Analytical Chemistry Final Revision Sheet
15.4: Strength of Acids and Bases
 Strength of acids
Questions:
Q) Explain what is meant by strength of an acid?
Q) Create a list of strong acids and weak acids?
Q) Classify each of the following species as a weak or strong acid:
(a) HNO3
(b) HF
(c) H2SO4
(d) HSO4
(e) H2CO3
(f) HCO3
(g) HCl
(h) HCN
(i) HNO2
Q) Investigate the dissociation of strong acids and weak acids.
Q) Calculate the pH of a 1.8x10-7 M Ba(OH)2 solution.
Q) Calculate the pH of
(a) a 1.0 x 10-3 M HCl solution
(b) 0.020 M Ba(OH)2 solution
Solution
Keep in mind that HCl is a strong acid and Ba(OH)2 is a strong base. Thus, these species are
completely ionized and no HCl or Ba(OH)2 will be left in solution.
(a) The ionization of HCl is
HCl(aq)
H+(aq) + Cl-(aq)
The concentrations of all the species (HCl, H+, and Cl-) before and after ionization can be
represented as follows:
A positive (+) change represents an increase and a negative (-) change indicates a decrease in
concentration. Thus,
[H+] = 1.0 x 10-3 M
pH = -log (1.0 x 10-3)
= 3.00
(b) Ba(OH)2 is a strong base; each Ba(OH)2 unit produces two OH- ions:
Ba(OH)2(aq)
Ba2+(aq) + 2OH-(aq)
The changes in the concentrations of all the species can be represented as follows:
Thus,
[OH-] = 0.040 M
pOH = -log 0.040 = 1.40
Therefore, from Equation (15.8),
pH = 14.00 - pOH
= 14.00 - 1.40 = 12.60
Check Note that in both (a) and (b) we have neglected the contribution of the autoionization of
water to [H+] and [OH-] because 1.0 x 10-7 M is so small compared with 1.0 x 10-3 M and
0.040M.
 Strength of bases
Properties of conjugate acid-base pairs
Consider the following equations
 HCl + H2O
H3O+ + ClIf an acid is strong, its conjugate base has no measurable strength. Thus the Cl- ion, which is
the conjugate base of the strong acid HCl, is an extremely weak base.
 F- + H2O
OH- + HF
OH- is the strongest conjugate base of an acid H2O and HF is the conjugate acid of F- base.
Question:
Q) Create a list of strong and weak bases?
Q) Predict the direction of the following reaction in aqueous solution:
HNO2(aq) + CN-(aq)
HCN(aq) +
(aq)
Solution:
The problem is to determine whether, at equilibrium, the reaction will be shifted to the right,
favoring HCN and
, or to the left, favoring HNO2 and CN-. Which of the two is a stronger
acid and hence a stronger proton donor: HNO2 or HCN? Which of the two is a stronger base
and hence a stronger proton acceptor: CN- or
? Remember that the stronger the acid, the
weaker its conjugate base.
In Table 15.2 we see that HNO2 is a stronger acid than HCN. Thus, CN- is a stronger base
than
. The net reaction will proceed from left to right as written because HNO2 is a better
proton donor than HCN (and CN- is a better proton acceptor than ).
Q) Predict the direction of the given reaction:
F- + H2O
HF + OH-
15.8: Diprotic and Polyprotic Acids
Questions:
Q) Guess What are some diprotic and Polyprotic acids?
Q) Compare and contrast diprotic and triprotic or Polyprotic acids?
Q) Carbonic acid is a diprotic acid. Explain what that mean?
Q) How does the dissociation of an acid determines its strength?
Q) Give some examples of diprotic and Polyprotic acids?
15.9: Molecular Structure and Strength of Acids
Hydrohalic Acids
The Halogens form a series of binary acids called the Hydrohalic acids. Analyse
the table below:
Table 15.6 shows that HF has the highest bond enthalpy and HI has the lowest.
Based on bond enthalpy HI should be the strongest acid because it is easiest to
break the bond and form the H+ and I- ions.
The fact that HI is a strong acid and HF is a weak acid indicated that bond enthalpy
is the main factor in determining the acid strength of binary acids. The weaker the
bond, the stronger the acid so that the strength of the acids increases as follows:
HF < HCl < HBr < HI
Oxoacids
It contains hydrogen, oxygen and one other element which occupies a central
position.
To compare their strength, divide the oxoacids into two groups.
 Oxoacids having different central atoms that are from the same group of the
periodic table and have the same oxidation number. Within this group, acid
strength increases with increasing electronegativity of the central atom, as
HClO3 and HBrO3
Cl and Br have the same oxidation number +5. However because Cl is more
electronegative than Br, thus the relative strengths are
HClO3 > HBrO3
 Oxoacids having the same central atom (Z) but different numbers of attached
groups. Acid strength increases as the oxidation number of Z increases
HClO4 > HClO3 > HClO2 > HClO
Carboxylic acids
The strength of carboxylic acids depend on the nature of the R group. Consider for
example acetic acid and chloroacetic acid:
The presence of the electronegative Cl atom in chloroacetic acid shifts electron
density toward the R group, thereby making the OH bond more polar.
Consequently there is a greater tendency for the acid to ionize.
Questions:
Q) List four factors that affect the strength of an acid?
Q) Oxoacids: Predict the relative strengths of the oxoacids in each of the
following groups:
 HClO, HBrO, and HIO
 HNO3 and HNO2
Solution:
Examine the molecular structure. In (a) the two acids have similar structure but
differ only in the central atom (Cl, Br, and I). Which central atom is the most
electronegative? In (b) the acids have the same central atom (N) but differ in the
number of O atoms. What is the oxidation number of N in each of these two acids?
(a) These acids all have the same structure, and the halogens all have the same
oxidation number (+1). Because the electronegativity decreases from Cl to
I,the Cl atom attracts the electron pair it shares with the O atom to the
greatest extent. Consequently, the O−H bond is the most polar in HClO and
least polar in HIO. Thus, the acid strength decreases as follows:
HClO > HBrO > HIO
(b) The structures of HNO3 and HNO2 are shown in Figure 15.5. Because the
oxidation number of N is +5 in HNO3 and +3 in HNO2, HNO3 is a stronger
acid than HNO2.
Q) Predict the acid strength of the following compounds: H2O, H2S, H2Se.
Q) Which of the following acids is weaker: HClO2, HClO3.
Q) Compare the strengths of the following pairs of acids:
(a) H2SO4 and H2SeO4
(b) H3PO4 and H3AsO4
Q) Which of the following is the stronger acid: CH2ClCOOH or CHCl2COOH?
Explain your choice.
15.10: Acid-Base Properties of Salts
 Salt that produces Neutral solutions:
A salt that is derived from the reaction of a strong acid with a strong base forms a
solution that has a pH of 7. An example is sodium chloride, formed from the
neutralization of HCl by NaOH. A solution of NaCl in water has no acidic or basic
properties.
Another example
A salt formed by the reaction of NaOH with HNO3, dissolves in water, it
dissociates completely as follows
NaOH + HNO3 ________ NaNO3 +H2O
NaNO3 ___H2O____ Na+
+ NO3 Salt that produces Basic solutions:
Salts that are derived from the neutralization of a weak acid (HF) by a strong base
(NaOH) will always produce salt solutions that are basic.
HF + NaOH ________ NaF + H2O
Another example
The dissociation of sodium acetate(CH3COONa) in water is given by
CH3COOH +NaOH __________ CH3COONa +H2O
CH3COONa ____H20___ Na+
+ CH3COOThe hydrated Na+ ion has no acidic or basic properties. The acetate ion however is
the conjugate base of weak acid CH3COOH and therefore has an affinity for H+
ions.
CH3COO- + H2O ________CH3COOH +OH Acidic Solutions:
Salts derived from a strong acid and a weak base.
When a salt derived from a strong acid such as HCl and a weak base such as NH3
dissolves in water, the solution becomes acidic. For example, consider the process
HCl +NH3 ______ NH4Cl
Questions:
Q) Predict the acid strength of the following compounds: H2O, H2S, H2Se.
Q) Which of the following acids is weaker: HClO2, HClO3.
Q) Compare the strengths of the following pairs of acids:
(a) H2SO4 and H2SeO4,
(b) H3PO4 and H3AsO4.
Q) Which of the following is the stronger acid: CH2ClCOOH or
CHCl2COOH? Explain your choice.
Q) Predict whether the following solutions will be acidic, basic, or nearly
neutral:
(a) NH4I
(b) NaNO2
Solution
We first break up the salt into its cation and anion components and then
examine the possible reaction of each ion with water.
(a) The cation is NH4+, which will hydrolyze to produce NH3 and H+. The Ianion is the conjugate base of the strong acid HI. Therefore, I- will not
hydrolyze and the solution is acidic.
(b) The Na+ cation does not hydrolyze. The NO2- is the conjugate base of
the weak acid HNO2 and will hydrolyze to give HNO2 and OH-. The solution
will be basic.
15.11: Acid-base properties of Oxides and Hydroxides
 Acidic Oxides
An acidic oxide reacts with water and produces an acid. Usually, it is the oxide of
non-metals. Examples include SO2, CO2, SO3, Cl2O7, P2O5, and N2O5.
 Basic Oxides
Basic oxides are oxides that show basic properties, in opposition to acidic oxides.
A basic oxide can either react with water to form a base, or with an acid to form a
salt and water in a neutralization reaction. Examples include: Sodium oxide, which
reacts with water to produce sodium hydroxide.
 Amphoteric Oxides
“Amphoteric oxides are the oxides that behave as both acidic and basic oxides.
Amphoteric Oxides have features of acidic as well as basic oxides that neutralize
both acids and bases.” Amphoteric oxides dissolve in water to form alkaline
solutions. Alkaline solutions contain hydroxide ions.
Acidic and basic oxides
Reaction between acidic oxides and bases and those between basic oxides and
acids resemble normal acid-base reactions in that the products are a salt and
water.
Examples CO2 + 2NaOH ______ Na2CO3 + H2O
BaO + 2HNO3 _______ Ba(NO3)2 + H2O
Amphoteric Oxides:
Consider Al2O3
Al2O3 +6HCl ___________2AlCl3 + 3H2O
Al2O3 + 2NaOH ___________2NaAlO2 + H2O
Aluminum oxide (Al2O3) is amphoteric. Depending on the reaction
conditions, it can behave either as an acidic oxide or as a basic oxide.
Hydroxides:
Alkali metals and alkaline earth metal hydroxides are basic in properties. The
following hydroxides are amphoteric: Be(OH)2 , Al(OH)3, Sn(OH)2,
Pb(OH)2.
Example
Al(OH)3 + 3H+ ____________ Al3+ + 3H2O
Al(OH)3 + OH- ____________Al(OH4)Aluminum hydroxide reacts with both acids and bases.
Questions:
Q) What are the representative elements of the periodic table?
Q) List some common representative elements?
Q) Analyse the figure below and compare acidic, basic and amphoteric oxides by
discussion.
Q) Write down the hydrolysis of basic metallic oxides like Na2O and BaO.
Q) Complete the following reactions between acidic oxides and water
CO2 + H2O
SO3 + H2O
N2O5 + H2O
P4O10 + H2O
Cl2O7 + H2O
Q) Classify the following oxides as acidic, basic, or amphoteric: (a) CO2 (b)
K2O (c) CaO (d) N2O5 (e) CO (f) NO (g) SnO2 (h) SO3 (i) Al2O3 (j) BaO.
Q) Write equations for the reactions between
(a) SO3 and NaOH (b) Na2O and HNO3.
15.12: Lewis Acids and Bases
Arrhenius acid is a substance that produces H+ (H3O+) in water
A Brønsted acid is a proton donor
A Lewis acid is a substance that can accept a pair of electrons
A Lewis base is a substance that can donate a pair of electrons
The hydration of carbondioxide to produce carbonic acid carbonic acid.
CO2 + H2O _______H2CO3
The first step involves donation of a lone pair on the oxygen atom in the H2O to
the carbon atom in CO2. H2O is a Lewis base and CO2 is a Lewis acid.
Questions:
Q) Draw Lewis structure for the following compounds?
a. H20
b. CH3OH
c. CH4
d. HCl
Q) Identify Lewis acid and Lewis base in the following reactions:
AlCl3 + Cl-_______
Ag+ + 2NH3 ______
O2- + SO3_______
F- + BF3_______
Q) Classify each of the following species as a Lewis acid or a Lewis base:
CO2 b. H2O c. I- d. SO2 e. NH3 f. OHg. H+ h. BCl3
Q) Analyse the structure below
17.2: Spontaneous and non-spontaneous processes
Spontaneous Physical and Chemical Processes Examples:
 A waterfall runs downhill
 A lump of sugar dissolves in a cup of coffee
 At 1 atm, water freezes below 0 oC and ice melts above 0 oC
 Heat flows from a hotter object to a colder object
 A gas expands in an evacuated bulb
 Iron exposed to oxygen and water forms rust
The figure shows an isolated system consists of an ideal gas in one flask that is connected by a
closed valve to a second flask containing a vacuum. Once the valve is opened, the gas
spontaneously becomes evenly distributed between the flasks. Processes have a natural tendency
to occur in one direction under a given set of conditions. Water will naturally flow downhill, but
uphill flow requires outside intervention such as the use of a pump. Iron exposed to the earth’s
atmosphere will corrode, but rust is not converted to iron without intentional chemical treatment.
A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous
process, on the other hand, will not take place unless it is “driven” by the continual input of
energy from an external source.
Differentiate enthalpy and entropy
Enthalpy (in) Delta H
 The amount of heat or internal energy
 it was delta U
 Part of system, system energy
decrease
 Delta H is negative
 Delta S is positive
 Reactants have more potential energy
 Favors product side
 Spontaneous exothermic
 Example:
 combustion
 iron rusting to oxygen forming ater
 Fe2 + o2 = Fe2O3
 Cellular respiration
 H2+ + Cl2- = 2HCl
 Spontaneous reactions have –ve
Kj/mol
Entropy (out) Delta S
 External energy, free energy, or gibbs
free energy
 Part of the surrounding
 Delta H is positive
 Delta S is negative
 Products have more potential energy
 Favors reactants side
 Non- spontaneous endothermic
 Example: decomposition and
photosynthesis
 HCl = H+ + Cl Non Spontaneous reactions have +ve
Kj/mol
Questions:
Q) Analyze the figure below and share your understanding
The first person exerts more effort and energy going
up than the person going down. The person going
down is spontaneous (exothermic) and need no energy,
The person going up is nonspontaneous and needs
energy.
Q) The conversion of diamond into graphite: C(s, diamond)⟶C(s, graphite)
Q) Explain what is meant by a spontaneous processes. Give two examples each of spontaneous
and non spontaneous processes.
Q) Does a decrease in enthalpy mean a reaction proceeds spontaneously?
17.3: Entropy
In a coffee cup the entropy of the particles are increasing as the liquid in the cup turns to steam.
The entropy increases because at higher temperature the particles have more energy and so there
is a greater probability of places that the particles may be.
What is entropy?
It is a measure of disorder or randomness of a particular
system to surrounding. The spreading of molecules is
called microstates the energy can’t be located or predicted
like the collision theory with sufficient energy and correct
orientation causing energy to move from a place to the
other causing all in or particle to be equally distributed.
More disorder means less predictable and more entropy of
molecules. Like particles moving in gas it is a
nonbconserved property as energy released entropy called
Gibbs free energy to atmosphere or environment
transforms but can’t be located and hence isn’t useful for
work doing. It is –ve as particles come closer and –ve as
they move away from each other.
Entropy change of a system
Processes that lead to an increase in entropy of the system:
(a) melting: S Liquid > S solid
(b) vaporization: S vapor >
S1;quid;
(c) dissolving: Ssoln > Ssolute:
Melting, vaporization, dissociation
dissolving are all process that cause
entropy increase with more
randomness and disturbance.
Synthesis reactions have less
entropy. We determine entropy by
checking the state.
In1868 Boltzmann showed that the entropy of a system is related to the natural log of
the number of microstates (W):
S = k In W…………….1
Entropy and microstates are directly proportional
Where k is called the Boltzmann constant (1.38 X 10-23 J/K). Thus, the larger the W the greater
is the entropy of the system. Consider a certain process in a system. The entropy change for the
process, AS, is
Where Wi and Wf are the corresponding numbers of microstates in the initial and final states.
Thus, if Wf > Wi, AS > 0 and the entropy of the system increases. Ln is log, K is boltzmen
constant=1.38*10^-23. In J/K
Microstate:
Gases have more entropy than liquids that has more entropy than solids. The more
intermolecular spaces between molecules the increase in randomness and mictrostate. Microstate
is the space between the particles or the arrangement of molecules and change in state associated
with entropy.
Questions:
Q) What do you know about Enthalpy?
Q) Compare enthalpy with entropy?
Q) Define entropy. What are the units of entropy?
Q) How does the entropy of a system change for each of the following processes?
A solid melts. S increase
A liquid freezes. S decrease
A liquid boils. S increase
A vapor is converted to a solid. S decrease
A vapor condenses to a liquid. S decrease
A solid sublimes (solid to gas directly). S Increase
Urea dissolves in water .S increase
17.4: The second and third law of thermodynamics
The minus sign is used because if the process is exothermic, ΔHsys is negative and Δssurr is a
positive quantity, indicating an increase in entropy. On the other hand, for an endothermic
process, ΔHsys is positive and the negative sign ensures that the entropy of the surroundings
decreases. When heat moves from the system to the surroundings (exothermic) entropy increases
(heat released delta H is negative and delta s is positive – they are inversely proportional). When
heat moves from surropunding to the system (endothermic) the entropy decreases and heat is
absorbed (delta h increase to positive and delta s is now negative).
There is an inverse relationship between Δssurr and temperature of surrounding because of more
randomness and disorder or less impact of the delta S surroundings that is , the higher the
temperature , the smaller the Δssurr and the lower the temperature, the greater the Δssurr .
It is more than 0 because it is a spontaneous process and occurs naturally
with no energy needed.
The process in which all reactants and products reach to equilibrium and
this is the case of the 2nd law of thermodynamics.
According to second law of thermodynamics: The entropy of the universe increases in a
spontaneous process and remains unchanged in an equilibrium process.
For a spontaneous process S universe must be greater than zero.
Entropy Changes in the System (DSsys)
The standard entropy of reaction (DS0rxn ) is the entropy change for a reaction carried out at 1
atm and 250C.
Any coefficient is multiplied with the entropy but not the subscripts and the entropy will be provided in
table. Positive entropy means there is an increase in entropy, negative entropy means there is a decrease
in entropy.
Third law of thermodynamics
According to third law of thermodynamics, the entropy of a perfect crystalline substance is zero
at the absolute zero of temperature. As the temperature increases, the freedom of motion
increases. Thus the entropy of any substance at a temperature of above 0 K is greater than 0.
Figure 17.6 shows the change (increase) in entropy of a substance with temperature. At absolute
zero, it has a zero entropy value. As it is heated, its entropy increases gradually because of
greater molecular motion. At the melting point, there is a sizable increase in entropy as the liquid
state is formed. Further heating increases the entropy of the liquid again due to enhanced
molecular motion. At the boiling point there is a large increase in entropy as a result of the
liquid-to-vapor transition. Beyond that temperature, the entropy of the gas continues to rise with
increasing temperature. Each phase of same molecules in different phases have different
entropies. As temperature increase, the entropy increases as the particles are heated and move
quicker in motion the entropy increases more boiling in liquids cause even more increase.
(according to 3rd law only the system). In 2nd law –Delta H , heat increases and Positive Delta S (here
they are inversed). An exception is the crystalline substance in which 0S at 0 temperature if temperature
increases then S increases.
Entropy is inversely proportional to temperature because entropy and enthalpy are inversely
related, in exothermic reactions temperature of surrounding increase, delta H decrease and hense
delta S increase in positive so temperature decreases. Ion endothermic reactions enthalpy delta H
in positive so temperature of the surrounding decreases and enthalpy decreases and entropy delta
S increases so temperature decreases.
Questions:
Q) From the standard entropy values in Appendix 3, calculate the standard entropy changes
for the following reactions at 25°C.
(a) CaCO3(s)
= CaO(s) + CO2(g)
(b) N2(g) + 3H2(g)
=
2NH3(g)
(c) H2(g) + Cl2(g)
=
2HCl(g)
d. H2(g) + CuO(s) _________ Cu(s) + H2O(g)
e. 2Al(s) + 3ZnO(s) _________ Al2O3(s) + 3Zn(s)
f. CH4(g) + 2O2(g)___________ CO2(g) + 2H2O(l)
Solution:
To calculate the standard entropy of a reaction, we look up the standard entropies of reactants and
products in Appendix 3 and apply Equation (17.7). As in the calculation of enthalpy of reaction [see
Equation (6.18)], the stoichiometric coefficients have no units, so ΔS°rxn is expressed in units of
J/K·mol.
(a)
ΔS°rxn = [S°(CaO) + S°(CO2)] - [S°(CaCO3)]
= [(39.8 J/K·mol) + (213.6 J/K·mol)] - (92.9 /K·mol)
= 160.5 J/K·mol
Thus, when 1 mole of CaCO3 decomposes to form 1 mole of CaO and 1 mole of gaseous CO2, there is
an increase in entropy equal to 160.5 J/K·mol.
(b) ΔS°rxn = [2S°(NH3)] - [S°(N2) + 3S°(H2)]
= (2)(193 J/K·mol) - [(192 J/K·mol) + (3)(131 J/K·mol)]
= -199 J/K·mol
This result shows that when 1 mole of gaseous nitrogen reacts with 3 moles of gaseous hydrogen to
form 2 moles of gaseous ammonia, there is a decrease in entropy equal to -199 J/K·mol.
(c) ΔS°rxn = [2S°(HCl)] - [S°(H2) + S°(Cl2)]
= (2)(187 J/K·mol) - [(131 J/K·mol) + (223 J/K·mol)]
= 20 J/K·mol
Thus, the formation of 2 moles of gaseous HCl from 1 mole of gaseous H2 and 1 mole of gaseous Cl2
results in a small increase in entropy equal to 20 J/K·mol.
17.5: Gibbs free energy
The second law of thermodynamics says that the entropy of the universe always increases for a
spontaneous process:
In chemistry, a spontaneous process is one that occurs without the addition of external energy. A
spontaneous process may take place quickly or slowly, because spontaneity is not related to
kinetics or reaction rate.
When a process occurs at constant temperature (T) and pressure (P) we can rearrange the second
law of thermodynamics and define a new quantity known as Gibbs free energy
For convenience, we can change the preceding equation by multiplying it throughout by -1 and
replacing the > sign
with<:
In order to express the spontaneity of a reaction more directly, we introduce another
thermodynamic function called Gibbs free energy (G), or simply free energy.
where H, is enthalpy, T is temperature (in kelvin, K), S, end is the entropy. Gibbs free energy is
represented using the symbol G and typically has units of kJ/mol
The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it
occurs under standard-state conditions.
Questions:
Q) Define free energy. What are its units?
Ans: Gibbs free energy G, is the energy available in a system to do useful work and is different
from the total energy change of a chemical reaction.
It is measured in joules
Q) Why is it more convenient to predict the direction of a reaction in terms of ΔGsys instead of
ΔSuniv? Under what conditions can ΔGsys be used to predict the spontaneity of a reaction?
Ans: Predicting the direction of a reaction in terms of ΔGsys is more convenient because it only
considers the system, whereas ΔSuniv considers both the system and the surroundings. This
makes calculations and predictions simpler and more focused on the system of interest.
Q) Determine whether the reactions with the following ΔH and ΔS values are spontaneous or
non-spontaneous. State whether the reactions are exothermic or endothermic.
(a) ΔH=−110kJ, ΔS=+40JK−1 at 400K
(b) ΔH=+40kJ, ΔS=−120JK−1 at 250K.
Solution
(a) Given: ΔH=−110kJ
ΔS=40Jk −1
=0.04KJk −1
Temperature, T=4000K
ΔG=?
Since ΔH is −Ve, the reaction is exothermic
ΔG=ΔH−TΔS
=−110−400×0.04
=−110−16
=−126kJ
Since, ΔG is negative, the reaction is spontaneous and exothermic.
(b) Given: ΔH=40kJ,
ΔS=−120Jk −1
=−0.12KJk −1
Temperature, T=250K
ΔG=?
Since ΔH is +ve, the reaction is endothermic.
ΔG=ΔH−T.ΔS
=40−250×(−0.12)
=40+30
=70kJ.
Since ΔG>0, the reaction is non-spontaneous
ΔG=70kJ; The reaction is endothermic and non-spontaneous.
18.1: Redox Reactions
 Balancing Redox Equations in Acidic Solution
Agree or disagree
 Redox reactions doesn't involve the transfer of electrons.
 Equations representing redox processes can be balanced by using ion-electron method.
 All electrochemical reactions involve the transfer of electrons and therefore are redox
reactions.
Electrochemical processes are oxidation-reduction reactions in which:
the energy released by a spontaneous reaction is converted to electricity or
electrical energy is used to cause a nonspontaneous reaction to occur
Oxidation number:
The charge the atom would have in a molecule (or an ionic compound) if electrons were
completely transferred.
 Balancing Redox Equations in Basic Solution
Steps to add for basic solution
For each H+, add one OH- to both sides.
Combine H+ and OH- to make H2O.
Subtract H2O from both sides if possible.
Questions:
Q) Write the steps for balancing redox equations in acidic solutions?
Q) Balance the following redox equations in acidic solution:
Q) Write a short note on redox reactions?
Q) Write a balanced ionic equation to represent the oxidation of iodide ion (I-) by
MnO-
4 ) in basic solution to yield molecular iodine (I2) and
permanganate ion (
manganese(IV) oxide (MnO2).
Strategy
We follow the preceding procedure for balancing redox equations. Note that the reaction takes
place in a basic medium.
Solution
Step 1: The unbalanced equation is
Step 2: The two half-reactions are
Step 3: We balance each half-reaction for number and type of atoms and charges. Oxidation halfreaction: We first balance the I atoms:
To balance charges, we add two electrons to the right-hand side of the equation:
Reduction half-reaction: To balance the O atoms, we add two H2O molecules on the right:
To balance the H atoms, we add four H+ ions on the left:
There are three net positive charges on the left, so we add three electrons to the same side to
balance the charges:
Step 4: We now add the oxidation and reduction half reactions to give the overall reaction. In
order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 3
and the reduction half-reaction by 2 as follows:
Finally, combining the H+ and OH- ions to form water, we obtain
Step 5: A final check shows that the equation is balanced in terms of both atoms and charges.
Q) Balance the following redox equations in basic solution:
18.2: Galvanic Cells
Overview
1. Definition:
o A galvanic cell (or voltaic cell) is an electrochemical cell that derives electrical
energy from spontaneous redox reactions occurring within the cell.
2. Components:
o Anode: The electrode where oxidation occurs (loss of electrons).
o Cathode: The electrode where reduction occurs (gain of electrons).
o Electrolyte: Ionic substance that allows ions to move between the electrodes.
o Salt Bridge: A pathway that allows the transfer of ions to maintain electrical
neutrality.
3. Working Principle:
o The anode and cathode are connected by an external circuit, allowing electrons to
flow from the anode to the cathode.
o The salt bridge contains a salt solution that allows ions to flow between the halfcells to balance the charge as electrons move through the external circuit.
Key Reactions:
1. Oxidation at the Anode:
o
o
Zinc metal is oxidized to zinc ions, releasing electrons.
Zn (s)→Zn2+(aq)+2e−\text{Zn (s)} \rightarrow \text{Zn}^{2+} \text{(aq)} + 2e^Zn (s)→Zn2+(aq)+2e−
2. Reduction at the Cathode:
o
o
Copper ions in solution are reduced to copper metal by gaining electrons.
Cu2+(aq)+2e−→Cu (s)\text{Cu}^{2+} \text{(aq)} + 2e^- \rightarrow \text{Cu
(s)}Cu2+(aq)+2e−→Cu (s)
Electrical Flow:

Electrons flow from the anode (zinc electrode) to the cathode (copper electrode) through an
external circuit, powering any connected devices (e.g., a light bulb or voltmeter).
Salt Bridge:

The salt bridge typically contains a salt solution (e.g., KCl or Na₂SO₄) and allows the transfer of
ions to maintain charge balance. It prevents the solutions from mixing but allows ion flow, which
is essential for the cell's operation.