the nucleus: a chemist’s view. nuclear symbols element symbol mass number, a (p + + n o ) atomic...

The Nucleus: A Chemist’s View

Nuclear Symbols

23592U

Element symbol

Mass number, A (p+ + no)

Atomic number, Z(number of p+)

Balancing Nuclear Equations

nKrBanU 10

9136

14256

10

23592 3

Areactants = Aproducts

Zreactants = Zproducts

235 + 1 = 142 + 91 + 3(1)

92 + 0 = 56 + 36 + 3(0)

Balancing Nuclear Equations #2

4222688Ra

226 = 4 + ____222

222

88 = 2 + ___86

86

Atomic number 86 is radon, Rn

Rn

Balancing Nuclear Equations #3

nInU 10

13953

10

23592 2

235 + 1 = 139 + 2(1) + ____95

3992 + 0 = 53 + 2(0) + ____

39

95

Atomic number 39 is yttrium, Y

Y

Alpha Decay

Alpha production (a): an alpha particle is ahelium nucleus

ThHeU 23490

42

23892

Alpha decay is limited to heavy, radioactivenuclei

ThU 23490

42

23892

242

242 orHe

Alpha Radiation

Limited to VERY large nucleii.

Beta Decay

Beta production (b):A beta particle is an electron ejected from the nucleus

ePaTh 01

23491

23490

Beta emission converts a neutron to a proton

0123491

23490 PaTh

0101 ore

Beta Radiation

Converts a neutron into a proton.

Gamma Ray Production

Gamma ray production (g):

92238

24

90234

002U He Th

Gamma rays are high energy photons produced in association with other forms of decay.

Gamma rays are massless and do not, by themselves, change the nucleus

Deflection of Decay Particles

Opposite charges_________ each other.

Like charges_________ each other.

attract

repel

Positron Production

Positron emission:Positrons are the anti-particle of the electron

1122

10

1022Na e Ne

Positron emission converts a proton to a neutron

e01

Electron Capture

Electron capture: (inner-orbital electron is captured by the nucleus)

80201

10

79201

00Hg e Au

Electron capture converts a proton to a neutron

Types of Radiation

NuclearStability

Decay will occur in such a way as to return a nucleus to the band (line) of stability.

The most stable nuclide is Iron-56

If Z > 83, the nuclide is radioactive

A Decay Series

A radioactive nucleus reaches a stable state by a series of steps

Half-life Concept

Decay KineticsDecay occurs by first order kinetics (the rate of decay is proportional to the number of nuclides present)

ktN

N

0

lnN = number of nuclides remaining at time t

N0 = number of nuclides present initially

k = rate constant

t = elapsed time

Calculating Half-life

kkt

693.0)2ln(2/1

t1/2 = Half-life (units dependent on rate constant, k)

Sample Half-Lives

Nuclear Fission and Fusion

Fusion: Combining two light nuclei to form a heavier, more stable nucleus.

01

92235

56142

3691

013n U Ba Kr n

23

11

24

10He H He e

Fission: Splitting a heavy nucleus into two nuclei with smaller mass numbers.

Energy and MassNuclear changes occur with small but measurable losses of mass. The lost mass is called the mass defect, and is converted to energy according to Einstein’s equation:

DE = Dmc2

Dm = mass defect DE = change in energy

c = speed of light

Because c2 is so large, even small amounts of mass are converted to enormous amount of energy.

Fission

Fission Processes

Event

NeutronsCausingFission Result

subcritical < 1 reaction stopscritical = 1 sustained reactionsupercritical > 1 violent explosion

A self-sustaining fission process is called a chain reaction.

A Fission Reactor

Fusion

ReviewOxidation reduction reactions involve a

transfer of electrons.OIL- RIGOxidation Involves LossReduction Involves GainLEO-GER Lose Electrons OxidationGain Electrons Reduction

Solid lead(II) sulfide reacts with oxygen in the air at high temperatures to form lead(II) oxide and sulfur dioxide. Which substance is a reductant (reducing agent) and which is an oxidant (oxidizing agent)?  

A. PbS, reductant; O2, oxidant 

B. PbS, reductant; SO2, oxidant 

C. Pb2+, reductant; S2- oxidant  D. PbS, reductant; no oxidant 

E. PbS, oxidant; SO2, reductant

ApplicationsMoving electrons is electric current.

8H++MnO4-+ 5Fe+2 +5e-

® Mn+2 + 5Fe+3 +4H2OHelps to break the reactions into half

reactions.

8H++MnO4-+5e- ® Mn+2 +4H2O

5(Fe+2 ® Fe+3 + e- ) In the same mixture it happens without

doing useful work, but if separate

H+

MnO4-

Fe+2

Connected this way the reaction startsStops immediately because charge builds

up.

e-e- e-

e-e-

H+

MnO4-

Fe+2

Galvanic Cell

Salt Bridge allows current to flow

H+

MnO4-

Fe+2e-

Electricity travels in a complete circuit

H+

MnO4-

Fe+2

Porous Disk

Instead of a salt bridge

Reducing Agent

Oxidizing Agent

e-

e-

e- e-

e-

e-

Anode Cathode

Cell PotentialOxidizing agent pulls the electron.Reducing agent pushes the electron. The push or pull (“driving force”) is called

the cell potential EcellAlso called the electromotive force (emf) Unit is the volt(V) = 1 joule of work/coulomb of chargeMeasured with a voltmeter

Zn+2 SO4-

2

1 M HCl

Anode

0.76

1 M ZnSO4

H+

Cl-

H2 in

Cathode

1 M HCl

H+

Cl-

H2 in

Standard Hydrogen ElectrodeThis is the reference

all other oxidations are compared to

Eº = 0 º indicates standard

states of 25ºC, 1 atm, 1 M solutions.

Cell PotentialZn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s)The total cell potential is the sum of the

potential at each electrode.

Eºcell = EºZn® Zn+2 + EºCu+2 ® Cu

We can look up reduction potentials in a table.

One of the reactions must be reversed, so change it sign.

Cell PotentialDetermine the cell potential for a galvanic

cell based on the redox reaction.Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq)

Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 V

Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V

Cu(s) ® Cu+2(aq)+2e- Eº = -0.34 V

2Fe+3(aq) + 2e-® 2Fe+2(aq) Eº = 0.77 V

Reduction potentialMore negative Eº

–more easily electron is added–More easily reduced–Better oxidizing agent

More positive Eº –more easily electron is lost–More easily oxidized–Better reducing agent

Line Notation solid½Aqueous½½Aqueous½solidAnode on the left½½Cathode on the

rightSingle line different phases.Double line porous disk or salt bridge. If all the substances on one side are

aqueous, a platinum electrode is indicated.

Cu2+ Fe+2

For the last reactionCu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)

In a galvanic cell, the electrode that acts as a source of electrons to the solution is called the __________; the chemical change that occurs at this electrode is called________.  

a.  cathode, oxidation  b.  anode, reduction  c.  anode, oxidation  d.  cathode, reduction

Under standard conditions, which of the following is the net reaction that occurs in the cell?

Cd|Cd2+ || Cu2+|Cu  a.  Cu2+ + Cd → Cu + Cd2+  b.  Cu + Cd → Cu2+ + Cd2+  c.  Cu2+ + Cd2+ → Cu + Cd  d.  Cu + Cd 2+ → Cd + Cu2+ 

Galvanic Cell The reaction always runs

spontaneously in the direction that produced a positive cell potential.

Four things for a complete description.1) Cell Potential2) Direction of flow3) Designation of anode and cathode4) Nature of all the components-

electrodes and ions

PracticeCompletely describe the galvanic cell

based on the following half-reactions under standard conditions.

MnO4- + 8 H+ +5e- ® Mn+2 + 4H2O

Eº=1.51 V

Fe+3 +3e- ® Fe(s) Eº=0.036V

Potential, Work and DGemf = potential (V) = work (J) / Charge(C)E = work done by system / chargeE = -w/qCharge is measured in coulombs. -w = q E Faraday = 96,485 C/mol e-

q = nF = moles of e- x charge/mole e-

w = -qE = -nFE = DG

Potential, Work and DG DGº = -nFEº if Eº > 0, then DGº < 0 spontaneous if Eº< 0, then DGº > 0 nonspontaneous In fact, reverse is spontaneous.Calculate DGº for the following reaction:Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq)

Fe+2(aq) + e-® Fe(s) Eº = 0.44 V

Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V

Cell Potential and Concentration

Qualitatively - Can predict direction of change in E from LeChâtelier.

2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)Predict if Ecell will be greater or less than Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M

if [Al+3] = 1.0 M and [Mn+2] = 1.5M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

The Nernst EquationDG = DGº +RTln(Q) -nFE = -nFEº + RTln(Q)

E = Eº - RTln(Q)

nF2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)

Eº = 0.48 V Always have to figure out n by balancing. If concentration can gives voltage, then

from voltage we can tell concentration.

The Nernst EquationAs reactions proceed concentrations of

products increase and reactants decrease.

Reach equilibrium where Q = K and Ecell = 0

0 = Eº - RTln(K) nF

Eº = RTln(K) nF

nF Eº = ln(K)

RT

Batteries are Galvanic CellsCar batteries are lead storage batteries.

Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O

Batteries are Galvanic CellsDry Cell

Zn + NH4+ +MnO2 ®

Zn+2 + NH3 + H2O + Mn2O3

Batteries are Galvanic CellsAlkaline

Zn +MnO2 ® ZnO+ Mn2O3 (in base)

Batteries are Galvanic CellsNiCad

NiO2 + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2

CorrosionRusting - spontaneous oxidation.Most structural metals have reduction

potentials that are less positive than O2 .

Fe ® Fe+2 +2e- Eº= 0.44 V

O2 + 2H2O + 4e- ® 4OH-Eº= 0.40 V

Fe+2 + O2 + H2O ® Fe2O3 + H+ Reactions happens in two places.

Water

Rust

Iron Dissolves-

Fe ® Fe+2

e-

Salt speeds up process by increasing conductivity

O2 + 2H2O +4e- ® 4OH-

Fe2+ + O2 + 2H2O ® Fe2O3 + 8 H+

Fe2+

Preventing CorrosionCoating to keep out air and water.Galvanizing - Putting on a zinc coatHas a lower reduction potential, so it is

more easily oxidized.Alloying with metals that form oxide

coats.Cathodic Protection - Attaching large

pieces of an active metal like magnesium that get oxidized instead.

Running a galvanic cell backwards.Put a voltage bigger than the potential

and reverse the direction of the redox reaction.

Used for electroplating.

Electrolysis

1.0 M

Zn+2

e- e-

Anode Cathode

1.10

Zn Cu1.0 M

Cu+2

1.0 M

Zn+2

e- e-

AnodeCathode

A battery >1.10V

Zn Cu1.0 M

Cu+2

Calculating platingHave to count charge.Measure current I (in amperes)1 amp = 1 coulomb of charge per secondq = I x tq/nF = moles of metalMass of plated metalHow long must 5.00 amp current be

applied to produce 15.5 g of Ag from Ag+

Calculating plating1. Current x time = charge2. Charge ∕Faraday = mole of e-

3. Mol of e- to mole of element or compound

4. Mole to grams of compoundOr the reverse if you want time to plate

Calculate the mass of copper which can be deposited by the passage of 12.0 A for 25.0 min through a solution of copper(II) sulfate.

How long would it take to plate 5.00 g Fe from an aqueous solution of Fe(NO3)3 at a current of 2.00 A?

Other usesElectrolysis of water.Separating mixtures of ions.More positive reduction potential means

the reaction proceeds forward. We want the reverse.Most negative reduction potential is

easiest to plate out of solution.

RedoxKnow the table2. Recognized by change in oxidation

state.3. “Added acid”4. Use the reduction potential table on the

front cover.5. Redox can replace. (single replacement)

6. Combination Oxidizing agent of one element will react with the reducing agent of the same element to produce the free element.

I- + IO3- + H+ ® I2 + H2O

7. Decomposition.a) peroxides to oxidesb) Chlorates to chloridesc) Electrolysis into elements.d) carbonates to oxides

69

Examples1. A piece of solid bismuth is heated

strongly in oxygen.2. A strip or copper metal is added to a

concentrated solution of sulfuric acid.3. Dilute hydrochloric acid is added to a

solution of potassium carbonate.

70

23. Hydrogen peroxide solution is added to a solution of iron (II) sulfate.

24. Propanol is burned completely in air.25. A piece of lithium metal is dropped into

a container of nitrogen gas.26. Chlorine gas is bubbled into a solution

of potassium iodide.

71

Examples5. A stream of chlorine gas is passed

through a solution of cold, dilute sodium hydroxide.

6. A solution of tin ( II ) chloride is added to an acidified solution of potassium permanganate

7. A solution of potassium iodide is added to an acidified solution of potassium dichromate.

72

70. Magnesium metal is burned in nitrogen gas.

71. Lead foil is immersed in silver nitrate solution.

72. Magnesium turnings are added to a solution of iron (III) chloride.

73. Pellets of lead are dropped into hot sulfuric acid

74. Powdered Iron is added to a solution of iron(III) sulfate.

A way to rememberAn Ox – anode is where oxidation occursRed Cat – Reduction occurs at cathodeGalvanic cell- spontaneous- anode is

negativeElectrolytic cell- voltage applied to make

anode positive

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.

(a) Draw a diagram of this cell. (b) Describe what is happening at the

cathode (Include any equations that may be useful.)

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.

(c) Describe what is happening at the anode. (Include any equations that may be useful.)

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.

(d) Write the balanced overall cell equation.

(e) Write the standard cell notation.

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.(f) The student adds 4 M ammonia to the copper sulfate solution, producing the complex ion Cu(NH3)+ (aq). The student remeasures the cell potential and discovers the voltage to be 0.88 volt. What is the Cu2+ (aq) concentration in the cell after the ammonia has been added?