AP Chemistry Super Saturday Review

AP Chemistry Super Saturday Review
I tried to include as much review material as possible in this session. Work on practice tests and review the Bozeman videos for other material.

AP Chemistry Super Saturday Review
5 Essentials Know the basics – writing formulas, writing and balancing equations, dimensional analysis, atomic theory, acid-base theories (Arrhenius and Bronsted Lowry), VSEPR, kinetic molecular theory, and collision theory.

5 Essentials 2. Atomic and Molecular Structures Atomic structures (like electron configurations) will help explain relationships on the periodic table which explains many physical and chemical properties. Molecular structures involves Lewis structures and VSEPR to determine shapes which describes polarities thus describing intermolecular forces which describes many physical properties.

5 Essentials 3. Stoichiometric Calculations Basic stoichiometry, limiting reactants, titration calculations, and empirical formulas. 4. Principles of chemical kinetics, equilibrium, And thermodynamics Kinetics- describes the speed in which substances react Equilibrium – used to determine the extent of a reaction or the composition at equilibrium Thermodynamics – explains why chemical reactions happen in terms of kinetic and potential energies

5 Essentials Representation and Interpretation Be able to draw what is happening at the molecular level and read and interpret graphs and data tables

Net Ionic Equations Graphic: Wikimedia Commons User Tubifex

Solubility Rules – AP Chemistry
All sodium, potassium, ammonium, and nitrate salts are soluble in water. Memorization of other “solubility rules” is beyond the scope of this course and the AP Exam. What dissociates (“breaks apart”) – Aqueous solutions of the following: All strong acids (HCl, HBr, HI, HNO3, H2SO4, and HClO4) Strong bases (group I and II hydroxides) Soluble salts

Write the net ionic for the following: 1
Write the net ionic for the following: 1. Solutions of lead nitrate and potassium chloride are mixed 2. Solutions of sulfuric acid and potassium hydroxide are mixed. 3. Solid sodium hydroxide is mixed with acetic acid

Big Idea #6:Chemical Equilibrium
2NO2(g)  2NO(g) + O2(g) Sketch a graph of change in concentration vs. Time for the reaction above

2NO2(g)  2NO(g) + O2(g) Be able to explain the variance in slope

jA + kB  lC + mD Law of Mass Action For the reaction:
Where K is the equilibrium constant, and is unitless

Product Favored Equilibrium
Large values for K signify the reaction is “product favored” When equilibrium is achieved, most reactant has been converted to product

Reactant Favored Equilibrium
Small values for K signify the reaction is “reactant favored” When equilibrium is achieved, very little reactant has been converted to product

Writing an Equilibrium Expression
Write the equilibrium expression for the reaction: 2NO2(g)  2NO(g) + O2(g) K = ???

Conclusions about Equilibrium Expressions
The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse 2NO2(g)  2NO(g) + O2(g) 2NO(g) + O2(g)  2NO2(g)

Conclusions about Equilibrium Expressions
When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. 2NO2(g)  2NO(g) + O2(g) NO2(g)  NO(g) + ½O2(g)

If the equilibrium constant for A + B C is 0
If the equilibrium constant for A + B C is then the equilibrium constant for 2C A + 2B is A) 0.584 B) 4.81 C) 0.416 D) 23.1 E) 0.208 Answer: D

Equilibrium Expressions Involving Pressure
For the gas phase reaction: 3H2(g) + N2(g)  2NH3(g)

Heterogeneous Equilibria
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present Write the equilibrium expression for the reaction: PCl5(s)  PCl3(l) + Cl2(g) Pure solid Pure liquid

jA + kB  lC + mD The Reaction Quotient
For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action. jA + kB  lC + mD

Significance of the Reaction Quotient
If Q = K, the system is at equilibrium If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

If you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a
one-liter container, which direction would the reaction initially proceed? A) To the left. B) To the right. C) The above mixture is the equilibrium mixture. D) Cannot tell from the information given.

LeChatelier’s Principle
When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress and restore a state of equilibrium. Henry Le Chatelier

Le Chatelier Translated:
When you take something away from a system at equilibrium, the system shifts in such a way as to replace some of what you’ve taken away. When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

Do FRQ #1

Acid Equilibrium and pH
Søren Sørensen

Acid/Base Definitions
Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors

Acid Dissociation Acid Proton Conjugate base HA  H+ + A-
Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)

Dissociation Constants: Strong Acids
Formula Conjugate Base Ka Perchloric HClO4 ClO4- Very large Hydriodic HI I- Hydrobromic HBr Br- Hydrochloric HCl Cl- Nitric HNO3 NO3- Sulfuric H2SO4 HSO4- Hydronium ion H3O+ H2O 1.0

Dissociation Constants: Weak Acids
Formula Conjugate Base Ka Iodic HIO3 IO3- 1.7 x 10-1 Oxalic H2C2O4 HC2O4- 5.9 x 10-2 Sulfurous H2SO3 HSO3- 1.5 x 10-2 Phosphoric H3PO4 H2PO4- 7.5 x 10-3 Citric H3C6H5O7 H2C6H5O7- 7.1 x 10-4 Nitrous HNO2 NO2- 4.6 x 10-4 Hydrofluoric HF F- 3.5 x 10-4 Formic HCOOH HCOO- 1.8 x 10-4 Benzoic C6H5COOH C6H5COO- 6.5 x 10-5 Acetic CH3COOH CH3COO- 1.8 x 10-5 Carbonic H2CO3 HCO3- 4.3 x 10-7 Hypochlorous HClO ClO- 3.0 x 10-8 Hydrocyanic HCN CN- 4.9 x 10-10

Reaction of Weak Bases with Water
The base reacts with water, producing its conjugate acid and hydroxide ion: CH3NH2 + H2O  CH3NH3+ + OH- Kb = 4.38 x 10-4

Kb for Some Common Weak Bases
Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you? Base Formula Conjugate Acid Kb Ammonia NH3 NH4+ 1.8 x 10-5 Methylamine CH3NH2 CH3NH3+ 4.38 x 10-4 Ethylamine C2H5NH2 C2H5NH3+ 5.6 x 10-4 Diethylamine (C2H5)2NH (C2H5)2NH2+ 1.3 x 10-3 Triethylamine (C2H5)3N (C2H5)3NH+ 4.0 x 10-4 Hydroxylamine HONH2 HONH3+ 1.1 x 10-8 Hydrazine H2NNH2 H2NNH3+ 3.0 x 10-6 Aniline C6H5NH2 C6H5NH3+ 3.8 x 10-10 Pyridine C5H5N C5H5NH+ 1.7 x 10-9

Reaction of Weak Bases with Water
The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H2O  BH+ + OH- (Yes, all weak bases do this – DO NOT try to make this complicated!) Ex. Write the reaction of ammonia with water

Self-Ionization of Water
H2O + H2O  H3O+ + OH- At 25, [H3O+] = [OH-] = 1 x 10-7 Kw is a constant at 25 C: Kw = [H3O+][OH-] Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

Relationship between pH and pOH
Calculating pH, pOH pH = -log10(H3O+) pOH = -log10(OH-) Relationship between pH and pOH pH + pOH = 14 Finding [H3O+], [OH-] from pH, pOH [H3O+] = 10-pH [OH-] = 10-pOH

Calculate the pH of a 0.1M HCl solution? (answer =1)

A Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? (answer=4.52)

A Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? (answer=9.48)

Acid-Base Properties of Salts
To determine if a salt is acidic or basic, determine the stronger parent. Examples: KCl NH4Cl NaC2H3O2 NaCl KNO3

Acid-Base Properties of Salts If both parents are weak:
Type of Salt Examples Comment pH of solution Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid NH4C2H3O2 NH4CN Cation is acidic, Anion is basic See below IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral

Buffered Solutions A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt

Acid/Salt Buffering Pairs
The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH) Weak Acid Formula of the acid Example of a salt of the weak acid Hydrofluoric HF KF – Potassium fluoride Formic HCOOH KHCOO – Potassium formate Benzoic C6H5COOH NaC6H5COO – Sodium benzoate Acetic CH3COOH NaH3COO – Sodium acetate Carbonic H2CO3 NaHCO3 - Sodium bicarbonate Propanoic HC3H5O2 NaC3H5O2 - Sodium propanoate Hydrocyanic HCN KCN - potassium cyanide

Base/Salt Buffering Pairs
The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3) Base Formula of the base Example of a salt of the weak acid Ammonia NH3 NH4Cl - ammonium chloride Methylamine CH3NH2 CH3NH2Cl – methylammonium chloride Ethylamine C2H5NH2 C2H5NH3NO3 - ethylammonium nitrate Aniline C6H5NH2 C6H5NH3Cl – aniline hydrochloride Pyridine C5H5N C5H5NHCl – pyridine hydrochloride

Titration of an Unbuffered Solution
A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

Titration of a Buffered Solution
A solution that is 0.10 M CH3COOH and 0.10 M NaCH3COO is titrated with 0.10 M NaOH

Comparing Results Buffered Unbuffered

Henderson-Hasselbalch Equation
This is an exceptionally powerful tool that can be used in your problem solving.

Title: Endpoint is above pH 7 A solution that is 0.10 M CH3COOH
is titrated with 0.10 M NaOH

Title: Endpoint is at pH 7 A solution that is
0.10 M HCl is titrated with 0.10 M NaOH

Title: A solution that is 0.10 M NaOH is titrated with 0.10 M HCl
Endpoint is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.

Title: A solution that is Endpoint is below
0.10 M HCl is titrated with 0.10 M NH3 Endpoint is below pH 7

Solubility Equilibria
Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate. Graphic: Wikimedia Commons user PRHaney

Solving Solubility Problems
For the salt AgI at 25C, Ksp = 1.5 x 10-16 Answer = solubility of AgI in mol/L = 1.2 x 10-8 M

Solving Solubility Problems
For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5 Answer = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

Solving Solubility with a Common Ion
For the salt AgI at 25C, Ksp = 1.5 x 10-16 What is its solubility in 0.05 M NaI? Answer = solubility of AgI in mol/L = 3.0 x M

Big Idea #5: Spontaneity, Entropy and Free Energy

Spontaneity, Entropy and Free Energy
G = H - TS Spontaneity, Entropy and Free Energy

Spontaneous Processes and Entropy
First Law “Energy can neither be created nor destroyed" The energy of the universe is constant Spontaneous Processes Processes that occur without outside intervention Spontaneous processes may be fast or slow Many forms of combustion are fast Conversion of diamond to graphite is slow

Which of the following reactions is spontaneous?
H2(g) + I2(g) ↔ 2HI Kc=49 Br2 + Cl2 ↔ 2BrCl Kc=6.9 HF + H2O ↔ F- + H Kc=6.8x10-4

Second Law of Thermodynamics
"In any spontaneous process there is always an increase in the entropy of the universe" Ssolid < Sliquid << Sgas

For reactions at constant temperature:
Calculating Free Energy Method #1 For reactions at constant temperature: G0 = H0 - TS0

Calculating Free Energy: Method #2
An adaptation of Hess's Law: Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ Cgraphite(s) + O2(g)  CO2(g) G0 = -394 kJ Cdiamond(s) + O2(g)  CO2(g) G0 = -397 kJ CO2(g)  Cgraphite(s) + O2(g) G0 = +394 kJ Cdiamond(s)  Cgraphite(s) G0 = -3 kJ

Calculating Free Energy Method #3
Using standard free energy of formation (Gf0): Gf0 of an element in its standard state is zero

Free Energy and Equilibrium
Equilibrium point occurs at the lowest value of free energy available to the reaction system At equilibrium, G = 0 and Q = K G0 K G0 = 0 K = 1 G0 < 0 K > 1 G0 > 0 K < 1

Bond Energy Breaking bonds require energy (+)
Forming bonds releases energy (-) If the reaction A + B → C is exothermic, which is larger – the energy needed to break the bonds or the energy released when forming the bonds?

TRY FRQ #2

Big Idea #4: Kinetics, Rates, and Rate Laws

Reaction Rate The change in concentration of a reactant or product per unit of time

Reaction Rates: 1. Can measure disappearance of reactants
2NO2(g)  2NO(g) + O2(g) Reaction Rates: 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically

Reaction Rates: 4. Are equal to the slope tangent to that point
2NO2(g)  2NO(g) + O2(g) Reaction Rates: 4. Are equal to the slope tangent to that point 5. Change as the reaction proceeds, if the rate is dependent upon concentration [NO2] t

Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. The differential rate law is usually just called “the rate law.” Integrated rate laws express (reveal) the relationship between concentration of reactants and time

Writing a (differential) Rate Law
Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: 2 NO(g) + Cl2(g)  2 NOCl(g) Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 11.4 x 10-6

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: R = k[NO]x[Cl2]y Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6 2 0.500 5.72 x 10-6 3 2.86 x 10-6 4 1.14 x 10-5 In experiment 1 and 2, [Cl2] is constant while [NO] doubles. The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO]2[Cl2]y

Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: R = k[NO]2[Cl2] Experiment [NO] (mol/L) [Cl2] Rate Mol/L·s 1 0.250 1.43 x 10-6

Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO]2[Cl2] 2 + 1 = 3  The reaction is 3rd order Overall order is the sum of the exponents, or orders, of the reactants

Determining Order with Concentration vs. Time data
(the Integrated Rate Law) Zero Order: First Order: Second Order:

Solving an Integrated Rate Law
Time (s) [H2O2] (mol/L) 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050 Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!! (Click here to download my Rate Laws program for theTi-83 and Ti-84)

Time vs. [H2O2] Regression results: y = ax + b a = -2.64 x 10-4
Time (s) [H2O2] 1.00 120 0.91 300 0.78 600 0.59 1200 0.37 1800 0.22 2400 0.13 3000 0.082 3600 0.050 Regression results: y = ax + b a = x 10-4 b = 0.841 r2 = r =

Time vs. ln[H2O2] Regression results: y = ax + b a = -8.35 x 10-4
Time (s) ln[H2O2] 120 300 600 1200 1800 -1.514 2400 -2.04 3000 -2.501 3600 -2.996 Regression results: y = ax + b a = x 10-4 b = -.005 r2 = r =

Time vs. 1/[H2O2] Regression results: y = ax + b a = 0.00460
Time (s) 1/[H2O2] 1.00 120 1.0989 300 1.2821 600 1.6949 1200 2.7027 1800 4.5455 2400 7.6923 3000 12.195 3600 20.000 Regression results: y = ax + b a = b = r2 = r =

And the winner is… Time vs. ln[H2O2]
1. As a result, the reaction is 1st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

Finding the Rate Constant, k
Method #2: Obtain k from the linear regresssion analysis. Regression results: y = ax + b a = x 10-4 b = -.005 r2 = r = Now remember:  k = -slope k = 8.35 x 10-4s-1

Rate Laws Summary Rate = k Rate = k[A] Rate = k[A]2 [A] = -kt + [A]0
Zero Order First Order Second Order Rate Law Rate = k Rate = k[A] Rate = k[A]2 Integrated Rate Law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 Plot that produces a straight line [A] versus t ln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life

Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs. The sum of the elementary steps must give the overall balanced equation for the reaction The mechanism must agree with the experimentally determined rate law

Rate-Determining Step
In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step

Identifying the Rate-Determining Step
For the reaction: 2H2(g) + 2NO(g)  N2(g) + 2H2O(g) The experimental rate law is: R = k[NO]2[H2] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g) Step #2 N2O(g) + H2(g)  N2(g) + H2O(g) Step #1 agrees with the experimental rate law

Identifying Intermediates
For the reaction: 2H2(g) + 2NO(g)  N2(g) + 2H2O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g) Step #2 N2O(g) + H2(g)  N2(g) + H2O(g) 2H2(g) + 2NO(g)  N2(g) + 2H2O(g)  N2O(g) is an intermediate

Collision Model Key Idea: Molecules must collide to react.
However, only a small fraction of collisions produces a reaction. Why?

Collision Model Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). Colliding particles must be correctly oriented to one another in order to produce a reaction.

Factors Affecting Rate
Increasing temperature always increases the rate of a reaction. Particles collide more frequently Particles collide more energetically Increasing surface area increases the rate of a reaction Increasing Concentration USUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

TRY FRQ #3

Big Idea #2 Intermolecular Forces

Relative Magnitudes of Forces
The types of bonding forces vary in their strength as measured by average bond energy. Strongest Weakest Covalent bonds (400 kcal/mol) Hydrogen bonding (12-16 kcal/mol ) Dipole-dipole interactions (2-0.5 kcal/mol) London forces (less than 1 kcal/mol)

London Dispersion Forces
The temporary separations of charge that lead to the London force attractions are what attract one nonpolar molecule to its neighbors. London forces increase with the size of the molecules. Fritz London Synonyms: “Induced dipoles”, “dispersion forces”, and “dispersion-interaction forces”

London Dispersion Forces

Dipole-Dipole Forces of attraction between two polar molecules Permanent dipoles

Hydrogen Bonding Special type of dipole-dipole in which hydrogen bonds with N, O, F. Hydrogen bonding between ammonia and water

Hydrogen Bonding in DNA
Thymine hydrogen bonds to Adenine T A

Boiling point as a measure of intermolecular attractive forces

TRY FRQ #4

Atomic Radius Big Idea #1: Periodic Trends
Definition: Half of the distance between nuclei in covalently bonded diatomic molecule Radius decreases across a period Increased effective nuclear charge due to decreased shielding Radius increases down a group Each row on the periodic table adds a “shell” or energy level to the atom

Table of Atomic Radii

Period Trend: Atomic Radius

Ionization Energy Definition: the energy required to remove an electron from an atom Increases for successive electrons taken from the same atom Tends to increase across a period Electrons in the same quantum level do not shield as effectively as electrons in inner levels Irregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to remove Tends to decrease down a group Outer electrons are farther from the nucleus and easier to remove

Ionization Energy: the energy required to remove an electron from an atom
Increases for successive electrons taken from the same atom Tends to increase across a period Electrons in the same quantum level do not shield as effectively as electrons in inner levels Irregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to remove Tends to decrease down a group Outer electrons are farther from the nucleus

Table of 1st Ionization Energies

Periodic Trend: Ionization Energy

Electronegativity Definition: A measure of the ability of an atom in a chemical compound to attract electrons Electronegativity tends to increase across a period As radius decreases, electrons get closer to the bonding atom’s nucleus Electronegativity tends to decrease down a group or remain the same As radius increases, electrons are farther from the bonding atom’s nucleus

Periodic Table of Electronegativities

Periodic Trend: Electronegativity

Summary of Periodic Trends

Ionic Radii Cations Anions Positively charged ions formed when
an atom of a metal loses one or more electrons Cations Smaller than the corresponding atom Negatively charged ions formed when nonmetallic atoms gain one or more electrons Anions Larger than the corresponding atom

Table of Ion Sizes

Determine the element Answer: Hg

Determine the element Answer: Na and Mg

TRY FRQ #5

Electrochemistry

Electrochemistry Terminology #1
Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na+ + e- Reduction – A process in which an element attains a more negative oxidation state Cl2 + 2e-  2Cl-

An old memory device for oxidation and reduction goes like this… LEO says GER Lose Electrons = Oxidation Gain Electrons = Reduction

Anode The electrode where oxidation occurs Cathode The electrode where reduction occurs Memory device: Reduction at the Cathode

Galvanic (Electrochemical) Cells
Spontaneous redox processes have: A positive cell potential, E0 A negative free energy change, (-G)  G=-nFE

Zn - Cu Galvanic Cell From a table of reduction potentials:
Zn2+ + 2e-  Zn E = -0.76V Cu2+ + 2e-  Cu E = +0.34V

Zn - Cu Galvanic Cell Zn  Zn2+ + 2e- E = +0.76V
The less positive, or more negative reduction potential becomes the oxidation… Cu2+ + 2e-  Cu E = +0.34V Zn  Zn2+ + 2e E = +0.76V Zn + Cu2+  Zn2+ + Cu E0 = V

Line Notation Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) | || |
An abbreviated representation of an electrochemical cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Anode material Anode solution Cathode solution Cathode material | || |

Calculating G0 for a Cell
G0 = -nFE0 n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e- Zn + Cu2+  Zn2+ + Cu E0 = V

Both sides have the same components but at different concentrations.
??? Concentration Cell Both sides have the same components but at different concentrations. Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

Both sides have the same components but at different concentrations.
??? Concentration Cell Both sides have the same components but at different concentrations. Anode Cathode The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentration Zn2+ (1.0M) + 2e-  Zn (reduction) Zn  Zn2+ (0.10M) + 2e- (oxidation) Zn2+ (1.0M)  Zn2+ (0.10M)

Electrolytic Processes
Electrolytic processes are NOT spontaneous. They have: A negative cell potential, (-E0) A positive free energy change, (+G)

Solving an Electroplating Problem
Q: What mass of copper is plated out when a current of 10 amps is passed for 30 minutes through a solution of Cu2+? (Amp=C/sec) Cu2+ + 2e-  Cu 10 C 1800sec 1 mol e- 1 mole Cu 63.5 g Cu 2 mol e- 1 mole Cu C sec = 5.94g Cu

Good Luck on the Exam. Try your best
Good Luck on the Exam!! Try your best. You have worked hard and will do great! I am very proud of each one of you. Mrs. L

AP Chemistry Super Saturday Review

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