FCH 152: General Chemistry II
Final Exam Review
Thursday, May 4th
8:00 - 10:00 AM
Marshall Auditorium / 140 Baker
Additional Survey Questions
14. Based on my Instructor’s recommendation, I
spent 2-3 hours after every lecture reviewing
the class notes and working problems from the
text.
15. My expectation for a challenging course such
as FCH152 is that most of my “learning” is done
in class.
16. Baker 140 is a significant improvement over
Marshall Aud. for General Chemistry classes.
17. I frequently used the tutorials and quizzes
available on the CD accompanying the text to
prepare for exams.
18. Weekly quizzes would be a good way to improve
student performance in FCH152.
FCH 152: General Chemistry II
Disclaimer !!!
This study guide does not contain all of the
material likely to be on the final exam …..
It is just a compendium of subjects that
will form the basis of Exam questions.
Problems from Exams, Homework's and
Workshops might show up again !
1
Exam Questions
(Non-Binding Instructors Estimate !)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Kinetic Theory of Gases
No Nukes !
Ideal Gas Law
Ionic Compounds
Stoichiometry / Balancing Equations
Empirical Formulae
Solutions / Solubility Product (2)
Colligative properties of solutions (2)
Thermodynamics (2)
Kinetics (2)
Reaction Equilibria – Equilibrium Constant
Acids and Bases (2)
Electrochemistry
More on Kinetic Molecular Theory...
• Gases: More K.E. than liquids or solids.
• K.E. is directly proportional to
absolute Temperature (T, K).
• T increases, K.E. increases
• At a given T, every gas has the same
K.E. but different speed.
• K.E. = 1/2 Mu2 , where M = molar mass
and u = “root mean square” speed of
the molecule.
Temperature in °K
Influence of Temperature
on Molecular Velocity in
the Gas Phase
u2 = (3.R.T)/M
(Speed ->)
2
(M = 32)
(M = 28)
K.E. = 1/2Mu2 so…
u2 = 2K.E./M
M = molar mass
u = “root mean square”
speed (rms)
(M = 4)
(M = 2)
Speed
Review Ideal Gases…
PV = nRT
If T and n are fixed, then: P1V1=P2V2
If P and n are fixed, then V1/T1=V2/T2
If V and n are fixed, then P1/T1=P2/T2
R is the Gas Constant =
0.082 atm.L/mol.K
1 mole of ideal gas = 22.4 L
Ideal Gases… Not on Exam
Henry’s Law (Solubility on liquids)
Raoult’s Law (Partial Pressure and Mole
Fraction)
3
Ionic Compounds
Polyatomic Ions and Their Charges……
Sulfate, carbonate, etc.
Know: Lattice Energy (U) and what
atomic factors are responsible for
higher or lower energy.
Relationship between lattice energy and
ionic material properties (melting point ,
density, etc)
Balancing the Automobile
Air Bag Equation
Active Ingredient: Sodium Azide (NaN3)
NaN3(s) → Na(s) + N2(g)
NaN3(s) → Na(s) +
[NaN3(s)
2
→ Na(s) +
NaN3(s) →
2
1.5
1.5
N2(g)
N2(g)] x
Na(s) +
3
2
N2(g)
2 x 65 g → 67.2 L of nitrogen gas !
Empirical Formulae for Dummies
Analytical data is usually in weight %
Assume you have 100 g of a sample, then %
of each element corresponds to grams of that
element.
• For example: Compound W is 60% N, then
100 grams of W contains 60 g of N
Calculate the moles of each element by
dividing its weight in grams by its molar mass.
Divide the moles of each element by smallest
value of moles in the compound.
Adjust results to end up with whole numbers
for each element….. Voila!
4
Analytical Data
Calculating Reaction Yield…
The combustion of 58 g of butane (C4H10) yields 158 g of
CO2. What is the yield of this reaction ?
2 C4H10 + 13 O2 → 10 H2O + 8 CO2
Theoretical yield is 8 moles of CO2 for each 2 moles of
butane or, 4 moles of CO2 for each 1 mole of butane.
1 mole of butane = 58 g ; 4 moles of CO2 = 176 g
Actual Yield (Given) = 158 g CO2
Theoretical yield for 1 mole of butane (58 g) = 176 g CO2
% Yield = Actual/Theoretical = 158/176 =
.897 = 89.7%
Solution Concentrations
Molarity (M; moles / Liter of solvent)
Molality (m; moles / kg of solvent)
Percent (%):
• Grams / 100 g of solvent (w/w)
• Grams / 100 ml of solvent (w/v)
• Milliliters / 100 ml (v/v)
Parts per million (ppm): grams per
1,000,000 grams or ml of solvent
ppb, ppt, etc. (Very dilute)
5
Concentration Conversions
Solution Dilution !
For Dilutions:
Voriginal x Moriginal =
Vdiluted x Mdiluted
50 ml = 0.050 L
Say it’s 5.0M
= 0.250 mol
250 ml = 0.250 L
0.250 mol / 0.250 L = 1.0 M
Types of Solutes
Electrolyte: Solute that dissociates into
ions in solution, enhancing the electrical
conductivity of the solvent (water mostly).
Ex. NaCl → Na+ + ClWeak Electrolyte: A solute that only
partially dissociates in water and weakly
enhances conductivity. Ex. Acetic Acid:
CH3COOH ↔ CH3COO- + H+
Nonelectrolyte: None of the above. No
dissociation, no conductivity. Ex. Scotch
and soda (ethanol in water)
6
Colligative Solution Properties
Properties that depend on the number of
moles of solute present in solution and
not solute composition (1 mole of sugar
has the same effect as one mole of
ethanol, for example).
1mol NaCl 1 mol Na+ + 1 mol Cl- = 2
moles total in solution !
Typical “colligative” properties:
•
•
•
Boiling Point Elevation
Freezing Point Depression
Osmotic Pressure
Calculating Boiling Point Elevation
ΔTb = Tb - Tbo = Kb.m, where:
•
•
•
•
•
ΔTb = Change in Boiling Point
Tb = Observed Boiling Point of solution
Tbo = Pure Solvent Boiling Point
Kb = 0.52°C/m - water boiling point elevation constant.
m = Molality = moles of Solute / kg of Solvent
For chloride ion in seawater:
0.558M or 0.565m - Almost the same!
Example: For a 1.14m seawater solution, its change in
boiling point is: (0.52 °C / m).(1.14m) = 0.60°C
Tb = ΔTb + Tbo
Tb = 0.60°C + 100°C = 100.6 °C
Calculating Freezing Point Depression
ΔTf = T0f - Tf = Kf.m, where:
•
•
•
•
•
ΔTf = Change in Freezing Point
Tf = Observed Freezing Point
Tfo = Pure Solvent Freezing Point
Kf = 1.86°C/m - water freezing point depression constant.
m = Molality = moles of Solute / kg of Solvent
Example: For a 1.14m seawater solution, its change in
freezing point is: (1.86 °C / m).(1.14m) = 2.12°C
Tf = Tfo - ΔTf
Tf = 0°C - 2.12 °C = -2.12 °C
7
Osmotic Pressure:
π
Calculating Osmotic Pressure (π)
π
= MRT where:
•M = Molarity (mol/L)
•R = 0.082 atm.L / mol.K
•T = Temperature (K)
Units of
π: atm
Note similarity to
PV=nRT
π
across a membrane
Example: Calculate
separating pure water from seawater at 25ºC.
π
= MRT = 1.14 mol/L x 0.0821(L.atm)/(mol.K) x
(25+273 K) = 27.9 atm
“Solubility Product” (Ksp)
More Generally:
Ionic Compound Dissolves: MmXx ↔ mM+ + xXDefinition of Ksp = [M]m . [X]x
Ksp values are tabulated for many
compounds ….. See Appendix Table A5.3
of text book.
CaF2 in water, Ksp = [Ca++] x [F-]2 = 3.9 x 10-11
8
More on “Solubility Product” (Ksp)
Caution, Beware, This Gets Tricky Here !!
Dissolving CaF2 in water ↔ Ca++(aq) + 2F- (aq)
If you start with 1 mol of CaF2 you get one mole
of Ca++ and 2 moles of F-, or
[s] ↔ [s] + [2s] (Recall [ ] = mol / L)
If Ksp = [Ca++] . [F-]2 = 3.9 x 10-11
Then: s . [2s]2 = 4s3 = 3.9 x 10-11
So, s = 2.1 x 10-4
And [s] = 2.1 x 10-4 M = [CaF2 ] = It’s “solubility”
Aqueous Solutions of Acids and Bases
Bronsted-Lowry Definition (circa 1923)
Acid: A proton donor
Base: A proton acceptor or OH- producer
Lewis Definition
Acid: Accepts a pair of electrons
Base: Donates a pair of electrons
More About Acids / Bases
A strong acid or base completely
dissociates to ions in solution.
Strong Acid: HCl (aq) → H+ (aq) + Cl- (aq)
Strong Base: NaOH (aq) → Na+ (aq) + OH- (aq)
A weak acid or base does not completely
dissociate. (Example to follow)
Strong acids can form other strong
acids:
HNO3 (aq) → H+ (aq) + NO3- (aq)
HNO3 (aq) + Cl- → NO3- (aq) + HCl (aq)
9
Water: An Acid and a Base
Water can both accept and donate a
proton… It’s “amphoteric”.
H20 (l) + H20 (l) ↔ OH- + H3O+ (Auto-ionization)
Acid + Base ↔ Conjugate Base + Conjugate Acid
At auto-ionization “equilibrium”:
Kw = [H3O+ ] [OH-] = 1 x 10-14
Where: [H3O+ ] = [OH-] = 1 x 10-7 (Very Low Conc. !)
So, in a solution where [H3O+ ] > [OH-], it’s acidic
And, in a solution where [H3O+ ] < [OH-], it’s basic
Nitric Acid: Strong
Nitrous Acid: Weak
Degree of Acid Ionization: Ka
A strong acid completely dissociates to
ions in solution.
HCl (aq) → H3O+ (aq) + Cl- (aq)
[0.1 M] → [0.1M] + [0.1M]
Weak acids only partially dissociate to
ions….
Ka values reflect the degree of
ionization. Ex: HNO2 (Nitrous Acid; Weak)
HNO2 (aq) → H+ (aq) + NO2- (aq)
Ka = [NO2- ] [H+] / [HNO2]= 4.0 x 10-4
10
Sample Exercise 16.1: Calculate the value of [H+] of
0.100 M Acetic Acid. (Ka = 1.76x10-5).
C2H3OOH (aq)
C2H3OO- (aq) + H+ (aq)
Ka = [C2H3OO-][H+]
[C2H3OOH]
= 1.76x10-5
(X)(X) = 1.76x10-5
(0.100 – X)
1. Since Ka is so small and acid concentration relatively large, the value
of X will be small compared to 0.100 M. Assume (0.100 – X) = 0.100.
(X)(X)
(0.100)
= 1.76x10-5
2. After some simple algebra, we get:
X = [H+] = 1.33 x 10-3
Conjugate pairs !
16_08.jpg
What is pH ?
Very simply……the “potential of the
hydrogen ion” (SØren SØrenson, 1868-1939).
pH = - log [H3O+ ] in water
or, generally, pH = - log [H+]
So, at “equilibrium”:
[H3O+ ] = [OH-] = 1 x 10-7
Then the pH of “neutral” water is: 7 !
11
What’s a Titration ?
An experiment where you add a measured
amount of one reactant (for example, an acid or
base solution in water) to another reactant (like a
solution of acid or base of unknown concentration)
to determine the point of equivalence (where equal
moles of each reactant are consumed or
“neutralized”).
Useful for determining the Molarity of solutions
of unknown concentration.
Titration of a Strong
and Weak Acid
Original Solutions:
20 ml of 0.10M Acid
In General, For Titrations….
Assuming “Monoprotic” Acids
Problem: Molarity is unknown for the acid or base …
Volume x Molarity = Volume x Molarity
Acid
Base
Volume x Molarity = L x mol / L = moles
moles of acid = moles of base
For diprotic acids: 0.5 mole acid can neutralize one mole of base
12
Thermodynamics
Definition: The study of energy and
its transformations.
To Begin : Review Dr. Teece’s
discussion of Energy, Spontaneity,
Free Energy, Enthalpy and Entropy!!
Energy of phase transitions
Reaction energies
• No Calorimetry
Internal Energy
Let’s call a gas trapped in a vessel a “Chemical
System”.
Internal energy of a system = Sum of kinetic
and potential energies = E
Change in Internal Energy going from state A to
state B is ΔE = q + w, where q = heat flow into
or out of the system and w = work done on or
by the system. (Unit of work is the joule (J)).
This is the First Law of Thermodynamics:
Click on Image
Summarizing the Confusion ……
• q > 0 if heat is transferred from the surroundings
to the system (the system is endothermic)
• q < 0 if heat is transferred from the system to the
surroundings (the system is exothermic)
• w > 0 if work is done by the surroundings on the
system .
• w < 0 if work is done by the system on the
surroundings.
Remember
These !
• Therefore: The value of ΔE depends on the signs
and magnitude of q and w.
13
Internal Energy and Enthalpy
For a chemical system at constant
pressure, a new “State” function is defined
where: H = E +PV, where H = Enthalpy
If this system experiences a change (say
work is done on it or by it), then ΔH = ΔE
+ PΔV.
However, since ΔE = q +w = q - PΔV.
Then ΔH = q -PΔV + PΔV.
Then, at constant pressure, ΔH = qp, where
qp = heat flow at constant pressure.
Click on Image
Require input of heat
ΔH positive11_06.jpg
ΔH negative
Some Definitions…..
Exothermic: If a chemical reaction or physical
change results in heat flow from a system to its
surroundings, its exothermic
Endothermic: Reactions and physical changes that
absorb heat from their surroundings are
endothermic.
Example: H2O(l) →H2O(g) ?
q is heat flow: By definition, if q < 0, reaction /
change is exothermic, and if q> 0, its endothermic.
14
Entropy
Entropy (S): measurement of disorder
• More disordered, larger entropy
• Disorder increases = Entropy increases
• Entropy changes during reactions, phase changes
and other chemical processes:
ΔS = Sfinal - Sinitial
ΔS = Sproducts - Sreactants
ΔS is positive, increased disorder
ΔS is negative, increased order
Entropy Changes (ΔS)
Examples: (a) If the reaction of a solid and a liquid yields a gas
for a product, then the product molecules have more freedom
(kinetic theory !) and more disorder than the reactants…
Entropy increases. (ΔS is positive)
(b) If two gases react to form a liquid, then entropy
decreases (ΔS is negative)
(c) If a ice melts (solid – liquid phase change) entropy
increases (ΔS is positive)
(d) If one mole of a compounds dissociates into two moles of
product, entropy increases. (ΔS is positive)
(e) Dissolving ammonia, a gas, in water decreases its entropy
(ΔS is negative)
S
solid
< S
liquid
< S
gas
Gibbs Free Energy ΔG
For just the “system”, a new function called Gibbs
Free Energy (ΔG) can be defined at a constant
pressure (1 atm) and a specific temperature:
ΔG = ΔH – TΔS
For a reaction/event (just the system) to be
spontaneous, ΔG must be negative (-ve)
S(s) + O2(g) → SO2(g); ΔH = -300 kJ
This is an Exothermic Reaction: ΔH is negative (–ve)
and Entropy increases: ΔS is +ve
ΔG = ΔH - TΔS
= (-) - [T(+)] = -ve (SPONTANEOUS
!!)
ΔH -ve and ΔS +ve = Reaction is SPONTANEOUS !
15
Gibbs Free Energy: A Review
ΔG = ΔH – TΔS
ΔH -ve and ΔS +ve = SPONTANEOUS
ΔH +ve and ΔS -ve = NON-SPONTANEOUS
Always TRUE at all Temperatures
If ΔH and ΔS have SAME SIGN, then
temperature is critical
Phase Transitions of Water
D
B
C
E
•Heat flow (q), enthalpy
and its changes (ΔH; exo
= -; endo =+)
•qp = ΔH = n Cp ΔT
•ΔH of phase changes; qp
= n ΔH fusion or vaporization
A
Molar Heat Capacity vs.
Specific Heat
Molar Heat Capacity
q = n . Cp .ΔT, where:
• n = number of moles
• Cp = molar heat capacity of water (=75.3 J/mol.°C)
• ΔT = Tfinal - Tinitial = temperature change
Specific Heat:
q = mass (grams) . C .ΔT, where:
C is in J / g-°C or K
16
Heat Flow (q) required to:
Stage B → C
Melt ice at 0°C;
q = n . ΔHfusion , where:
• n = number of moles
• ΔHfusion = molar heat of fusion of water
(heat of fusion is the enthalpy change that
takes place when one mole of a substance is
melted) = 6.01 kJ/mol for water
Ex.: q = (15.0 mol)(6.01 kJ/mol)
= 90.0 kJ
Enthalpy Changes During
Chemical Reactions
Bonds break ! (energy required)
Bonds form !! (energy released)
For a specific bond, like C-H, the energy of
bond breaking is the same as the energy
required for formation.
ΔHreaction = Energy required + Energy liberated (-)
Bond Energy = Energy required to break a mole of
bonds
• Bond Breaking: ΔH is positive (+)
• Bond Formation: ΔH is negative (-)
Changed from
Online Version
17
Enthalpy of Reaction (ΔHrxn)
from Standard Enthalpies
ΔH0 reaction = Σ nΔH0formation, products - Σ mΔH0formation, reactants
n, m = number of moles in the balanced equation
ΔH0formation values for a compound or element in
its ionic non-standard state are found in
“tables” like Appendix A4.3 in your text.
Example: Calculate the Heat of Reaction
(ΔHrxn) for the combustion of propane (C3H8
(g)) from standard enthalpies of formation…
Table 11.3: Standard Enthalpies
of Formation at 298 K
Compound
Compound
CH4 (g)
ΔHf0
(kJ/mol)
-74.8
CO (g)
ΔHf0
(kJ/mol)
-110.5
C2H6 (g)
-84.7
CO2 (g)
-393.5
C3H8 (g)
-103.8
H2O (g)
-241.8
C4H10 (g)
-125.7
H2O (l)
-285.8
O2 (g)
0.0
H2 (g)
0.0
Thermodynamics Refresher … Let’s
“wake up” those resting brain cells
For a chemical reaction or a process, like freezing or boiling:
ΔG = ΔH -TΔS.
ΔG = The change in Gibbs Free Energy, ΔH = enthalpy (heat)
change, and ΔS = entropy change.
If ΔG < O, reaction is spontaneous (One possibility: ΔH < 0 and ΔS is
>0)
If ΔG > O, reaction is non-spontaneous (ΔH > 0, ΔS < 0)
If ΔG = O, reaction or process is in “equilibrium” (ΔH= -TΔS)
If ΔS > O, More system disorder: S solid < S liquid < S gas
ΔH = heat flow = qp at constant pressure
•
•
Endothermic: ΔH is positive (+)
Exothermic: ΔH is negative (-)
ΔH = qp = n Cp ΔT where Cp = molar heat capacity or = mass (g) x
C x ΔT where C = specific heat in kJ / g ; (n =moles)
qp for a phase transition = ΔH
molar heat of fusion or vaporization
xn
18
Chemical Kinetics
Definition: The area of chemistry that deals
with speed of reactions, or the reaction rate,
is called “Chemical Kinetics”.
Reaction rates typically depend on:
The concentration of reactants
Physical state (gas > liquid > solids )
Temperature
The presence of a catalyst (enzymes in natural
systems)
• The surface area of solid or liquid reactants and
catalysts
•
•
•
•
Effect of Concentration on Rate
• Reaction: 2 NO + O2 (g) → 2 NO2
• Rate ∝ [O2]
• Rate ∝ [NO]2
• Rate ∝ [O2].[NO]2
• Rate = k [O2].[NO]2 (Rate Law)
• Where k = Rate Constant
• Generalized Rate Law: k [O2]m
Reaction
Intermediate →
is formed
(“Activated
Complex”)
.
[NO]n
Activation
Energy (Ea)
Exothermic →
19
Arrhenius Equation
Doing the “math”:
ln k = - Ea/R (1/T) + ln A
is in the form of y = mx + b,
so, y intercept = ln A
and slope =-Ea / R
Ea and A can then
be used to calculate
k at any other
temperature ….
Slower
Catalyst Effect:
Homogeneous Catalysis
Faster
Chemical Equilibrium
Chemical Equilibrium: The state in which the
rate of a reaction in the forward direction
matches its rate in the reverse direction…..
And… the concentrations of the reactant(s)
and product(s) do not change
a.k.a “Dynamic Chemical Equilibrium”
A + B ↔ C + D
20
Reaction: N2 (g) + O2 (g) ↔ 2 NO
15_01.jpg
Equilibrium has been
reached before all oxygen
has been consumed !
Generic “Equilibrium Constant”
General Reaction: aA + bB ↔ cC + dD
Kc = [C]c [D]d / [A]a [B]b (For a specific
temperature)
If the products and reactants are all gases,
then [concentration] = partial pressure (P) of
each gas.
Then, Kp = [PC]c [PD]d / [PA]a [PB]b
• K >> 1: Lots of Product
• K<< 1: Little product, lots of un-reacted
reactants
“Equilibrium Constant”
• 2 NO2 (g; brown) ↔ N2O4 (gas; colorless)
• Rate Forward = kf [NO2]2
• Rate Reverse = kr [N2O4]
• At Equilibrium: Rate Forward = Rate Reverse
• kf [NO2]2 = kr [N2O4]
• Or, kf / kr = [N2O4] / [NO2]2 = K
• K = Equilibrium Constant
21
Impact of K on ΔG: What Does K
Mean ???
ΔG0 = -RT ln K, R = 8.314 J/mol.K, T = 298 K
K = e-ΔG/RT
ΔG
K
ln K
17.1 kJ
10-3
-6.9
10-1
-2.3
Not much product
5.7 kJ
More Product
0
0
1
Equilibrium
-5.7 kJ
10
2.3
Lots of Product
☺
Equilibrium
LeChâtelier’s
Principle
A system at equilibrium
responds to a stress in
such a way that it
relieves the stress !
Stress:
Temperature,
Pressure, Concentration,
Precipitation, pH
☺
Acid / Base Equilibria…..
Let’s Look at Nitrous Acid:
HNO2 (aq) ↔ H+ (really, H3O+) + NO2Ka = [NO2- ] [H+ ] / [HNO2] = 4.0 x 10-4 (Book value)
0.1M HNO2 (aq) ↔ 0.0061M H+ + 0.0061M NO2Degree of ionization or Degree of dissociation = 0.0061/0.1 = 6.1%
Small values of Ka suggests that the equilibrium
favors the “reactant” and not products.
For nitric acid (a strong acid)
HNO3 (aq) → H+ + NO3-
0.1M HNO3 (aq) → 0.1M H+ + 0.1M NO3-
22
So What’s Happening ?
• Acetic Acid in Water (A Weak Acid):
– Initially: CH3COOH ↔ (←) CH3COO- + H30+
– Ka = [ H3O+] [CH3COO- = conjugate base] / [CH3COOH]
– Finally: CH3COOH + NaOH → CH3COO-Na+ + H20
(Products sodium acetate plus water)
• At the “mid-point” (pH = 4.75):
– [CH3COOH] = [CH3COO- ]
– [Acid] = [ Conjugate Base]
So What’s Happening II ?
• From: Ka = [ H3O+] [conjugate base] / [acid]
• Then pKa = pH – log [conjugate base] / [acid]
• Henderson-Hasselbach Equation:
pH = pKa + log [conjugate base] / [acid]
• At the “mid-point” (pH = 4.75):
–
–
–
–
Where [CH3COOH] = [CH3COO-]
[CH3COO- ] / [CH3COOH] = 1
Log 1 = 0
Then, pH = pKa
So Who Cares ?
• pKa (or pKb) is a useful property to know
about since it reflects how strong or weak
an acid or base is.
• Lower pKa means a stronger acid.
• Lower pKb means a stronger base.
• Higher pKa means a more basic.
23
Buffer Composition
• Typical Buffer Composition: A solution
containing a weak acid (acetic) and a salt of its
conjugate base. Example: Acetic Acid plus
Sodium Acetate (pH range near 4.75)
• Or,: A weak base and a salt of its conjugate
acid (like ammonia and ammonium ion, say
ammonium chloride). pH ~ 9
• Blood Buffer: Carbonic Acid / bicarbonate ion
(H2CO3 and HCO3 –)
H3O+ (aq) + HCO3 – (aq) ↔ H2CO3 (aq) ↔ H2O + CO2 (g)
CO2 is then expelled !
No Nukes ……
Redox Reactions
Definition: Reactions where there is a simultaneous
transfer of electrons from one chemical species to
another….
Oxidation: Loss of electrons
Reduction: Gain of electrons
Half Reactions: Two reactions
which combine to give an
overall “Redox” reaction
24
Oxidation - Reduction
Neither can take place without the other.
Consider the reaction between zinc metal and
an aqueous copper (II) sulfate solution:
Zn(s) + Cu2+
Zn2+ + Cu(s)
The overall reaction may be decomposed into
two half-reactions:
Zn(s)
Zn2+ + 2e- (oxidation – loss of electrons)
2+
Cu + 2eCu(s) (reduction – gain of electrons)
How much voltage
does this cell
generate ??
E°cell = E°cathode - E°anode
= 0.34 V – (-0.76 V)
= 1.10 Volts !
Which is Anode, Which is Cathode !!
Reductions: Gain Electrons
Reactions above
hydrogen, will
occur at the
cathode, and
hydrogen electrode
becomes the anode
Preference
For Being
Reduced
2 Cu+2 + 2 e- → Cu
2 H+ + 2 e- → H2
2
2
0.342 (cathode)
0.000 (anode)
25