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UA / Chemistry / CH 102 / Which attractive force is the weakest?

Which attractive force is the weakest?

Which attractive force is the weakest?

Description

School: University of Alabama - Tuscaloosa
Department: Chemistry
Course: General Chemistry
Professor: Paul rupar
Term: Fall 2016
Tags:
Cost: 50
Name: CH 102 | Final Study Guide
Description: This is the study guide for the final on December 7, 2016.
Uploaded: 12/02/2016
33 Pages 76 Views 6 Unlocks
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Final Exam Study Guide


Which attractive force is the weakest?



CH 102

Lectures 1, 2, & 3: Phases of Matter and Intermolecular Forces ∙ Describe the differences between the phases of matter (solids, liquids,  and gasses).

o Solids: high density, fixed shape, fixed volume, intermolecular  forces strong.

 A solid is a solid because the molecules are being held  together and the thermal motion/energy of the molecules  cannot overcome intermolecular forces

o Gasses: low density, shape not fixed, volume not fixed,  intermolecular forces weak.  If you want to learn more check out What are professional goals for nurses?

 A gas is a gas because the molecules are not being held  together and thermal motion/energy of the molecules does overcome the intermolecular forces

o Liquids: density moderately high, shape not fixed, fixed volume, intermolecular forces moderately strong.


What is a crystal lattice easy definition?



Don't forget about the age old question of Which immortal entity ate his children?
We also discuss several other topics like What is hypercapnia?

 A liquid’s intermolecular strength is in between that of a  gas and a solid

∙ Changing between states

heat

SOLI

heat or reduce  pressure

LIQUI 

GA

D cool cool or increase  

D

∙ What causes intermolecular attraction?

pressure

S

o All attractions between molecules are due to Coulombic  attractions

o Caulombic attractions are just charges

o If particles are close, then the attraction is strong We also discuss several other topics like Why is the qibla wall in mosques so heavily adorned?

o If particles are far away, then the attraction is weak

o Larger opposite charges = more attraction


How much does ph change when an acid or a base is added to a buffer solution?



∙ List the types of intermolecular forces in order of weak to strong and  describe them at the molecular level

o Dispersion Forces (London Forces):  

 Attractive forces that arise as a result of temporary dipoles  induced in atoms or molecules

 all molecules are subject to dispersion forces

 the instantaneous dipole of one molecule can induce a  dipole in neighboring atoms (look at my “Lecture 1” notes  to see an illustration of this)

 it is the attraction of these dipoles that cause dispersion  forces

 as molar mass increases, so does the boiling point  Don't forget about the age old question of What is the meaning of adjacent nodes?

 this is because the magnitude of the dispersion force will  increase with more electrons that can move around easily  larger electron cloud = larger dispersion forces

 generally, higher molecular weights = larger dispersion  forces

 Polarizability: how easy the electron distribution in the  atom or molecule can be distorted (or “pushed around”) ∙ Increases with: greater number of electrons, more  

diffuse the electron could

o Dipole-Dipole Forces:

 Attractive forces between polar molecules

 Many molecules have permanent dipoles

 Molecules with permanent dipoles have a stronger  

attraction to each other

o Hydrogen Bonding:

 A special dipole-dipole interaction between the hydrogen  atom in a polar N-H, O-H, or F-H bond and an  

electronegative O ,N, or F atom

o Ion-Dipole Forces:

 Attractive forces between an ion and a polar molecule o Ion-Ion Forces:

 Attractive forces between ions

Lecture 3: Surface Tension, Viscosity, and Capillary Action ∙ Define and explain molecular factors that influence tension, viscosity,  and capillary action

o Surface Tension:  

 the tendency of liquids to minimize surface area Don't forget about the age old question of What is the purpose of the treasury dept in the federal government?

 measured in mJ/ m^2

 how much surface area is takes to increase the surface  area of a liquid

 think about how you see bugs running on top of a pond  generally, the higher the surface tension, the  

stronger the intermolecular forces

o Viscosity:  

 the resistance of a liquid to a flow

 measured in units of centipoise (cP)=g*cm^-1*s^-1)  think of the thickness of honey and how it takes a long  time to pour

 Decreases with increasing temperature

o Capillary action:

 combination of intermolecular attraction of molecules in a  liquid and the attraction of a liquid to a surface

 think of what happens when you touch a paper towel to  water

 this also causes water to travel up narrow tubes

∙ the narrower the tube, the farther it travels

Lecture 4: Vapor Pressure, Boiling Point, Critical Point, Sublimation,  Fusion, and the Heating Curve of Water

∙ Describe the process of vaporization and condensation at the  molecular level

o Vaporization: molecules leave the liquid state and become a  gas

o Condensation: molecules leave the gas state and become a  liquid

∙ Vaporization is an endothermic process

o What does endothermic mean?

 A system that absorbs energy from the surroundings

∙ Condensation is an exothermic process

o What does exothermic mean?

 A system that releases energy into the surroundings  

∙ Water is held together by hydrogen bonds

o A certain amount of energy is required to overcome the  hydrogen bonds

o At any temperature, some molecules have more kinetic energy  than others

o Those with enough energy can escape and form a vapor ∙ Heat of Vaporization (enthalpy of vaporization)

o Amount of energy needed to vaporize 1 mol of a liquid to a gas o Always exothermic

o Know how to calculate this

 This is a dimensional analysis problem

 You will be given the ΔHvap  

 Your answer will be in kJ

 You can see example problems in my “Lecture 4” notes ∙ Apply the concept of dynamic equilibrium to the vapor pressure in a  closed system

o Dynamic Equilibrium: the rate of evaporation is equal to the  rate of condensation

o So in a closed system in dynamic equilibrium, the number of  water molecules that are vaporizing is equal to the number of  molecules condensing

o Think about pulling a piston up: volume increases, pressure falls,  more gas vaporizes, and then the pressure is restored

o Then think about pushing a piston down: volume is decreased,  pressure rises, more gas condenses, and then pressure is  restored

o There is an illustration in this in our powerpoint and in my  “Lecture 4” notes

o This piston example perfectly explains Le Chatelier’s  Principle:

 When a system in a dynamic equilibrium is disturbed, the  system responds so as to minimize the disturbance and  return to a state of equilibrium

o Temperature Dependence

 As you increase temperature, the vapor pressure increases  until it reaches its boiling point.

∙ Use the Clausius-Clapeyron equation (both forms)

o The Clausius-Claperion Equation relates vapor pressure to  vaporization

o This is in the form of y=mx+b

o You can plot and solve for this

o ������������ = (−∆��������/ ����) + ln ��

o ������������ = (−∆��������/ ��) (1/ ��) + ln ��

o You can see a worked out example in my “Lecture 4” notes ∙ Describe the critical point

o A point on a phase diagram at which both the liquid and gas  phases of a substance have the same density, and are therefore  indistinguishable 

∙ What is sublimation? 

o Direct transition of a solid to a gas 

o Think of dry ice 

∙ What is fusion? 

o Fancy word for melting 

o Heat of fusion (∆��fus)

 The energy required to melt one molecule of a substance ∙ The Heating Curve of Water

o You can see a good diagram of this in my “Lecture 4” notes  o Or you can google it and look at those diagrams (those may  exclude some of the info Dr. Rupar provided in class and expects  us to know)

Lecture 5: Phase Diagrams

∙ Calculate the quantify the heat needed for a substance to undergo  phase transitions  

o You can see a worked out example in my “Lecture 5” notes ∙ What is a phase diagram?

o a diagram representing the limits of stability of the various  phases in a chemical system at equilibrium, with respect to  variables such as composition and temperature 

o You can see a good diagram of this in my Lecture 5 notes taken  on September 1, 2016

o Or you can google it and look at those diagrams. Just make sure  the picture you study includes all of the “key parts” listed below ∙ Describe key parts of a phase diagram

o y-axis: pressure

o x-axis: temperature

o Gas: low pressure, high temperature

o Liquid: increase pressure

o Solid: increase pressure, lower temperature

o Critical Point: the point where the distinction between liquids  and gasses cease to exist

o Fusion Curve: liquid/solid barrier, also known as  

freezing/melting

o Vaporization Curve: liquid/vapor barrier, also known as  evaporation/condensation

o Sublimation Curve: solid/vapor barrier

o Triple Point: the temperature and pressure at which the three  phases (gas, liquid, and solid) of that substance coexist in  thermodynamic equilibrium

∙ Use a phase diagram to predict phase changes in a substance ∙ What makes the phase diagram of water special?

o Water expands when it freezes

∙ What is regelation?

o Melting ice by pressure

Lecture 6: Crystalline Solids

∙ Qualitatively describe x-ray crystallography

o X-ray particles go under constructive (dot) and destructive (not  dot) interference

o Constructive interference occurs when Bragg’s Law is satisfied ∙ What is Bragg’s Law?

o there is a definite relationship between the angle at which a  beam of x-rays must fall on the parallel planes of atoms in a  crystal in order that there be strong reflection 

∙ Use Bragg’s law to calculate spacing at the atomic level  o nλ=2dsinθ

o Look in my “Lecture 6” notes for a worked out example ∙ What is a crystal lattice?

o the symmetrical three-dimensional arrangement of atoms inside  a crystal

o a crystal lattice is made from unit cells 

∙ What is a unit cell?

o A particular arrangement of atoms that repeat over and over and over again to form a crystal lattice

o Tiles on a bathroom floor are the unit cells in a 2D space ∙ Describe the three cubic unit cells (way more notes on this in “Lecture  6” notes and in “Recitation 2” notes)  

o Simple Cubic Unit Cell:

 atoms at each corner

 coordination number of 6, packing efficiency of 52%, and  contains 1 atom per unit cell 

o Body-Centered Unit Cell:

 Atoms at each corner and one in the center

 Coordination number of 8, packing efficiency of 68%, and  contains 2 atoms per unit cell

o Face-Centered Unit Cell:

 Atoms at each corner and one on each face

 Coordination number of 12, packing efficiency of 74%, and  contains 4 atoms per unit cell

∙ What is a coordination number?

o The number of atoms with which each atom is in contact with ∙ What is packing efficiency?

o The % of volume of the unit cell occupied by the atoms o The higher the coordination number, the higher the packing  efficiency

Lecture 7: Classifying Crystalline Solids | Solutions and Solubility ∙ Describe the three crystalline solids

o Molecular Solids:

 Made from molecules

 Geld together by dipole-dipole bonds, hydrogen bonds, and dispersion forces

 They have a low melting point

 CO2 and H2O are good examples

o Ionic Solids:

 Made up of cations and anions

 They have a high melting point

 Interactions are much stronger

 NaCl is a good example

o Atomic Solids:

 Non-Bonding Atomic Solids:

∙ Held together by dispersion forces

∙ Very low melting points

∙ He and Ne are good examples

 Metallic Atomic Solids:

∙ Held together by metallic bonds

∙ Melting points have a huge range: from less than  

room temperature to 2000°C

 Network Covalent Atomic Solids:

∙ Held together by covalent bonds

∙ Tightly woven network of atoms

∙ High melting points

∙ Typically very hard

∙ Diamond (carbon) and SiO2 (quartz) are good  

examples

∙ List the components of a solution

o Solvent: the majority component of a solution

o Solute: the minority component of a solution

∙ Qualitatively describe entropy

o The pervasive tendency for energy to spread out

o Related to an increase in disorder

o And important role in solution creation

o Entropy drives a huge number of processes

o Entropy is always increasing

∙ Describe the energetics of the dissolution process  

∙ Using the concept of polarity, identify whether a compound will  dissolve in a solvent

o Solvents are broadly categorized as being polar or non-polar o Like dissolves like

 Polar solvents dissolve in polar compounds

 Non-polar solvents dissolve in non-polar compounds  Polar solvents do not dissolve in non-polar compounds  Non-polar solvents do not dissolve in polar compounds

Lecture 8: Ionic and Metallic Atomic Solids

∙ In Lecture 8, the three crystalline solids were described in even more  detail. In order to prevent redundancy, I’m going to leave this out of  my notes. You can look at my “Lecture 8” notes if you want to read  about the crystalline solids in more detail. However, I will provide an  outline and describe the new information given in the lecture: o Molecular Solids

o Ionic Solids

o Atomic Solids

 Non-Bonding Atomic Solids

 Metallic Atomic Solids

 Network Covalent Atomic Solids

∙ Graphite

o Sp2 hybridized carbon atom that forms fused  six-membered rings arranged in sheets

o The sheets can slide past each other which  makes graphite a good lubricant (used on  

automobiles)

o Trigonal planar geometry for each C atom o The sheets are held together by dispersion  forces

o Graphite’s melting point is around 3800°C o It’s hard to melt because you have to disrupt  covalent bonds between each individual sheet o Forms hexagonal crystals

o Graphite is an electrical conductor parallel to  the sheets (only one direction)

∙ Diamond

o Also made up of carbon atoms but sp3  

hybridized

o Tetrahedral

o Every carbon is bonded to 4 other carbon  atoms

o A diamond crystal is basically just one carbon  molecule- that’s why it’s so strong

o Diamond’s melting point is around 3800°C o Very hard and rigid

o Diamond is an electrical insulator- it doesn’t  conduct electricity

o But it is the best know thermal conductor ∙ Buckminsterfullerene: Buckyball

o Forms 60 carbon atoms in the shape of a  soccer ball

o C36-C100 but C60 is the most common form  The whole family of structures are called  “Fullerenes”

o If stacked, the balls are held together by  dispersion forces

o Individual balls are held together by covalent  bonds

o Because the stacked balls are held together by  dispersion forces, they can be dissolved

∙ Nanotubes

o These are like rolled up graphite layers

o Very strong

∙ Quartz

o SiO2

o Similar in structure to diamond

o 3D array of each Si atom covalently bonded to  

4 oxygen atoms

o Tetrahedral

o Melting point is around 1600°C

o Very hard

∙ Silicates

o SiO2 units but sometimes Al in place of Si

 The ones with aluminum are called  

“aluminosilicates”

 Aluminosilicates make up 90% of Earth’s  

crust

o Glass is SiO2

 Amorphous = not crystallized

 Melts at much lower temperature

∙ Know how to classify crystalline solids

∙ Describe the Cesium Chloride, rock salt, zinc blend, and fluorite  structures

o Cesium Chloride (CsCl)

 Body-Centered Cubic

 Coordination number = 8

 Cs+ in the center

 Cl- on the edges

o Rock Salt (NaCl)

 Face-Centered Cubic

 Coordination number = 6

 4 Cl- anions

 4 Na+ cations

o Zinc Blend

 Face-Centered Cubic

 Coordination number = 4

 4 S anions

 4 Zn cations

o Fluorite

 Face-Centered Cubic

 4 Ca cations  

 8 F anions

 Coordination number = 8

Lecture 9: Solutions and Solubility Continued | Solution Concentrations

∙ Energetics of solution formation

o ΔHsolution = ΔHsolute + ΔHsolvent + ΔHmix

o Δ Hmix determines the whether the solution is endothermic or  exothermic

o Typically, ΔHsolvent + ΔHmix are combined and called ΔHhydration ∙ Exothermic: energy cost of breaking solute-solute and solvent-solvent interactions is less than energy released upon mixing

∙ Endothermic: energy cost of breaking solute-solute and solvent solvent interactions is greater than energy released upon mixing ∙ Heats of Solution

o Because lattice energy is always exothermic, (therefore ΔHsolution  is always endothermic) the sign and the size of ΔHsolution tells us  something about ΔHhydration

o ΔHhydration < ΔHlattice, then dissolution will be endothermic o ΔHhydration > ΔHlattice, then dissolution will be exothermic ∙ Solution Equilibrium

o Imagine you have a glass of water and you pour salt (NaCl) in. At  a certain point, the salt will stop dissolving. This is called  saturation.

o Saturation: max amount of solute in a solvent

o Solution Equilibrium: rate of dissolution and rate of  recrystallization (un-dissolving) is equal

o Supersaturated Solution: solutions that contain more than  equilibrium saturated solution (very rare)

 solution with more dissolved solute than the solvent would  normally dissolve in its current conditions 

∙ The effect of temperature on the solubility of solids and gasses o The solubility of solvents increases with temperature 

o For gasses this is opposite: low temperature = higher solubility  This is why little bubbles form as you heat water (below  100°C) 

∙ Gas Solubility: Henry’s Law

o The higher the gas pressure, the more soluble a gas is in a liquid o Sgas = KH x Pgas

o Sgas: solubility in mol/L

o KH: Henry’s Law gas constant

o Pgas: pressure of the gas

∙ Perform calculations with Henry’s Law

o Look in my “Lecture 9” notes for a worked out example ∙ Concentrations of solutions:

o mol/L

o Molality=(mols solute)/(mass of solvent in kg)

o % mass=(mass solute)/(mass solvent)

o ppm=(mass solute)/(mass solvent) x 10^6

o ppb=(mass solute)/(mass solvent) x 10^9

∙ Calculate the concentration of a solution using molarity, molality, ppm  and ppb

o Look in my “Lecture 9” notes for worked out examples

Lecture 10  

∙ Colligative Properties: properties of a solution that depends on the  concentration

∙ Vapor Pressure Lowering: the vapor pressure of a solution is lower  than the vapor pressure of the pure solvent

∙ Raoult’s Law: the partial vapor pressure of each component of an  ideal mixture of liquids is equal to the vapor pressure of the pure  component multiplied by its mole fraction in the mixture o Psolvent=Xsolvent ×P0solvent

 Psolvent is the vapor pressure of the solvent above the  solution

 Xsolvent is the mole fraction of the solvent in the solution  P0solvent is the vapor pressure of the pure solvent

∙ Freezing Point Depression: the freezing point of a liquid is  depressed by a solute

o ∆Tf = mKf

 m = molality (moles of solute / kg of solvent)

 Kf = molal freezing point depression constant

∙ Boiling Point Elevation: the boiling point of a liquid is elevated by a  solute

o ∆Tb = mKb

 m = molality (moles of solute / kg of solvent)

 Kb = molal boiling point elevation constant

∙ Colligative properties do not depend on the identity of the solute ∙ Osmosis: the flow of solvent from a solution of low concentration to a  solution of high concentration

∙ Osmotic Pressure: the pressure required to stop osmotic flow o ∏=MRT

 ∏= osmotic pressure (atm)

 M= Molarity (mol/L)

 R= ideal gas constant (0.08206)

 T= temperature (K)

∙ We assume that strong electrolytes completely dissociate in solution  but that’s not always true

∙ Van’t Hoff Factor (i): a measure of the effect of a solute upon  colligative properties such as osmotic pressure, relative lowering in  vapor pressure, elevation of boiling point and freezing point depression o i= moles of particles in a solution/moles of formula units

o i depends on concentration, temperature, solvent, and solute  o You can incorporate i into all equations to account for Van’t Hoff  Factors

 ∆Tf = imKf

 ∆Tb = imKb

 ∏=iMRT

Lecture 11

∙ Rates of Reactions: the measure of the change in concentration of  the reactants or the change in concentration of the products per unit  time

o Reaction rate can be defined either as the increase in the  concentration of a product per unit time or as the decrease of the concentration of a reactant per unit time. By definition, reaction  rate is a positive quantity. 

o Ex.) X �� 2Y

o Rateof X=12(rate of Y)

∆ t=12(∆[ Y ]

o ¿−∆[ X ]

∆ t)

o So basically, rate = concentration change/time change o Factors that influence rate

 Concentration: increases the chances of collision

 Temperature: causes particles to excite and smash into  each other very fast so they can be successful in breaking  bonds

 Structure and orientation

o The rate of a chemical reaction is defined by either the loss of a  reagent or the formation of a product

∙ Instantaneous Rate of Reaction: can be measured at any time  between the moment at which the reactants are mixed and the  reaction reaches equilibrium. Extrapolating these data back to the  instant at which the reagents are mixed gives the initial instantaneous  rate of reaction

∙ Rate Law: an equation that links the reaction rate with concentrations or pressures of reactants and constant parameters

o rate = k [ A]n[B]m 

o k = the rate constant (specific to each reaction and conditions) o [A] = the concentration of the reagent

o n and m = reaction order

 must be found through experiment

 n+m= the order of the reaction

∙ For A+B �� C+D

o The rate = −∆[ A]

∆t=−∆[B]

∆t=∆[C]

∆t=∆[D]

∆t

∙ Reaction Orders: found by doing experiments

o Zero Order: rate is independent of [A]

 n = 0

 Rate = k

o First Oder: rate is proportional to [A]

 n = 1

o Second Order: rate is proportional to [A]^2

 n = 2

Lecture 12

∙ Integrated Rate Law: relates the concentration of the reactants and  time

o Zero Order: rate=k [ A]0=k

o First Order: [ A]=[ A]0e−kt 

o Second Order: 1[ A]=kt+1[ A0]

 It would be helpful to look up pictures of the zero order,  first order, and second order graphs

 You plot your equation to see what order it fits into

 If the line is straight, that means that it is the correct graph  If it is curved, then it is a different order

∙ Ex.) If you plot your equation and the graph of the  

second order is a straight line and the graphs of the  

zero order and first order are curved, then you have  

a second order reaction

∙ Reaction Half-Lives: the amount of time needed for the  concentration to fall to ½ (50%)

A

o Zero Order: t ½ =  

¿¿

¿0 ¿¿

o First Order: t ½ = 0.693 

k

o Second Order: t ½ = 1

k [ A]0 

∙ The Arrhenius Equation: relates the rate of a reaction with the  temperature

o k = Ae−Ea

RT

 k = rate constant

 Ae = frequency factor  

 Ea = activation energy

 R = gas constant (8.3145)

 T = temperature in kelvin

o ln(k1

k2 )=(Ea 

R)(1T2−1T1)

 k 1 = rate constant at temperature T1 

 k 2 = rate constant at temperature T2 

 Ea = activation energy

 R = gas constant (8.3145)

 T = temperature in kelvin

∙ Activation Energy: the minimum energy which must be available to a chemical system with potential reactants to result in a chemical  reaction

o You must put energy into molecules for them to be able to  rearrange

o For a reaction to occur, chemicals typically form a high energy  “activated complex” or “transition state”

o Activation energy = the difference in energy between the  reactants and the activated complex

 ALWAYS a positive number

o High Ea = small k = slow reaction

o Low Ea = big k = faster reaction

∙ Frequency Factor (Ae): can be thought of as the number of times  the reagents approach the activation barrier

o Most reactions require molecules to collide with specific  orientations for a reaction to occur

∙ Exponent Factor (-Ea/RT): the fraction of molecules that have  enough energy to overcome the activation barrier and form a transition state

o Always between 0 and 1  

o If Ea = 0, the exponent factor =1

 Therefore, every approach of molecules results in a  

reaction

 High Ea = slower reaction

 Low Ea = faster reaction

 High temperature = fast reaction

 Low temperature = slow reaction

Lecture 13

∙ The Collision Model: explains why most collisions between molecules do not result in a chemical reaction

o According to the collision model, a chemical reaction can occur  only when the reactant molecules, atoms, or ions collide with  more than a certain amount of kinetic energy and in the proper  orientation 

o Has two components:

 The orientation factor (p)

 The collision frequency (z)

o A=pz

 Where A = frequency factor

∙ Orientation Factor: represents the fraction of collisions with an  orientation that allows the reaction to occur

o If p = 1, this means that orientation does not matter and every  collision results in a reaction

o For most reactions 0≤ p≤1

o For reactions with complex collision requirements,

p=1×10−5∨1×10−6 

o Reactions that involve atoms (just atoms, not molecules) p = 1 or higher

 Some reactions with atoms have p>1

o Small p = more complex reaction

 Lower symmetry would make it more complex

o Reaction Mechanisms

 Most reactions occur through multiple steps

 Each step is called an elementary step

∙ Can’t be broken down any further

o Rate Laws for Elementary Steps

 The rate law for elementary steps can be determined from  the chemical equation for that step

∙ A �� product

o Rate = k [A]

o This is called a unimolecular step  

∙ A + A �� product

o Rate = k [ A]2 

o This is called a bimolecular step

∙ A + B �� product

o Rate = k [ A] [B]

o This is also called a bimolecular step

 Molecularity: number of reactant particles in an  

elementary step

Rate Determining Step (RDS): the slowest elementary step that the  overall rate of a reaction is often approximately determined by ∙ This is why some reactions are missing a reagent in the overall rate law o Ex.) NO2+CO→ NO+CO2 

o Rate = k [N O2]2

o This tells us that CO is not part of the RDS

Lecture 14

∙ Catalysts: increase the rate of the reaction but aren’t consumed o Catalysts lower the activation barrier (Ea)

∙ Dynamic Equilibrium: exists once a reversible reaction ceases to  change its ratio of reactants/products, but substances move between  the chemicals at an equal rate, meaning there is no net change  

o Equilibrium does not imply that the concentrations are the same  on both sides

o Equilibrium reactions are described by the constant K (big K) o Ex.) aA + bB ↔ cC + dD

 K= [C]c[D]d 

[ A]a[B]b 

 This relationship between chemical equation and  

concentration of reactants and products is called the Law  of Mass Action

o What does K mean?

 If K>>1, at equilibrium, there will be much more product  than starting materials  

∙ Forward reaction goes essentially to completion

 If K<<1, then the reactants are favored over the products ∙ The forward reaction does not go very far

 If K is close to 1 then the reaction goes about halfway  o Relationship Between K and Chemical Equations  If you add reaction equations to get a new equation, the Keq of the new equation is the product of Keq of the old equations

Lecture 15

∙ K in Terms of Pressure

o The concentration of a gas in a mixture is proportional to its  partial pressure

o For Keq=[C]c+[ D]d 

[ A]a+[ B]b 

o K p=PCcx PDd

PAax PBb

o K p depends on the units of the ideal gas constant “R”  R = 0.08206 L∗atm

mol∗K

o So partial pressure will be in atm

o K p=Keq(RT )Δ N

o R = the ideal gas constant

o T = temperature in K

o Δ N = the different between the number of moles of reactants  and the number of moles of products

∙ Heterogeneous Equilibrium: the concentration of a pure liquid and  a pure solid do not change during the course of a reaction ∙ The Reaction Quotient: aids in figuring out which direction a  reaction is likely to proceed, given either the pressures or the  concentrations of the reactants and the products

o If a reaction mixture is not at equilibrium, how can we determine  which direction it will proceed?

 We will compare the current concentration ratios to Keq ∙ Predicting the Direction of Change

o Q>K the reaction will go in the reverse direction

o Q<K the reaction will go in the forward direction

o Q=K the reaction is at equilibrium

∙ Le Chatelier’s Principle

o When a chemical system at equilibrium is disturbed, the system  shifts in a direction to minimize the disturbance

o Ex.) aA + bB ↔ cC + dD

 If you add more A or more B, the reaction will shift to the  right to lower [A] and [B]

 If you add more C or more D, the reaction will shift to the  left to lower [C] and [D]

 If you remove some A or B, the reaction will shift to the left   If you remove some C or D, the reaction will shift to the  right

Lecture 16

∙ The Effects of Volume and Pressure on Equilibrium o Decreasing the volume of a container increases  

pressure/concentration

o Via Le Chatelier’s principle, if pressure increases, the equilibrium  will shift to reduce pressure

∙ Effects of Temperature on Equilibrium

o Exothermic reaction releases heat ( −Δ H ¿

o Endothermic reaction absorbs heat ( +Δ H ¿

o If you add heat, the reaction shifts right

o If you remove heat, the reaction shifts left

∙ What is an acid?

o A chemical that causes +¿

H3O¿ (also referred to as +¿H¿ to form  

an aqueous solution

o Sour taste

o Tend to dissolve metals

o Neutralize bases

o Ex.) HCl, acetic acid (vinegar), and HF

∙ Organic Acids

o Most organic acids have a carboxylic acid function  You can google this to see what it looks like

∙ What is a base?

o A chemical that causes −¿

O H¿ to form an aqueous solution

o Tastes bitter

o Neutralize acids

o Feel very slippery

o Ex.) NaOH ,KOH ,N H3(ammonia)

∙ Definitions of Acids and Bases

o Arrhenius Definition (too narrow of a definition) (typically  rejected by chemists)

 Acids: chemicals that make +¿∈H2O

H3O¿ 

 Bases: chemicals that produce −¿∈H2O

O H¿ 

o Br ? nsted-Lowry Equation (preferred definition)  Acids: chemicals that donate +¿∈H2 O

H3O¿ 

 Bases: chemicals that accept +¿∈H2O

H3O¿ 

+¿

−¿+H−B¿ 

H−A+ B ↔ A¿ 

 H−A = acid

 B = base

−¿A¿ = conjugate base

+¿

H−B¿ = conjugate acid

 Be able¿ lable these∈achemical equation!

o Lewis Definition (most general definition) (typically rejected by chemists)

 Acids: chemicals that accept and electron pair

 Bases: chemicals that donate and electron pair

Lecture 17

∙ Binary Acids: certain molecular compounds in which hydrogen is  combined with a second nonmetallic element

o Ex.) HF, HCl, HBr, HI

o H-Y is the general formula

o The weaker the H-Y bond, the stronger the acid

o When looking at a periodic table: acid strength increases doing  down a column and increases from left to right

 Explained via electronegativity

∙ Oxyacids: an acid that contains an oxygen atom bonded to a  hydrogen atom and at least one other element

o Ex.) H2S O4, H3PO4, HN O3 

o HYO is the general formula

o The more electronegative Y is, the stronger the acid o Usually, the more oxygen atoms, the stronger the acid ∙ Strong Acid/Base vs. Weak Acid/Base

o A strong acid/base is a strong electrolyte

 100% ionization

o A weak acid/base is a weak electrolyte

 Only a small % ionization

 Equilibrium is left

∙ Acid Ionization Constant ( Ka )

o The bigger the Ka, the stronger the acid ( Ka >>1) o For weak acids, Ka <<1

∙ The Autoionization of Water

o H2O is amphoteric (it can be both an acid or a base) o This process is called autoionization

o ALL aqueous solutions have +¿

H3O¿ and −¿

O H¿ 

H

o In pure H2O at 25°C,  

¿+¿¿¿] = [ −¿

O H¿ ] = 10−7 M

o You do not include liquids or solids in Keq equations, only  aqueous solutions

o If +¿

H3O¿ > −¿

O H¿ , this is acidic

o If +¿

H3O¿ < −¿

O H¿ , this is basic

∙ pH: a measure of how acidic/basic something is. The range goes from  0 - 14, with 7 being neutral

H

o pH = -log

¿+¿¿¿]

o In pure water,  

H

¿+¿¿¿]= 1 x 10−7 M

o p HH2O=−log ?(1 x10−7)

o p HH2O=7

o Most things have a pH between 0 and 14

∙ pOH: another way of expressing the strength of an acid or base o the negative of the logarithm of the OH ion concentration o pH = -log[−¿

O H¿ ]

o [−¿

O H¿ ] = 10−pOH 

o In pure water, [ −¿

O H¿ ]= 1 x 10−7 M

o pOH=−log ?(1 x10−7)

o pOH=7

o If pOH > 7, the solution is acidic

o If pOH < 7, the solution is basic

o pH + pOH = 14.00 at 25°C

∙ p Ka : another way of expressing the strength of an acid or base o p Ka = -log Ka 

o Ka = 10−p Ka

o The stronger the acid, the smaller the p Ka 

∙ Finding pH of a strong acid solution

o With a strong acid, ionization is 100%

−¿

o o

+¿+C l¿ 

HCl+H2O → H3O¿ +¿

H3O¿ 

¿¿

+¿

o

H 3O¿ 

¿¿

pH=−log ¿

∙ Finding pH of a weak acid solution

o In weak aqueous solutions, there are two sources of +¿ 

H3O¿ 

 Weak acid

 H2O (tend to ignore because H2O is a very weak acid) o For a weak acid, finding +¿ 

H3O¿ requires solving an equilibrium  

equation

Lecture 18

∙ Percent Ionization

o The higher the percent ionization, the stronger the acid −¿

A¿ 

¿+¿

o

H3O¿ ¿¿¿¿

+¿

o Increasing [HA] will always increase  

H3O¿ ¿

o However, increasing [HA] will decrease % ionization +¿

o The increase in  

H3O¿ ¿

 is slower than the increase in [HA]

o The more dilute the concentration, the more dissociated ∙ Finding the pH of mixtures of acids

o If you have a mixture with a strong acid and a weak acid, the  strong acid dominates. For pH calculations, you can ignore the  weak acid

∙ Kb and pKb

−¿

o For  

+¿+O H¿ MOH ↔ M¿ +¿

M¿ 

 Kb=

¿−¿

O H¿ ¿¿¿

 pKb= −log ?[Kb]

o The higher the Kb, the stronger the base

o The lower the pKb, the stronger the base o The lower the Kb, the weaker the base

o The higher the pKb, the weaker the base ∙ Finding pOH and pH for a strong base solution o The easiest way to find pH is to first find the pOH o pOH = -log[OH]

o pH+pOH=14

∙ Acid-base properties of ions and salts

o Water soluble ionic compound

o Salt that has the cation of a strong base and the anion of a weak  acid is basic

o Salt that has the cation that is the conjugate acid of a weak base  is acidic

Lecture 19

∙ Relationship between Ka of an acid and Kb of its conjugate base o Ka x Kb = Kw

o Kw = 1 x 1 0−14 

o If Ka is big → strong acid

o If Ka is small → weak acid

o If Ka is big → weak base

o If Ka is small → strong base

o If Kb is big → weak acid

o If Kb is small → strong acid

o If Kb is big → strong base

o If Kb is small → weak base

o If pKa is small → strong acid

o If pKa is big → weak acid

o If pKa is small → weak base

o If pKa is big → strong base

∙ Cations as weak acids

o Cations of weak bases are weak acids

o The stronger the parent base, the weaker the conjugate acid ∙ Metal cations as weak acids

o Cations of small, highly charged metals are weakly acidic   Not group 1 or group 2 cations

∙ Classifying salt solutions as acidic, basic, or neutral

oN O3¿2, KBr

NaCl ,Ca ¿

 Neutral

oC2H3O2¿2,KN O3 

NaF ,Ca¿

 Basic

oN O3¿3 

N H4Cl , Al ¿

 Weak Acid

Lecture 20

∙ Polyprotic Acids

o Because polyprotic acids ionize in steps, each H has a separate  Ka value

o Generally, if Ka is >> than Ka2, you can ignore the 2nd ionization  when calculating pH

∙ Lewis acid-base theory

o Lewis acid: accepts and electron pair

 Electron deficient

∙ Incomplete → do not have a full octet

∙ Has lots of attached electronegative elements

o Lewis base: donates and electron pair

 Donates the electron pair to a lewis acid

 Generally, anions are better lewis bases than neutral  molecules

 Also, less electronegative molecules make better lewis  bases because they don’t hold on to their electron as  

tightly

o Chemists call lewis acids electrophiles

o Chemists call lewis bases nucleophiles

o Adduct: a lewis acid and base together

∙ Buffers

o Solutions that resist changes in pH upon addition of an acid or  base. They resist change by neutralizing the added acid or base o Two types:

 A solution of a weak acid and its conjugate base

 A Solution of a weak base and its conjugate acid

 BOTH must have a significant amount of the conjugate ∙ Buffers in action

o Adding a strong base to an acidic buffer → the strong base is  neutralized by the weak acid

o Adding a strong acid to an acidic buffer → the strong acid is  neutralized by the conjugate base of the buffer

o Buffers work by applying LeChatelier’s principle to weak acid  equilibrium

o Buffers contain a significant amount of weak acids

o The weak acid can react with a strong base to neutralize it  

Lecture 21

 What is a buffer?

o Solutions that resist changes in pH upon addition of an acid or  base

o They resist change by neutralizing the added acid or base o A solution of both a weak acid and its conjugate base

o You can calculate the pH of a buffer by either solving the usual  way (quadratic equation), or you can look at and if the numbers  are similar, you can approximate

 The short way: Henderson-Hasselbalch Equation

∙ pH = pKa + log [base]/[acid]

∙ This works well if Ka is small (like x 10−4 or smaller) ∙ It also works well if the [base] and [acid]  

concentrations are relatively high

∙ Ex.) If the initial acid and salt concentrations are  

100x higher than Ka, the Henderson-Hasselbalch  

method will work

 How much does pH change when an acid or a base is added to a buffer solution?

o Stoichiometry calculation

 If a strong acid is added, it will reduce the concentration of  the conjugate base and increase the concentration of a  weak acid

 If a strong base is added, the concentration of the  

conjugate base increases and the concentration of the  

weak acid decreases

o Perform an equilibrium calculation for +¿

H3O¿ 

 Basic buffers  

o Contain a weak base and its conjugate acid

 Ex.) N H3( weak base )∧N H4Cl(conjugate acid)

 Henderson-Hasselbalch equation for basic buffers

o pOH = pKb + log [acid]/[base]

Lecture 22

 Capacity and range

o Range: the pH at which the buffer can be most effective  Related to the pKa/pKb of the buffer components  

o Capacity: the amount of acid or base that can be neutralized  Related to the concentration of the buffer

 Buffering range

o A buffer is most effective if 0.1 < [base] : [acid] < 10 o The effective pH range of a buffer is ± 1 of the pKa

 Buffering Capacity

o Concentrated buffer = more effective

o Dilute buffer = less effective

 Effectiveness of buffers

o Buffers are more effective when [base] : [acid] = 1

o Buffers are more effective when [base] : [acid] is high  Titration

o In an acid base titration, a solution of unknown concentration is  added to a solution of known concentration

o An indicator is used that changes color when moles −¿

OH¿ =  

moles +¿ 

H3O¿ 

 Titration curve

o A plot of pH vs. amount of titrant added

o Equivalence point (point of infection): where the color change  occurs

 For a strong acid and a strong base, the equivalence point  is pH=7

Lecture 23

 Solubility product: Ksp

n−¿ ( aq)

o For  

m+¿ (aq)+m X¿ MnXm (s) ↔n M¿ n−¿ ¿m 

X¿ 

o Ksp =  

m+¿¿n¿ M¿ 

¿

 Molular solubility

o The number of moles of solute that dissolve in 1L of water o Ksp value can be directly compared for ionic compounds with the same dissociation stoichiometry

 The effect of common ions on solubility

−¿ (aq)

o For  

2+¿ (aq )+2C l¿ PbC l2(s) ↔ P b¿ 

o If you add more −¿

Cl¿ (from HCl), the equation will shift left

 The effect of pH on solubility

o For ionic hydroxide compounds, high pH reduces solubility o For ionic compounds that have anions of weak acids, the lower  the pH, the higher the solubility

 Precipitation

o If Q = Ksp, the solution is saturated

o If Q < Ksp, the solution is not saturated

o If Q > Ksp, the solution is above saturation and the reactant will  precipitate

 Reversibility of process

o A spontaneous process is irreversible because it is a net release  of energy

o A reversible process will proceed back and forth

 AKA an equilibrium

 No change in free energy

 Two factors affecting whether a reaction is spontaneous o Δ H : enthalpy

 Change in internal energy plus work (p x v) between  reactants and products

o Δ S : entropy

 Enthalpy change  

o Enthalpy is measured in kJ/mol

o Δ H is –ve: exothermic reaction

o Δ H is +ve: endothermic reaction

 Entropy and the 2nd law of thermodynamics

o 2nd law: entropy will always increase

o Entropy Δ S¿) is a thermodynamic function that increases the  number of energetically equivalent ways of arranging the  components of a system (J/mol)

 S = K ln W

 K: Boltman constant (1.38 x 10−23 J/K)

 W: number of energetically equivalent systems that can  coexist

Lecture 24

 Factors affecting whether a reaction is spontaneous

o Enthalpy change  

o Entropy change

 Second law of thermodynamics

o The total entropy change must be positive to be spontaneous o Δ S universe = Δ S system + Δ S surroundings

o If the entropy of the system decreases, then the entropy of the  surroundings must increase by a larger amount

o Entropy changes in the surroundings is proportional to the  amount of heat gained or lost

o qsurroundings=−qsystem 

o Entropy change in the surroundings is inversely proportional to  it’s temperature

o Δ S surroundings = −qsystem 

T=−Δ Hsystem

T

 Gibbs Free Energy and spontaneity

o Δ G system = Δ H system - T Δ Ssystem

o Gibbs free energy is the maximum amount of work energy that  can be released to the surroundings by a system at constant  temperature and pressure

∙ AKA the chemical potential of a system

o A process will be spontaneous when Δ G is negative, Δ H is  negative, and Δ S is positive

∙ Δ G will be negative when:

∙ Δ H is negative and Δ S is positive  

o Exothermic and more random

∙ Δ H is negative and large and Δ S is negative but  small

∙ Δ H is positive but small and Δ S is positive and  large  

o Or high temperature

∙ Δ G will be positive when:

∙ Δ H is positive and Δ S is negative  

o Never spontaneous are any temperature

 Standard conditions

o Gas: pure gas at exactly 1 atm

o Solid or liquid: pure solid or liquid at 1 atm at stable temperature  Third law of thermodynamics

o Absolute entropy

Lecture 25

∙ Δ G system = Δ H system - T Δ S system

∙ -T Δ S universe = Δ H system - T Δ S system

∙ Δ G system = -T Δ S system

∙ Δ S universe always > 0

∙ Standard free energies for formation

o Δ G° f: the change in free energy when 1 mol of compound x is  formed from its elements

 Δ G°reaction = Σ n Δ G° products - Σ n Δ G°reactants 

∙ Δ G Relationships

o If a reaction is reversed, the sign of Δ G flips

o If the amount of material in a reaction is multiplied by a factor,  the value of Δ G is also multiplied by the same factor

Lecture 26

∙ What’s free about free energy?

o The free energy is the maximum amount of energy released from a system that is available to do work on the surroundings o For many exothermic reactions ( Δ H system < 0), some of the  heat released increases the entropy of the surroundings, so  cannot do work

o | Δ Go| < ¿Δ Ho|, the difference in energy always increases  the heat of the surroundings and increases entropy of the  surroundings

∙ Free energy and reversible reactions

o The change in free energy is a theoretical limit as to the amount  of work that can be done

o If the reaction achieves its theoretical limit, it is a reversible  reaction

∙ Δ G under standard conditions:

∙ Δ G = Δ Go only when the reactants and products are in their standard states  

∙ Under nonstandard conditions:

∙ Δ G = Δ Go + RT lnQ

 Q = reaction quotient

∙ Δ Go and K

o Because Δ G reaction = 0 at equilibrium, Δ Go = -RT ln(K)  If K < 1, Δ Go is positive and the reaction is spontaneous  in the reverse direction

 If K > 1, Δ Go is negative and the reaction is spontaneous in the forward direction

∙ What happens to the entropy of liquid water when it evaporates? o The entropy increases

∙ Under what conditions can an endothermic reaction be spontaneous? o If T Δ S > Δ Hreaction

∙ Oxidation-Reduction (redox) reactions

o Reactions in which electrons are transferred from one reactant to the other

o To convert a free element to an ion, the atoms must gain or lose  electrons

o Atoms that lose electrons are being oxidized

o Atoms that gain electrons are being reduced

o OIL RIG

 Oxidation Is Losing

 Reduction Is Gaining

o Oxidation and reduction must occur simultaneously

Lecture 27

∙ Identifying redox reactions

o Oxidation: increase in oxidation state

o Reduction: decrease in oxidation state

∙ Redox Reactions

o The transfer of electrons does not need to be a complete transfer ∙ Oxidation States

o For reactions that are not metal and nonmetal, or do not involve  O2, a method for determining how the electrons are transferred  is required to determine which element is being oxidized number  to every atom in a molecule

o Chemists can assign an oxidation number to every atom in a  molecule

o In a redox reaction, the oxidation number of some atoms will  change

o Oxidation numbers are not the same as ionic charge

o Oxidation states are imaginary charges assigned based on rules ∙ Rules for assigning oxidation states

1.) Free elements have an oxidation state of zero  

2.) Monoatomic ions have an oxidation state equal to their charge 3.) The sum of the oxidation states of all the atoms in a compound is 0  and the sum of the oxidation states of all the atoms in a polyatomic  ion equals the charge on the ion

4.) Group I metals have an oxidation state of +1 in all their compounds 5.) Group II metals have an oxidation state of +2 in all their compounds 6.) In their compounds, nonmetals have oxidation states according to the table below:

Nonmetal

Oxidation State

Fluorine

-1

Hydrogen

+1

Oxygen

-2

Group 17

-1

Group 16

-2

Group 15

-3

∙ Assigning oxidation states

o+¿

N a¿ = +1

o Cl2 = 0

o LiF = Li: +1, F: -1

o C O2 = C: +4, O: -2,-2

oO43−¿

P¿ = P: +8, O: -2,-2,-2,-2

o K2O2 = K: +1, K: -1

∙ Half-Reactions

o The redox reaction is split into 2 separate half-reactions o Oxidation half-reactions have electrons as products

o Reduction half-reactions have electrons as reactants

∙ Balancing redox reactions by the half-reaction method o The reaction is broken down into 2 half-reactions, one for  oxidation and another for reduction

o Electrons are written are reagents

 Electrons are products for oxidation reactions  

 Electrons are reactants for reduction reactions

∙ Steps for balancing redox reactions by the half-reaction  method:

1.)Assign oxidation states by determining the element oxidized and  the element reduced.

2.)Write oxidation and reduction half-reactions, including electrons.  a. Oxidation electrons should be on the right and reduction  electrons on the left of the arrow.  

3.)Balance half-reactions by mass.  

a. First balance elements other than H and O.  

b. Add H2O where O is needed.  

c. Add H+ where H is needed.  

d. If the reaction is done in a base, neutralize H+ with OH− .  4.)Balance half-reactions:  

a. Balance charge by adjusting electrons.  

b. Balance electrons between half-reactions.  

5.)Add half-reactions  

6.)Check by counting atoms and total charge.

Lecture 28

∙ Electrochemistry

o The study of redox reactions that produce or require an electric  current

o Electrochemical cells

o Voltaic cells: spontaneous redox reaction that produces a current o You can use energy to reverse a reaction (charging the lithium ion battery in your cell phone)

∙ Voltaic (Galvanic) cells: spontaneous redox reactions o Electrical current: the amount of electrical charge that passes a  point in a given period of time (amps)

o Since redox reactions involve the transfer of electrons from one  substance to another, redox reactions can be used to generate  currents

o A voltaic cell produces an electrical current from a spontaneous  redox reaction

 For this to occur, there must be a separation of the  

oxidation and reduction reactions  

∙ Electrochemical cells

o Oxidation and reduction half-reactions are kept as separate in  half-cells in electrochemical cells

o Electrons will flow along a wire to complete the redox reaction o Ions flow between cells via salt bridge

o Need conducting solid electrodes

∙ Electrodes of electrochemical cells

o Anode: electrode where oxidation occurs

 Anions will be attracted to it

 Positive end of a battery

 Typically lose mass

o Cathode: electrode where reduction occurs

 Cations will be attracted to it

 Negative end of a battery

 Typically gain mass

∙ Voltage and current

o Voltage: the difference in potential energy between the  reactants and products

 Also called the potential difference

 Measured in volts

 1V = 1J of energy per coulomb

 think of this as the force pushing electrons through the  wire

o Current: the number of electrons that flow through the system  per second

 Measured in amps

∙ Cell potential

o The difference to potential energy between the anode and the  cathode in a voltaic cell is called the cell potential

o Depends in the reduction and oxidation reaction

o Measured under standard conditions (called emf or electromotive force, E°)

 25°C, 1 atm for gasses, 1 M concentration

∙ Cell Notation

o You can see a drawing of this in my notes or you can google  “voltaic cell notation”

o Make sure it distinguishes between anode and cathode, which  side is reduced and which is oxidized, where the salt bridge is,  and where the phase barriers are

∙ Electrodes

o Anode is made of the metal that is oxidized  

o Cathode is made of the metal that is produced by reduction  Can also use an inert electrode (platinum is often used) ∙ Standard reduction potential

o The absolute tendency of a half-reaction standard reduction  potential cannot be measured

o So chemists created the standard reduction potential o The reduction is assigned a potential of 0 V (SHE reduction  potential)

 Half-reactions with a stronger tendency towards reduction  than SHE have a positive E° reduction

 Half-reactions with a stronger tendency towards oxidation  than SHE have a negative E° reduction

 E°oxidation = -E°reduction

Lecture 29

∙ Calculating cell potentials under standard conditions

o Cell potentials are intensive properties or matter

 Because cell potentials are intensive physical properties,  when determining the cell potential do not multiply the  half-cell E° values, even if you need to multiply the half

reactions to balance the redox equation

o E°cell = E°cathode(reduction) - E°anode(oxidization)

∙ Predicting the spontaneity of redox reactions

o A Spontaneous reaction will take place when a reduction half reaction is paired with an oxidation half-reaction lower on the  table

o If E°cell is positive, the reaction is spontaneous (ΔG° < 0) o If E°cell is negative, the reaction is not spontaneous (ΔG° > 0) ∙ E°cell, ΔG°, and K equations

o E°cell = (0.0592 V / n) log k

o ΔG° = -nF E°cell

o ΔG° = -RT lnK

∙ Cell potential when ion concentration

o There is a relationship between the reaction quotient, Q, the  equilibrium constant, K, and the free energy change, ΔG° o If you change the concentration from 1M, ΔG will change  Therefore, Ecell will change

 Ecell = E°cell – (RT/nF) lnR

 If we are at 25°C, Ecell = E°cell – (0.0592/n) logQ

∙ Concentration cells

o Electrons will flow from the electrode in the less concentrated  solution to the more concentrated solution

o As long as electrode concentrations are different, it is possible to  get a spontaneous reaction

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