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 (): 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 DecayBeta production (b):A beta particle is an electron ejected from the nucleus
ePaTh 01
23491
23490
Beta emission converts a neutron to a proton
b0123491
23490 PaTh
b0101 ore
Beta Radiation
Converts a neutron into a proton.
Gamma Ray Production
Gamma ray production (g):
92238
24
90234
002U He Th g
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 CaptureElectron capture: (inner-orbital electron is captured by the nucleus)
80201
10
79201
00Hg e Au g
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)
ktNN
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 +4H2O
Helps 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 CellSalt 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 CuWe 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 VCu+2(aq)+2e- Cu(s) Eº = 0.34
VCu(s) Cu+2(aq)+2e- Eº = -0.34
V2Fe+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 VFe+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 VO2 + 2H2O + 4e- 4OH-Eº= 0.40 VFe+2 + O2 + H2O Fe2O3 + H+ Reactions happens in two places.
WaterRust
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 + H2O7. 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?