝑯 + 𝟔𝒆
−

Step 4#: Add the 2 half-reactions as if they were mathematical equations.

𝟑𝑯𝑵𝑶𝟐 + 𝟑𝑯𝟐 𝑶 → 𝟑𝑵𝑶− +
𝟑 + 𝟗𝑯 + 𝟔𝒆
−

𝟔𝒆− + 𝑪𝒓𝟐 𝑶𝟐− +

𝟕 + 𝟏𝟒𝑯 → 𝟐 𝑪𝒓
𝟑+
+𝟕 𝑯𝟐 𝑶
𝟐− 𝟑+ −
𝑪𝒓𝟐 𝑶𝟕 (𝒂𝒒) + 𝟑𝑯𝑵𝑶𝟐 (𝒂𝒒) + 𝟓 𝑯+ → 𝟐 𝑪𝒓(𝒂𝒒) + 𝟑 𝑵𝑶𝟑(𝒂𝒒) + 𝟒 𝑯𝟐 𝑶(𝒍)

Step 5#: Check that the elements and charges balance on both sides of the equation.
The atoms in our example balance and the sum of the charges is +3 on each side,
so our equation is correctly balanced.
𝟑+
𝑪𝒓𝟐 𝑶𝟕 𝟐− + −
(𝒂𝒒) + 𝟑𝑯𝑵𝑶𝟐 (𝒂𝒒) + 𝟓 𝑯 → 𝟐 𝑪𝒓(𝒂𝒒) + 𝟑 𝑵𝑶𝟑(𝒂𝒒) + 𝟒 𝑯𝟐 𝑶(𝒍)

❖ Balancing Oxidation-Reduction Equations in Basic Solution:

1. Complete steps 1-5 described above for an acidic solution.
− +
2. Add one 𝑂𝐻(𝑎𝑞) ion to both sides of the overall equation for every 𝐻(𝑎𝑞) ion
+ −
present. The 𝐻(𝑎𝑞) ions on one side combine with the added 𝑂𝐻(𝑎𝑞) ions to form
−
H2O(l), and 𝑂𝐻(𝑎𝑞) ions appear on the other side of the equation.
+ −
𝐻(𝑎𝑞) + 𝑂𝐻(𝑎𝑞) → H2 O(l)
3. If possible, simplify the resulting overall equation by cancelling like species.
4. Check that the elements and charges balance on both sides of the equation.

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Example (4): Balancing Redox Equations for Reactions Run in Basic
Conditions:
𝟑+
𝑪𝒓𝟐 𝑶𝟕 𝟐− −
(𝒂𝒒) + 𝑯𝑵𝑶𝟐 (𝒂𝒒) → 𝑪𝒓(𝒂𝒒) + 𝑵𝑶𝟑(𝒂𝒒) (Basic)
1. Complete steps 1-5 described above for an acidic solution.
𝟑+
𝑪𝒓𝟐 𝑶𝟕 𝟐− + −
(𝒂𝒒) + 𝟑𝑯𝑵𝑶𝟐 (𝒂𝒒) + 𝟓 𝑯 → 𝟐 𝑪𝒓(𝒂𝒒) + 𝟑 𝑵𝑶𝟑(𝒂𝒒) + 𝟒 𝑯𝟐 𝑶(𝒍)
2. For each H+ ion, add an OH– ion to both sides of the half-reaction.
𝟐− −
𝑪𝒓𝟐 𝑶𝟕 (𝒂𝒒) + 𝟑𝑯𝑵𝑶𝟐 (𝒂𝒒) + 𝟓 𝑯+ + 5𝑂𝐻(𝑎𝑞) → 𝟐 𝑪𝒓𝟑+ − −
(𝒂𝒒) + 𝟑 𝑵𝑶𝟑(𝒂𝒒) + 𝟒 𝑯𝟐 𝑶(𝒍) +5𝑂𝐻(𝑎𝑞)
Whenever H+ and OH– appear on the same side, combine them to make H2O as
shown below:
𝟐−
𝑪𝒓𝟐 𝑶𝟕 (𝒂𝒒) + 𝟑𝑯𝑵𝑶𝟐 (𝒂𝒒) + 𝟓 𝑯𝟐 𝑶(𝒍) → 𝟐 𝑪𝒓𝟑+ − −
(𝒂𝒒) + 𝟑 𝑵𝑶𝟑(𝒂𝒒) + 𝟒 𝑯𝟐 𝑶(𝒍) +5𝑂𝐻(𝑎𝑞)
3. Removing 4H2O (l) from both sides gives the final balanced equation in basic
medium:
𝟑+
𝑪𝒓𝟐 𝑶𝟕 𝟐− − −
(𝒂𝒒) + 𝟑𝑯𝑵𝑶𝟐 (𝒂𝒒) + 𝑯𝟐 𝑶(𝒍) → 𝟐 𝑪𝒓(𝒂𝒒) + 𝟑 𝑵𝑶𝟑(𝒂𝒒) + 5𝑂𝐻(𝑎𝑞)

Strengths of Oxidizing and Reducing Agents:

The relative strengths of oxidizing and reducing agents can be inferred from their
standard electrode potentials.

Electrochemical Cells:
• When two half reactions are
connected, we get an
electrochemical cell that can
generate a voltage potential and
electrical current
• Each half reaction has an
electrical potential, E
• Electrical potential is a measure of
how easily a species is reduced.
• The emf (electromotive force) of a
cell is determined by taking the
difference between the potentials of the cathode and the anode:
Ecell = Ecathode – Eanode
• The electromotive force of a galvanic cell built from a standard reference
electrode and another electrode to be characterized.
• By convention, the reference electrode is the Standard Hydrogen Electrode
(SHE).
❖ Standard Electrode Potentials:
• The standard electrode potential (E°half-cell) is the potential of a given half-
reaction when all components are in their standard states.
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 ✓ Gas phase species have a pressure of 1 atm
✓ Aqueous species have a concentration of 1 M
✓ Temperature 25℃
• All standard electrode potentials refer to the half-reaction written as a
reduction. For example, the standard reduction potentials for the 𝐹𝑒 2+ ∕ 𝐻 +
half reactions are:

Example (5):
In a galvanic cell consisting of a SHE and Cu2+ /Cu
half-cell determine the standard reduction
potential for Cu2+. In cell notation, the reaction is:
Pt(s)│H2 (g, 1 atm)│2H+ (aq,1 M) ║ Cu2+ (aq,
1M)│Cu(s)
Solution:
✓ Electrons flow from the anode to the cathode.
✓ The reactions, which are reversible, are:
Anode (oxidation): H2(g)⟶2H+(aq) + 2e− E°anode=0.00V
Cathode (reduction): Cu2+(aq)+2e−⟶Cu(s) E°cathode = +0.34 V
Overall: Cu2+ (aq) + H2 (g) ⟶ 2H+ (aq) + Cu(s)
E°cell = E°cathode - E°anode
E°cell = E° Cu2+/Cu - E°H+/H2 , +0.34 V= E° Cu2+/Cu - 0.00 V
E° Cu2+/Cu = +0.34
The electrochemical series:
• The electrochemical series is
built up by arranging various
redox equilibria in order of
their standard electrode
potentials (redox potentials).
• The most powerful oxidizing
agents lie at the top of
electrochemical series (High
positive 𝐸°red values) and
• The most powerful reducing
agents are present at the bottom
(High negative 𝐸°red values).

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Types of Chemical analysis
Analytical chemistry can be broken down into
two general areas of analysis:
1. Qualitative analysis
2. Quantitative Analysis
Each one of these two types can be either
classical or instrumental as we will see.

Qualitative Analysis:
This analysis detects (identify) the type of all or some of the substances present in
the sample (elements or ions or compounds).
Quantitative Analysis:
This analysis determines the amount of all or some of the substances present in the
sample.
Example of quantitative analysis:
➢ Titrimetric Methods of Analysis (Volumetric analysis):
The term titrimetric analysis refers to quantitative chemical analysis carried out by
determining the volume.
❖ Titration: a method to determine the quantity of a reagent (known
concentration) required to react with a known volume of sample (unknown
concentration).
❖ In titrimetric analysis the reagent of known concentration is called titrant and
the substance being titrated is termed the titrand.
Apparatus of Titrimetry analysis:

Measuring the Volume in the Burette

TERMS USED IN TITRIMETRY:
Equivalence point: a point in a titration when the mole of the titrant is equivalent
to the mole of the analyte
Indicator: reagent added to the analyte solution to produce an observable physical
change at the endpoint or near the equivalence point.
❖ Standard solution: a reagent of accurately known concentration used to carry
out a titrimetric analysis.
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Solution and their concentrations:
Concentration is a general measurement unit stating the amount of solute present
in a known amount of solution
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
Concentration =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
There is various expression of concentration as molarity, weight percent, volume
percent, weight-to-volume percent, parts per million, and parts per billion.
❖ Molarity (moles/L, or M):
• Molarity: The number of moles of solute per liter of solution (M).
𝒏𝒐. 𝒐𝒇 𝒎𝒐𝒍𝒆
𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 =
𝑳𝒊𝒕𝒓𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
Example (6):
Find the concentration in Molarity (M) of 12.00 g of benzene (C6H6) dissolved
up to a total volume of 250.00 ml in hexane.
Solution:
𝑛𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒 𝑤𝑡
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = , no. of mole =
𝐿𝑖𝑡𝑟𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑀𝑤𝑡
• Mwt of benzene=6*(12)+6*(1) =78 g/mol
𝑤𝑡 12 𝑔
• no. of mole = = 𝑔 = 0.154 𝑚𝑜𝑙
𝑀𝑤𝑡 78
𝑚𝑜𝑙
1𝐿
• Volume in Litre=250 𝑚𝐿 × = 0.0250 𝐿
1000 𝑚𝐿
0.154 𝑚𝑜𝑙
• Conc.𝐶6 𝐻6 = = 0.6114 𝑀
0.0250 𝐿

❖ Preparation of Solution:
weigh a calculated quantity of the reagent and dissolve it in a solvent then transferred
it to a needed volume of volumetric flask.

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Example (7):
Describe how to prepare 500 mL of approximately 0.20 M NaOH using solid
NaOH?
Solution:
The reagent is NaOH and solvent is H2O,
The molarity of solution is 0.20 M NaOH, the volume needed is 500 mL
𝑛𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = ,
𝐿𝑖𝑡𝑟𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒 = 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 × 𝐿𝑖𝑡𝑟𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 0.2mol/L x 0.5L = 0.1mol
𝑤𝑡
no. of mole = , The molecular weight of NaOH =23+16+1= 40 g/mol
𝑀𝑤𝑡
The weight of NaOH = no. of mole x 𝑀𝑤𝑡= 0.1 mol x 40 g/ mol = 4 g
Weight 4 g from NaOH and dissolve it in the water using 500 mL volumetric flask.

❖ Dilution of Solution:
• Dilution is the process whereby the concentration of a solution is lessened by
the addition of solvent.
• Dilute solution (with low concentration) can be prepared from a more
concentrated solution. A known volume of the concentrated solution can be
transferred into a new flask and diluted to the required volume or weight.
𝑴 𝟏 𝑽𝟏 = 𝑴 𝟐 𝑽𝟐

Example (8):
What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH?
Solution:
𝑉2 and 𝑀2 are given the volume and concentration of the diluted solution. The
concentration of a stock solution is 𝑀1 .The volume of a stock solution, 𝑉1 is
unknown. 𝐌𝟏 𝐕𝟏 = 𝐌𝟐 𝐕𝟐
𝑀2 𝑉2 (0.100𝑀)(5.00 𝐿)
𝑉1 = = = 0.314 𝐿. Thus, we would need 0.314 L of the 1.59 M
𝑀1 1.59𝑀
solution to prepare the desired solution.
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