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The Nucleus : A Chemist’s View. Nuclear Symbols. Mass number, A (p + + n o ). Element symbol. Atomic number, Z (number of p + ). Balancing Nuclear Equations. A reactants = A products. 235 + 1 = 142 + 91 + 3(1). 92 + 0 = 56 + 36 + 3(0).
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Nuclear Symbols Mass number, A (p+ + no) Element symbol Atomic number, Z (number of p+)
Balancing Nuclear Equations Areactants = Aproducts 235 + 1 = 142 + 91 + 3(1) 92 + 0 = 56 + 36 + 3(0) Zreactants = Zproducts
Balancing Nuclear Equations #2 222 226 = 4 + ____ 222 Rn 86 86 88 = 2 + ___ Atomic number 86 is radon, Rn
Balancing Nuclear Equations #3 95 235 + 1 = 139 + 2(1) + ____ 95 Y 39 39 92 + 0 = 53 + 2(0) + ____ Atomic number 39 is yttrium, Y
Alpha Decay Alpha production (a): an alpha particle is a helium nucleus Alpha decay is limited to heavy, radioactive nuclei
Alpha Radiation Limited to VERY large nucleii.
Beta Decay Beta production (b): A beta particle is an electron ejected from the nucleus Beta emission converts a neutron to a proton
Beta Radiation Converts a neutron into a proton.
Gamma Ray Production Gamma ray production (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 attract Opposite charges_________ each other. repel Like charges_________ each other.
Positron Production Positron emission: Positrons are the anti-particle of the electron Positron emission converts a proton to a neutron
Electron Capture Electron capture: (inner-orbital electron is captured by the nucleus) Electron capture converts a proton to a neutron
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 radioactive nucleus reaches a stable state by a series of steps A Decay Series
Decay Kinetics Decay occurs by first order kinetics (the rate of decay is proportional to the number of nuclides present) N0 = number of nuclides present initially N = number of nuclides remaining at time t k = rate constant t = elapsed time
Calculating Half-life t1/2 = Half-life (units dependent on rate constant, k)
Nuclear Fission and Fusion Fusion:Combining two light nuclei to form a heavier, more stable nucleus. Fission: Splitting a heavy nucleus into two nuclei with smaller mass numbers.
Energy and Mass Nuclear 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 Processes A self-sustaining fission process is called a chain reaction.
Review • Oxidation reduction reactions involve a transfer of electrons. • OIL- RIG • Oxidation Involves Loss • Reduction Involves Gain • LEO-GER • Lose Electrons Oxidation • Gain 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)? • PbS, reductant; O2, oxidant • PbS, reductant; SO2, oxidant • Pb2+, reductant; S2- oxidant • PbS, reductant; no oxidant • PbS, oxidant; SO2, reductant
Applications • Moving 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
e- e- e- e- e- • Connected this way the reaction starts • Stops immediately because charge builds up. H+ MnO4- Fe+2
Galvanic Cell Salt Bridge allows current to flow H+ MnO4- Fe+2
e- • Electricity travels in a complete circuit H+ MnO4- Fe+2
Instead of a salt bridge Porous Disk H+ MnO4- Fe+2
e- e- e- e- Anode Cathode e- e- Reducing Agent Oxidizing Agent
Cell Potential • Oxidizing agent pulls the electron. • Reducing agent pushes the electron. • The push or pull (“driving force”) is called the cell potential Ecell • Also called the electromotive force (emf) • Unit is the volt(V) • = 1 joule of work/coulomb of charge • Measured with a voltmeter
0.76 H2 in Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl
Standard Hydrogen Electrode • This is the reference all other oxidations are compared to • Eº = 0 • º indicates standard states of 25ºC, 1 atm, 1 M solutions. H2 in H+ Cl- 1 M HCl
Cell Potential • Zn(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 Potential • Determine 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 potential • More 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 • solid½Aqueous½½Aqueous½solid • Anode on the left½½Cathode on the right • Single line different phases. • Double line porous disk or salt bridge. • If all the substances on one side are aqueous, a platinum electrode is indicated.
For the last reaction • Cu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s) Fe+2 Cu2+
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. • Cell Potential • Direction of flow • Designation of anode and cathode • Nature of all the components- electrodes and ions
Practice • Completely 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 DG • emf = potential (V) = work (J) / Charge(C) • E = work done by system / charge • E = -w/q • Charge 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 Equation • DG = DGº +RTln(Q) • -nFE = -nFEº + RTln(Q) • E = Eº - RTln(Q) nF • 2Al(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.