MCATs_2

Solutions: Terms Solute, solvent, dissolution, precipitation, solubility, molarity, molality, mole fraction. Solute: substance in smaller proportion. Solvent: substance in greater proportion. Dissolution: process of dissolving. Precipitation: reverse of dissolution. Solubility: amount solute to saturate solvent. 1

Solutions: Terms Molarity: moles of solute per liter of solution. Molality: moles of solute per kg of solvent. Mole fraction: fraction of moles of given substance relative to total moles in solution. • Xa = na/Σn

What is an electrolyte? Free ions in a solution Ö conducts electricity. What is the van’t Hoff factor, i? Number of ions existing after dissolution of one unit of substance: NaCl, CaCl2

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Colligative Properties: Important! What does a colligative property depend upon? Number of solute particles in solution (not identity of solute) – van’t Hoff factor comes up big! There are three for the MCAT – name ‘em: • • •

vapor-pressure depression. boiling-point elevation. freezing-point depression. 3

Vapor Pressure Definition? Sometimes a tough concept – pressure exerted by the gaseous phase of a liquid that evaporated from exposed surface of liquid. My definition: pressure exerted by molecules leaving liquid phase on the atmosphere – let’s draw what I mean!

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Vapor Pressure

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Vapor Pressure Think about my definition…what property of the liquid will determine vapor pressure? Intermolecular forces! High ones will hold liquid in liquid form and decrease pressure exerted by molecules leaving liquid. Low ones will allow liquid to escape into gaseous form and increase pressure exerted by molecules as they change phase – volatile. 6

Vapor Pressure Depression Now for VP depression… Given a solution of two liquids, A and B: • Total VP is equal to sum of partial pressures of each liquid. • Partial pressure of A is equal to proportion of A in solution.

What is the law called that describes this situation? Raoult’s Law: Pa = XaPa*…where * is pure A. 7

Vapor Pressure Depression From Raoult’s Law, the presence of liquid B will lower Pa by lowering Xa. Vapor Pressure Depression: ∆Pa = –XbPa* Some deviations from Raoult’s Law are possible. Conceptually, what is going on? Generally, presence of liquid B will add intermolecular forces to solution A and make phase change to gas harder…interference! 8

Boiling Point Elevation What happens to a liquid when it boils? Input of energy overcomes intermolecular forces of liquid to cause phase change. Relation to VP Ö boiling point is temperature at which VP = Patmosphere • equilibrium between liquid pushing on air and vice versa…free movement of molecules between phases.

What causes BP change (elevation)? 9

Boiling Point Elevation Intermolecular forces! Addition of solute introduces extra intermolecular forces between solute and liquid, making it harder for phase change. Real life example – cooking pasta. What equation quantifies the temp change? ∆Tb = Kbim…define the terms! A constant, van’t Hoff factor, molality. Why does van’t Hoff factor matter? 10

Freezing Point Depression What happens to liquid when it freezes? Molecules assemble into an orderly, tightlypacked array. What might the addition of a different molecule have on this lattice formation? Interference! More difficult to achieve solid state Ö FP will go down (need to be colder to achieve the solid state). 11

Freezing Point Depression What is the equation that quantifies this concept? ∆Tf = –Kfim …define the terms! A constant, van’t Hoff factor, molality. What is the effect of van’t Hoff factor here? Higher number of ion species will cause more interference! 12

Kinetics: Some Terms Define kinetics (Don’t confuse with thermodynamics!!!). How fast a reaction occurs…this says absolutely nothing about spontaneity!! Mechanism, intermediates, rate-determining step… Mechanism is a sequence, intermediates are not reactants or products, rate-determining step puts limit on maximum speed. 13

Kinetics: Some Terms Rate-determining step: the slowest step in a process determines the overall reaction rate. In a two step reaction, if step two is slow, then the speed of step one is irrelevant: • 2NO N2O2 (fast) • N2O2 + O2 2NO2 (slow) • overall reaction: 2NO + O2

bimolecular 2NO2

termolecular unlikely!

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Kinetics: Reaction Rate What three factors determine reaction rate? They are… • frequency of collisions. • orientation of colliding molecules. • energy of molecules.

The concept of activation energy…what is it and why does it matter? 15

Activation Energy, Ea Activation energy is the extra “kick” of energy that a reaction needs to proceed. MCAT loves Ea diagrams, but first, what are exothermic and endothermic reactions? Two diagrams, one for each above type: Ea

Ea

R ∆H _

P

P R

∆H +

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Activation Energy, Ea From the activation energy concept, three statements can be made: • • •

low Ea Ö fast reaction rate. more reactants Ö fast rate (more collisions possible). higher temperature Ö fast rate (more reactant molecules have sufficient kinetic energy to overcome Ea).

Ea1

Ea2

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Catalysts What is a catalyst? Substance that makes reaction go faster by speeding up rate-determining step or providing an optimized route to products. Reactant vs. catalyst – catalyst unchanged at end of reaction. Catalysts change rate, not thermodynamics (∆G, ∆H, ∆S, etc.).

Ea

Ea,cat

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Rate Laws What does a rate law tell us? The rate at which a reactant disappears. Consists of rate constant, and concentrations of reactants in slow step only!! Rate laws determined experimentally. aA + bB ˆ cC + dD…rate law? 19

Rate Laws Rate = k[A]x[B]y x Ö order of rxn with respect to A. y Ö order of rxn with respect to B. x+y Ö overall order of reaction. k Ö rate constant. Note: cannot get orders of reactants or rate constant from balanced equation. Need to look at experiment to get rate law. 20

Rate Laws A+B+CˆD+E Experiment

[A]

[B]

[C]

1

0.2 M

0.1 M

0.05 M

Initial Rate, Ms–1 1 × 10–3

2

0.4 M

0.1 M

0.05 M

2 × 10–3

3

0.2 M

0.2 M

0.05 M

4 × 10–3

4

0.2 M

0.1 M

0.1 M

1 × 10–3 21

Rate Laws Look at two experiments where only one reactant concentration changes. Determine the factor by which it changes, and compare to factor by which rate changes. order 1. [A] doubles Ö rate doubles Ö [B] doubles Ö rate quadruples Ö order 2. [C] doubles Ö rate uneffected Ö order 0. 22

Rate Laws When order is 0, rate does not depend on concentration of reactant. Rate law for this example: k[A][B]2 Determining k…anyone remember the formula for finding k? Solve the rate law…k = rate/[A][B]2 Substitute numbers from any experiment for [A] and [B]. • k = 1 × 10–3 / 0.2 × (0.1)2 = 0.5 Ms–1 23

The Equilibrium Constant First, what is the definition of an equilibrium? The rate of the forward reaction equals reverse. Given this generic reaction, what is the Keq? • aA + bB ' cC + dD

Mass-action ratio Ö ratio of products to reactants: [C]c[D]d Keq = [A]a[B]b Gases and aqueous molecules only!! 24

The Equilibrium Constant The value of Keq for a reaction is a constant at a given temperature Ö temp change causes Keq change. Equilibrium constants, in general, may have different subscripts (eq, sp, a, b, etc.) but they all obey the same rules of calculation and temp dependence. The value of K Ö what values favor react. vs. prod. • • •

Keq < 1 favors reactants. Keq = 1 balances reactants and products. Keq > 1 favors products.

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The Reaction Quotient What is the reaction quotient? Symbol? Ratio of concentrations of products to reactants, Q, but NOT at equilibrium Ö differs from K. How do the values of K relate to values of Q? When Q < K Ö proceeding in forward direction. When Q = K Ö at equilibrium. When Q > K Ö proceeding in reverse direction. 26

Le Châtelier’s Principle What does Le Châtelier’s principle state? A system at equilibrium will try to neutralize any imposed change (stress) in order to reestablish equilibrium. The effects of different stresses: • • • •

Addition of product or reactant. Removal of product or reactant. Changing volume and temperature. Adding inert gas and adding catalyst. 27

Le Châtelier’s Principle N2 (g) + 3 H2 (g) ' 2 NH3 (g) + heat Adding or removing N2 Adding or removing NH3 Reducing volume? Increasing volume? • cut volume, raise pressure Ö favor fewer moles • raise volume, decrease pressure Ö favor greater moles.

Changing temp? Adding or removing heat. Inert gas and catalyst Ö no change (catalyst changes rate of forward and reverse equally). 28

Solubility Product How is the solubility product constant defined? The extent to which a salt will dissolve in water. Ksp is determined just like Keq. Solubility product constant is also temperature dependent. MnXm(s) ' nMx+(aq) + mXy–(aq) Ksp = [Mx+]n[Xy–]m 29

Solubility Product: Example The value of the solubility product for copper (I) chloride is Ksp = 1.2 × 10–6. Under normal conditions, the maximum concentration of an aqueous CuCl solution will be: • less than 10–6 M • greater than 10–6 M and less than 10–4 M • greater than 10–4 M and less than 10–2 M • greater than 10–2 M and less than 10–1 M 30

Ion Product and Common Ion Effect What is an ion product? Actually just the reaction quotient (Q) for solubility reactions. Allows us to make predictions just as before… When Qsp < Ksp Ö more salt can be dissolved. When Qsp = Ksp Ö solution is saturated. When Qsp > Ksp Ö excess salt will precipitate. Think of Ksp as the “saturation ceiling”. 31

Ion Product and Common Ion Effect What is the common ion effect? Disturbing of a solubility equilibrium by adding a common ion, one that already exists in the equilibrium. Follows Le Châtelier’s principle, causing a decrease in solubility (equilibrium favors the solid and not the aqueous). Mg(OH)2(s) ' Mg2+ (aq) + 2OH–(aq)…after OH– added (e.g. soluble NaOH(aq)).

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Acids and Bases: Definitions Arrhenius acids/bases, Brønsted-Lowry acids/bases, Lewis acids/bases…differentiate! Arrhenius: acids ionize in water to produce H+ and bases ionize to produce OH–. Brønsted-Lowry: acids are proton (H+) donors and bases are proton (H+) acceptors. Lewis: acids are electron pair acceptors and bases are electron pair donors. 33

Conjugate Acids and Bases What are conjugate acids and bases, or how would you recognize them? When Brønsted-Lowry acid donates an H+ the remaining structure is conjugate base. When Brønsted-Lowry base accepts an H+ the new species is the conjugate acid. Name the acid, base, and conjugates: • NH3 + H2O ' NH4+ + OH– base acid conj. acid conj. base

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Strengths of Acids and Bases Brønsted-Lowry acids/bases placed in two broad catagories: strong and weak. What is a strong acid? Strong acids dissociate nearly completely. Their Ka > 1, favoring products considerably. Examples (memorize!) For the MCAT, all others not listed here are considered weak: • HI, HBr, HCl, HClO4, HClO3, H2SO4, HNO3 35

Strengths of Acids and Bases What is a weak acid? Beyond the opposite of a strong acid, a weak acid has a Ka < 1 because reactants are favored. Ranking acid strength: higher Ka means stronger acid…same for base strength. How does this all work for conjugates? 36

Strengths of Conjugate Pairs Given a strong acid, is conjugate base strong or weak? More importantly, why? It is weak, because complete forward reaction means essentially no reverse (conjugate cannot take up H+) and therefore is weak. Same holds true for strong base. For weak acid? Conjugate is weak base, but the smaller the Ka, the stronger the conjugate base will be, relative to other weak bases.

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Strengths of Conjugate Pairs: Example Of the following anions, which is the strongest base? • I– • CN– • NO3– • Br–

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Strengths of Conjugate Pairs: Example Of the following, which acid has the weakest conjugate base? • HClO4 • HCOOH • H3PO4 • H2CO3

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Amphoteric Substances What is an amphoteric substance? A substance that can act as either an acid or a base. Where does this happen? Polyprotics! When a substance has more than one proton to donate, the conjugate base can either donate another or accept a free proton. H3PO4 ' H2PO4– + H+ ' HPO42– + 2H+ ' 40

Ion-Product Constant of Water Water is an amphoteric substance. H2O + H2O ' H3O+ + OH– • reacts with itself in Brønsted-Lowry reaction.

Anyone know the term for this equilibrium? Autoionization of water Ö Kw = [H+][OH–] Value of autoionization constant at 25 °C? Kw = 1.0 × 10–14…as in pH + pOH = 14 41

pH Calculations: A Shortcut Normal formula: pH = –log[H+]…also, pOH = –log [OH–] Great when [H+] = 1 × 10–2 Ö pH = 2 But how about when [H+] = 2.3 × 10–4 M? Use this shortcut… • if [H+] = y × 10–n M (where n is a whole number). • then pH is between (n–1) and n.

For above example, pH is between 3 and 4! 42

pH Calculations Remember the shortcut for estimating pH from [H+]. When looking at weak acid (base), given its initial concentration, use the ICE method to find [H+] and pH. What is pH of 0.2 M solution of HCN? • Ka = 4.9 × 10–10

The Ka < 10–4 rule…the usual on the MCAT. 43

pKa and pKb We know pH is –log [H+], so what do you think pKa or pKb is? Any “p” function represents the –log! pKa = –log Ka & pKb = –log Kb High Ka or Kb (closer to 1) means strength, so how about for pKa or pKb? Lower is stronger! Think of it like pH Ö lower is more concentrated. Don’t forget Ö pKa + pKb = 14 44

Neutralization Reactions What is a neutralization reaction? When an acid and base react to form a salt and water. Like when you take antacid (Tums, etc.). A handy formula for complete neutralization: • a × [A] × Va = b × [B] × Vb • a is # acidic H, b is # H’s base can accept. • [X] is acid/base concen. and V is volume. 45

Neutralization Reactions How much 0.1 M NaOH solution needed to neutralize 40 mL of 0.3 M HCl? HCl(aq) + NaOH(aq) ˆ NaCl(aq) + H2O(l) a × [A] × Va = b × [B] × Vb 1 × 0.3 × 40 = 1 × 0.1 × Vb Solve for Vb to get 120 mL. Remember the formula to get you out of jams…works with neutralization reactions only to save time over the ICE method. 46

Indicators An indicator marks the endpoint of a titration, but why does it change color? Indicator is actually weak acid whose protonated form is one color and deprotonated is another. How does this help us? Don’t forget that the indicator is a weak acid! 47

Indicators HIn ' H+ + In– Ka = [H+][In–]/[HIn]…now rearrange: [H+]/Ka = [HIn]/[In–]…and look at ratios: • If [H+] » Ka, [HIn] » [In–]…see color 1 • If [H+] = Ka, [HIn] = [In–]…mix of two colors • If [H+] « Ka, [HIn] « [In–]…see color 2

So indicator turns colors over short pH range near its pKa. 48

Picking an Indicator When picking an indicator, use one whose pKa value lies within the pH range that you want to detect. For indicator, pKa +/– 1 represents effective range for color change. So don’t expect to detect a pH change from 3 to 4 if your indicator has a pKa of 8 or 9! 49

Hydrolysis of Salts The reaction of a substance (salt/ion) with water is a hydrolysis reaction. Question…will the hydrolysis result in a neutral, acidic, or basic salt? NaCl in water Ö neutral because neither Na+ nor Cl– will react with water: • •

Cl– is conjugate base of a strong acid. Na+ is conjugate acid of a strong base.

NH4Cl in water…acidic because NH4+ will react with water (it is a weak acid).

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Hydrolysis of Salts NH4CN in water…NH4+ is weak acid, but CN– is good base, so which wins?? Need to know Ka/Kb values! Ka for NH4+ = 6.3 × 10–10 Kb for CN– = 1.6 × 10–5 Solution will be basic because CN– is better base than NH4+ is an acid. 51

What is a buffer? A solution that resists changing pH when a small amount of acid or base is added. The resistance comes from the presence of a weak acid or base and its conjugate in roughly equal concentrations. An expression to remember: Ka =

[conj. + [H ]

base] [acid] 52

Buffer Compensation A biochemical example: the blood. The main buffer: carbonic acid: CO2 + H2O ' H2CO3 ' H+ + HCO3–

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Buffer Compensation Compensation critical to prevent medical disorders like acidosis and alkylosis. Essentially Le Châtelier’s principle at work. Addition of small amount of acid like HCl causes added H+ to react with present HCO3– to form H2CO3. Equilibrium has shifted to the left, but existing H2CO3 prevents large change in pH (compensation reverses effects). 54

Buffer Compensation Provided that acid and its conjugate are in similar concentration, the pH change is minimal. Addition of small amount of base like KOH causes added OH– to react with present H2CO3 to form HCO3–. Equilibrium has shifted to the right, but existing HCO3– prevents large change in pH (compensation reverses effects). 55

Dealing with Buffers Henderson-Hasselbalch Equation: [A–] pH = pKa + log [HA]

[HA] pOH = pKb + log [A–]

If [A–] = [HA], then pH = pKa An ideal buffer works under these conditions. Small changes to the ratio are acceptable, but the ratio should stay near 1. 56

Dealing with Buffers – Example Which of the following compounds could be added to a solution of HCN to create a buffer? • HNO3 • CaCl2 • NaCN • KOH

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Acid-Base Titrations – Terms Titration Ö experimental technique to determine concentration or identity of unknown weak acid or base by determining pKa or pKb. Titrant Ö strong acid or base of known concentration that is added to solution of the unknown base or acid. Titration curve Ö a plot of pH vs. volume of titrant added. 58

Acid-Base Titrations – Terms Buffering domain Ö the section of the titration curve where the pH changes gradually before the equivalence point. Equivalence point Ö the point during a drastic pH change in which there is complete neutralization of acid and base. End point Ö one or two drops of titrant past equivalence point! • indicator color flip visually detectable:

HIn GF In–

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Titration Curves – The Basics Equivalence point and pH: not necessarily at pH of 7 Ö salt hydrolysis? Looking at the pH of the equivalence point allows you to determine if unknown is weak or strong. Some quick rules for determining the pH at the equivalence point. 60

Titration Curves – The Basics weak acid + strong base: • pH > 7 because product contains basic salt. • HF + NaOH p NaF + H2O

strong acid + weak base:

• pH < 7 because product contains acidic salt. • NH3 + HCl p NH4Cl

strong acid + strong base: • pH = 7 because product contains neutral salt. • HCl + NaOH p NaCl + H2O 61

Titration Curves – The Basics Half-equivalence? The volume of titrant at half-equiv. is half that at equivalence. More importantly, the concentration of unknown at half equiv is equal to the concentration of the unknown’s conjugate at half equivalence. • NH3 + HCl p NH4Cl

Buffer! 62

Titration Curves – The Basics In that case, the following must be true: • [HA]half equiv = [A–]half equiv = [HA]i

By H-H equation, the pH at half equiv. must be equal to pKa Ö this allows identification of unknown. Types of Curves: • • •

Slope to upper right Ö SA+SB, WA+SB Slope to lower right Ö WB+SA Polyprotic acids have multiple equiv points (why?)

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Titration Curves – SA + SB

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Titration Curves – WA + SB

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Titration Curves – WB + SA

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Titration Curves – Examples Methyl red is an indicator that changes from red to yellow in pH range 4.4 – 6.2. For which of the following titrations would methyl red be useful for indicating the equivalence point? • HCN with KOH • NaOH with HI • C6H5COOH with LiOH • C6H5NH2 with HNO3 67

Titration Curves – Examples Draw the following titration curves, paying particular attention to shape while labeling the relevant parts: • CH3COOH titrated with KOH • C6H5NH2 titrated with HNO3 • H3PO4 titrated with KOH

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System and Surroundings The system is the thing we’re looking at, for example a melting ice cube or a solid dissolving into water. Everything else is known collectively as the surroundings, for example the table the ice cube sits on and the surrounding air or the beaker that the solution sits in. System + surroundings = thermodynamic universe. 69

System and Surroundings: Energy Flow Remember conservation of energy? When energy flows into a system from the surroundings (+), the energy of the system increases and energy of surroundings decreases. When energy flows out of a system into the surroundings (–), the energy of the system decreases and energy of surroundings increases. 70

First Law of Thermodynamics The First Law Ö anybody know it? Total energy of the universe is constant, so energy can be transferred but not created or destroyed. Transfer of energy can occur as heat or work: ∆E = q + w Ö w = P∆V Energy input Ö positive q (heat absorbed) and w (work done on system). Energy out Ö negative q (heat released) and w (work done by system). 71

First Law of Thermodynamics Energy transfer in car air conditioner. Air taken in through hose and moves over fins of evaporator. This hot air evaporates refrigerant inside (heat transfer) which travels to condenser and is changed back into liquid (heat transfer). Air which was cooled when heating the fins of the evaporator flows into the car’s interior. No net change in energy for system…refrigerant heated but then cooled to its original state. 72

First Law of Thermodynamics Adiabatic and isothermic? Isothermic: expansion or compression of gas at constant temperature (heat input or output). Adiabatic: expansion or compression of gas without a heat exchange (temperature changes; q = 0 Ö ∆E = w ) Adiabatically expanding gases cool while adiabatically compressing gases warm. Adiabatic principle at work: snow-making machine.

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First Law of Thermodynamics Snow-making machines contain mixture of compressed air and water vapor (20 atm). When sprayed into the air, the mixture undergoes a huge pressure change and expands so rapidly that essentially no heat is exchanged between system and surroundings (adiabatic expansion). Adiabatic expanding gases cool, so the water vapor changes almost instantaneously to snow. 74

Second Law of Thermodynamics The Second Law Ö what does it state? All processes tend to run in a direction that leads to maximum disorder. If a process is spontaneous in one direction, then the reverse cannot be spontaneous Ö makes sense! From here, we can discuss entropy, enthalpy, and free energy. 75

Entropy: Disorder or Randomness ∆S = Sproducts – Sreactants

• Increasing randomness (decreasing order) results in a positive ∆S. • Decreasing randomness (increasing order) results in negative ∆S.

What are some situations in which entropy is predictable? Liquids have more than solids, gases more than liquids, particles in solution more than solids, two moles more than one mole. 76

Enthalpy: Heat Energy Two important principles Ö bond formation releases energy (not needed to ensure stability), energy must be added to break a bond. ∆H = Hproducts – Hreactants Exothermic: heat released because products of reaction have stronger, more stable bonds than reactants. Endothermic: heat absorbed because products of reaction have weaker bonds and energy needed to transition to this more unstable state.

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Enthalpy and Heat of Formation Amount of energy required to make one mole of a compound from pure elements in their natural or standard state. If –∆H, heat given off…if +∆H, heat required. What is standard state? T = 298 K, P = 1 atm, [X] = 1 M. Designated by ° superscript, standard state is condition for determining heat of formation, enthalpies, constants, etc. 78

More Enthalpy Each reaction has own ∆H Ö doubling reactants will double the heat required or released. Hess’s Law of Heat Summation, Remember? If a reaction occurs in several steps, the sum of the energies absorbed or released in all steps will be the same as the overall reaction. If reaction is reversed, the sign of ∆H reverses too. If equation multiplied by constant, so too is ∆H.

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Hess’s Law Example C(s) + 2H2(g) Direct path ∆Hnet Indirect path + 2O2(g)

= ∆H1 + 2∆H2 + (–∆H3)

CH4(g)

∆Hrx = ∆H1 + 2∆H2

– 2O2(g) ∆Hrx = –∆H3

CO2(g) + 2H2O(g)

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Gibbs Free Energy The energy available to do useful work from a chemical reaction. Spontaneity determined by enthalpy and entropy, so spontaneity can be accounted for by ∆G. ∆ G = ∆ H – T∆ S When… • ∆G < 0 Ö spontaneous • ∆G = 0 Ö at equilibrium • ∆G > 0 Ö non-spontaneous

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∆G and Temperature ∆G = ∆H – T∆S ∆H

∆S

∆G

Reaction is…

–

+

–

Spontaneous

+

+

Spontaneous Non-spontaneous

–

–

+

–

– at high T + at low T + at high T – at low T +

Non-spontaneous Spontaneous Non-spontaneous 82

Reaction Energy Diagrams Revisited Two different diagrams Ö endothermic and exothermic. ∆G ≈ ∆H when T∆S is small. Do not confuse ∆G, ∆H, or ∆S with rate or Ea! • catalysts lower Ea, but don’t touch ∆G!

Reversibility: the reverse of any reaction has the same magnitude for all thermodynamic values (opposite signs) and the same reaction pathway: • Coming from products side, Ea is now forward Ea plus ∆G…look at the graph. 83

Thermodynamics and Equilibrium The two concepts are related by this equation: • ∆G° = –2.3RTlogKeq

Should you know this equation? Probably not, but know this for sure… • • •

∆G° < 0 Ö spontaneous Ö Keq > 1 ∆G° = 0 Ö at equilibrium Ö Keq = 1 ∆G° > 0 Ö non-spontaneous Ö Keq < 1

Quick summary: for a spontaneous reaction, products are favored; for non-spontaneous reaction, reactants are favored. 84

Oxidation-Reduction Reactions Oxidation number: how many electrons an atom is donating or accepting in moleculaar bonding. Oxidation-reduction (redox) reaction: reaction in which oxidation numbers of reactants change. LEO says GER, LEO the GERman, OIL RIG • Oxidation is loss, Reduction is gain of electrons.

Reducing agent causes reduction and is oxidized. Oxidizing agent causes oxidation and is reduced. 85

Galvanic (Voltaic) Cells

Harnessing flow of electrons from redox reaction to generate an electric current. Red cat, an ox Ö Reduction @ cathode; oxidation @ anode. 86

Galvanic (Voltaic) Cells Half-reactions and potentials Ö positive E is spontaneous. ∆G = –nFE…where F is Faraday’s constant (96500 Cmol–1), n is number of electrons transferred, E is half-reaction potential. Flipping half-reaction changes sign on E. More negative reduction potential Ö product strong reducing agent…more positive reduction potential Ö reactant strong oxidizing agent. 87

Redox Electrode Potentials

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Nernst Equation 2.303RT 0.0592 V logQ = E° – logQ E = E° – nF n Electrode potentials for non-standard conditions, but T still = 298 K. Zn(s) + Cu2+(aq) Š Zn2+(aq) + Cu(s) [Zn2+]

0.0592 V log E = E° – 2 [Cu2+] 89

Electrolytic Cells Unlike a galvanic cell, electrolytic cells use an external voltage source like a battery to create an electric current that forces a non-spontaneous redox reaction to occur Ö electrolysis. Electroplating: plating a thin layer of metal on top of another material Ö application of electrolytic cells, often done in coin mints or with jewelry. Red cat, an ox still applies! 90

Faraday’s Law of Electrolysis Determine amounts of species produced at electrolytic electrodes using this law. The amount of chemical change is proportional to the amount of electricity flowing through cell. Three steps: • Determine amount of electricity in coulombs from Q = It • Convert Q to moles of electrons with Faraday constant using Q = nF. • Use stoichiometry of reaction to finish the calculation. 91

Faraday’s Law of Electrolysis A piece of steel is the cathode in a hot solution of chromic acid to electroplate it with chromium metal. How much Cr deposited after 48250 C of electricity was forced through cell? • 1/12 mol • 1/6 mol • 1 mol • 3 mol 92

Concentration Cells A galvanic cell that has identical electrodes but whose half-cells have different ion concentrations. Electric current from potential difference arising from unequal concentrations. Electrons will flow in direction of highest concentration of positive ions. When concentrations of solutions become equal, the reaction will stop. 93

Concentration Cells

[Ni2+]dilute

0.0592 V log E = E° – 2 [Ni2+]concentrated 94