Energy Part 3-Nuclear and Redox Reactions PDF

Energy: Nuclear Reaction and Redox Reaction Marcelino Dela Rama Tradio Jr., RCh, LPT, PhD-Chem Assistant Professor Chemistry Analytical Environmental Section Department of Chemistry...

Energy: Nuclear Reaction and Redox Reaction Marcelino Dela Rama Tradio Jr., RCh, LPT, PhD-Chem Assistant Professor Chemistry Analytical Environmental Section Department of Chemistry School of Arts and Sciences University of San Carlos Nasipit Talamban Cebu City 6000, Philippines Nuclear Reactions as Source of Energy Nuclear reactions involve the conversion of one chemical element into another and are typically accompanied by the emission of radiation in the form of high energy particles and photons. Some nuclear reactions occur naturally, whereas others can be induced by human intervention. The four main nuclear reaction types which can be used to harness energy 1. Fission 2. Fusion 3. Transmutation 4. Nuclear Decay/radioactive decay n.b. In nuclear reactions, the mass is converted into energy. Nuclear Reactions as Source of Energy Nuclear Particles involve in nuclear reactions Balancing Nuclear Reactions Nuclear reactions are balanced in two ways 1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products. 2. The sum of the charges of the reactants equals the sum of the charges of the products. n.b. balancing nuclear reactions can be used to identify a particular nuclide in a nuclear reaction Example 1: The reaction of an alpha particle with magnesium-25 (2512Mg) produces a proton and a nuclide of another element. Identify the new nuclide produced. 25 Mg + 4 ɑ ⟶ 1 H + A X ; A=? , Z=? , X =? 12 2 1 z 25 Mg + 4 ɑ ⟶ 1 H + 28 X 12 2 1 13 25 Mg + 4 ɑ ⟶ 1 H + 28 Al 12 2 1 13 Balancing Nuclear Reactions Example 2: 210 Po ⟶ 20682Pb + AzX ; A=? , Z=? , X =? 84 84 Po ⟶ 210 206 Pb + 42X 82 210 Po ⟶ 206 Pb + 42He 84 82 Example 3: 7N + 42ɑ ⟶ 11H + AzX ; A=? , Z=? , X =? 14 14 N + 42ɑ ⟶ 11H + 178X 7 14 N + 42ɑ ⟶ 11H + 178O 7 Balancing Nuclear Reactions Example 4: 9 Be + A X ⟶ 12 C + 1 n ; A=? , Z=? , X =? 4 z 6 0 9 Be + 4 X ⟶ 12 C + 1 n 4 2 6 0 9 Be + 4 He ⟶ 12 C + 1 n 4 2 6 0 Example 5: Nb-89 ejects a positron 89 Nb ⟶ A X + 0 e ; A=? , Z=? , X =? 41 z +1 89 Nb ⟶ 89 X + 0 e 41 40 +1 89 Nb ⟶ 89 Zr + 0 e 41 40 +1 Balancing Nuclear Reactions Example 6: Tc-94 undergoes electron capture 94 Tc + 0 e ⟶ A X ; A=? , Z=? , X =? 43 -1 z 94 Tc + 0 e ⟶ 94 X 43 -1 42 94 Tc + 0 e ⟶ 94 Mo 43 -1 42 Energetics of Nuclear Reactions Binding Energy of a nucleus: the energy that would be released if the nucleus were formed from a collection of free nucleons. Example: Calculate the binding energy during the formation of He-4 atom that has an experimental mass of 4.002603 amu. total mass of nucleons = total mass of protons + total mass of neutrons total mass of nucleons = 2(1.007825 amu) + 2(1.008665 amu) = 4.032980 amu ∆m = total mass of nucleons – experimental mass ∆m = 4.032980 amu – 4.002603 amu = 0.030377 amu E = ∆mc2 E = 0.030377 amu (1.66054 x 10-27 kg/amu) (2.99792 x 108 m/s)2 E = 4.5335 x 10-12 kg m2/s2 per atom E = 4.5335 x 10-12 J per atom Fission Reaction Splitting of heavy nucleus into two or more smaller nuclei to attain greater stability Not all nucleus are fissionable or fissile. Fission reaction happens spontaneously (large nucleus simply breaks into smaller nucleus); or through neutron bombardment. example: 23592U + 10n ⟶ 23692U* ⟶ 14156Ba + 9236Kr + 3 10n Sample problem: Calculate the energy released by a nucleus of uranium-235 if it splits into a barium-141 nucleus and a krypton-92 nucleus according to the equation above. Fission Reaction Sample problem 1: Calculate the energy released by a nucleus of uranium-235 if it splits into a barium-141 nucleus and a krypton-92 nucleus according to the equation above. Mass of reactants = 235.0439231 amu + 1.0086649 amu = 236.0525880 amu Mass of products = 141.9144064 amu + 91.9261528 amu + 3(1.0086649 amu) = 235.8665539 amu Mass defect, ∆m = mass of products – mass of reactants Mass defect, ∆m = 235.8665539 – 236.0525880 = − 0.1860341 E = ∆mc2 E = − 0.1860341 amu (1.66053886 x 10-27 kg / amu) (2.99792458 x 108 m/s)2 E = —2.776406 x 10-11 kg m2/s2 E = —2.776406 x 10-11 J n.b. 1 J = 1 kg m2/s2 Fission Reaction Nuclear Power Plant Fusion Reaction Small nuclei combine to form larger, and more stable nuclei. Though fusion reaction is relatively safe source of energy; however, the use of fusion to generate electricity is complicated due to the repulsion of positively charged nuclei that requires enormous energy to force the nuclei close enough to overcome the Coulombic forces (this require temperature at the order of 106 K) The energy of the sun originate in a fusion reaction. 4 11H ⟶ 42He + 2 01β + 2𝑣 + energy Transmutation One nucleus of a particular element changes to another, either by natural decay or in response to some outside intervention, such as neutron bombardment. 10 1 11 7 4 5B + 0n ⟶ 5B* ⟶ 3Li + 2He Reverse of nuclear decay. Transmutation reactions are used to produce a number of medically useful radioisotopes. Nuclear Decay or Radioactive Decay Process by which an unstable isotope of a particular element spontaneously transforms into a new element by emission of ionizing radiation. Isotopes: atoms of a given element with the same atomic number but different atomic mass; hence, they have different number of neutrons. Atomic mass of a given element: is equal to the weighted average atomic masses of all the naturally-occurring isotopes It is classified depending upon the type of emitted or absorbed ionizing radiation as: 1. Alpha Decay 2. Beta Decay 3. Gamma Decay 4. Electron Capture 5. Positron Emission Alpha Decay A nuclear decay of unstable isotopes which eject alpha particle that is accompanied with the decrease of mass number of 4 and decrease of atomic number of 2. 238 U ⟶ 234 Th + 4 He 92 90 2 Beta Decay A nuclear decay of unstable isotopes which emit beta particle, β- or 0-1β that is accompanied with the ejection of electron from the nucleus and emission of antineutrino, ν̄, that has no charge and no mass. 14 C ⟶ 14 N + 0 β + ν̄ 6 7 -1 Gamma Decay A nuclear decay that emits a high-energy photon that is accompanied with other types of decay In gamma decay, the nucleus will be in the excited states, and to return to its ground state, it emits a photon (a subatomic light particle that has discrete bundle of electromagnetic energy) Since, the nuclear energy level is very large, this photon will take the form of very high energy gamma radiation (wavelength is 10-12 m and frequency is 3 x 1020/s) Gamma decay will neither change the mass number nor the atomic number. 14 C ⟶ 14 N + 0 β + ν̄ + 0 γ 6 7 -1 0 Electron Capture Nucleus captures an electron from the first (n=1) shell in the atom. Proton in the nucleus is converted to neutron. 1 p+0 e⟶1 n+𝝼 1 -1 0 Reverse of beta emission Nuclear charge decreases by one 26 Al + 0-1e ⟶ 2612Mg + 𝝼 13 Positron Emission Decay Positron is a positively charged electron, β+ or 01β Collision of positron and electron results in the conversion of their combined masses into energy (511 keV gamma-ray photons Positron decay result to the decrease of one nuclear charge. 15 O ⟶ 15 N + 0 β + 𝝼 8 7 1 Kinetics of Nuclear or Radioactive Decay Recall: In a chemical reaction, the rate of reaction is expressed as the rate at which a reactant is consumed or the rate at which the product is formed. In radioactive decay, the rate is measured directly based on the count of decay in a given time period. Radioactive Activity (Decay Rate): the rate at which a sample decays and it is equal to the slope of radioactive decay equation. Ø A = # of decays/sec ; N(t) = Noe-kt Ø A = d N(t) dt d N(t) = d (Noe-kt ) dt dt dN(t) = N0 e-kt dt dN(t) = -kN(t) dt Kinetics of Nuclear or Radioactive Decay The SI unit of nuclear activity is Becquerel (Bq), defined as one nuclear disintegration per second) or 1 decay/sec The older unit of nuclear activity is Curie (Ci), which is 3.7 x 1010 decay/sec or 3.7 x 1010 Bq Kinetics of Nuclear or Radioactive Decay For a sample of N nuclei, the rate of disintegration is given by ∆N/∆t Since most activity is proportional to the number of nuclei present which decreases exponentially with time, the N also decreases exponentially. Equation can be re-written to monitor the number a sample decayed: ln N = ln N0 − kt k= decay constant or rate constant ln N0 − ln N = kt ln (N0 /N) = kt N = N0 e-kt Half-life, t1/2: the time it takes for N0 to decay to one-half its original value t1/2 = ln2/k = 0.693/k Kinetics of Nuclear or Radioactive Decay Sample Problem 2: The half-life of strontium-90, 9038Sr is 28.8 years. Find the activity (decay rate) of 1.00 g of the material. k= 0.693/t1/2 k= 0.693/28.8 year k= 0.0240625/year (1year/31,536,000 sec) k= 7.630 x 10-10/sec We need to find the number of nuclei in 1.00 g of the material: N = 1.00 g (1 mol/89.91 g) (6.022 x 1023 atom/mol) N = 6.70 x 1021 nuclei A = -kN A = 7.630 x 10-10/sec (6.70 x 1021 nuclei) A = 5.11 x 1010 decays/sec A= 5.11 x 1012 Bq Kinetics of Nuclear or Radioactive Decay Sample Problem 3: A sample of radon has an activity of 60,000 Bq. If the half-life of radon is 15 h, how long before the sample’s activity is 3,750 Bq? dN(t) = N0 e-kt dt 3,750 Bq = 60,000 Bq e-kt 3,750 Bq = 60,000 Bq e-(0.693/15h)t 3,750 Bq = e-(0.693/15h)t 60,000 Bq ln[0.0625= e-(0.693/15h)t] -2.77259 = - (0.693/15h) t t = 2.77259 (15h/0.693) t= 60.0 h Kinetics of Nuclear or Radioactive Decay Sample Problem 4: Cobalt-60 is used extensively in medicine as a source of gamma rays. Its half-life is 5.27 years. How long will it take a Co-60 source to decrease to 45% of its original activity? dN(t) = N0 e-kt dt 0.45 = 1 e-kt 0.45= 1 e-(0.693/5.27)t ln[0.45 = e-(0.693/5.27)t] -0.79851 = - (0.693/5.27 year) t t= 0.79851 (5.27 year/0.693) t= 6.07 year Kinetics of Nuclear or Radioactive Decay Sample Problem 5: Cobalt-60 is used extensively in medicine as a source of gamma rays. Its half-life is 5.27 years. What % of its activity remains after 29 months or 2.417 year? dN(t) = N0 e-kt dt dN(t) = 1 e-kt dt dN(t) = e-(0.693/5.27year)(2.417year) = 0.7277 or 72.8 % dt Oxidation—Reduction Reaction as Source of Energy Recall: Oxidation Reaction: involves the loss of electrons and the chemical specie that undergo oxidation are known as reducing agent. Na ⟶ Na+1 + 1e- Reduction Reaction: involves the gain of electrons, and the chemical specie that undergo reduction are known as oxidizing agent. F + 1e- ⟶ F-1 N.B. If the Oxidation—Reduction Reaction is spontaneous, energy is released. Oxidation—Reduction Reaction as Source of Energy Rules for assigning oxidation numbers to atom 1. Atoms in their elemental state are assigned an oxidation number of zero (0). 2. Atoms in monoatomic (i.e., one atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, which differentiates them from charges. 3. In compounds, fluorine is assigned a -1 oxidation number; oxygen is usually assigned a -2 oxidation number (except in peroxide compound, wherein its oxidation number is - 1) and in binary compounds with fluorine, wherein it has +1); and hydrogen is usually assigned a +1 oxidation number (except when it exists as the hydride ion, H-, in which case rule 2 prevails. HF; H has +1 and F has -1 H2O; O has -2 4. In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (where is zero if the species is neutral) Oxidation—Reduction Reaction as Source of Energy Calculating oxidation of an element in a compound (especially those elements that has more than one oxidation number) Problem 6: What is the oxidation number of manganese in KMnO4? 0 = 1 (K) + 1 (Mn) + 4(O) 0 = 1 (+1) + Mn + 4(-2) 0 = +1 + Mn - 8 0 = -7 + Mn Mn = + 7 Oxidation—Reduction Reaction as Source of Energy Calculating oxidation of an element in a compound (especially those elements that has more than one oxidation number) Problem 7: What is the oxidation number of Chromium in Cr2O7-2? -2 = 2 Cr + 7(O) -2 = 2 Cr + 7 (-2) -2 = 2 Cr - 14 2Cr = -2+14 2Cr = +12 Cr = +6 Electrochemical Cell Electrochemical Cell: a device that can generate electrical energy from the chemical reactions occurring in it. ü The chemical reactions are separated into two half reactions: the oxidation half cell reaction, and reduction half cell reaction. ü A wire or conducting solution is attached to both cells in order to drive the electrons from one side to the other. ü Electrochemical cell can be galvanic (also known as voltaic) or electrolytic. Galvanic or Voltaic Cell Galvanic or Voltaic Cell: an electrochemical cell that uses spontaneous redox reactions to generate electric current (or electricity). Terminologies in Galvanic cells: Electrodes: electrically conducting sites at which either oxidation or reduction occur. Anode: electrode where oxidation occurs Cathode: electrode where reduction occurs Cell Notation: shorthand notation representing specific electrochemical reactions that contains the list of metals and ions involved in the reaction. Anode l electrolyte of anode ll electrolyte of cathode l cathode Cu(s) l Cu2+ (aq)(1 M) ll Ag+(aq)(1 M) ll Ag(s) Salt Bridge: complete the circuit which contains strong electrolyte that allows either cations or anions to migrate into the solution where they are needed to maintain charge neutrality. Atomic Perspective of Galvanic or Voltaic Cell ü At the anode, oxidation occurs, and cations dissolve into solution, leaving behind negative charge on the anode. ü At the cathode, reduction occurs, brings positively charge ions to the cathode. ü The build-up of charge on the anode and cathode generate cell potential or electromotive force (EMF) ü EMF is related to the maximum electrical work that can be obtained from an electrochemical cell. wmax = qE q= charge; E=cell potential Cell Potential of Galvanic or Voltaic Cell (under Standard Condition) ü Cell Potentials in an Electrochemical Cell are measured based on standard electrode (i.e., hydrogen standard electrode, SHE) ü Cell Potentials are expressed as Standard Reduction Potentials ü Standard Reduction Potentials can be either positive or negative potentials Positive Voltage would mean that reduction occurs o Large, positive standard reduction potential implies that the substance is reduced readily and a good oxidizing agent Negative Voltage would mean that oxidation occurs. o Large, negative standard reduction potential implies that the substance is oxidize readily an a good reducing agent. Cell Potential of Galvanic or Voltaic Cell (under Standard Condition) ü In Galvanic or Voltaic Cell, the half-reaction with the more positive reduction potential will be the cathode. ü Hence, the standard cell potential, Eocell , in Galvanic cell for any pair of half-reactions can be determined using the equation below. Eocell = Eored − Eoox ü Eocell in Galvanic Cell must be positive value Eored – standard reduction potential of the cathode Eoox − standard reduction potential of the anode Cell Potential of Galvanic or Voltaic Cell (under Standard Condition) Sample Problem 6: Copper and iron (generally in the form of steel) are two of the many metals used in designing machine. (a) Using standard reduction potentials, identify the anode and the cathode. (b) determine the cell potential for a galvanic cell, Eocell. What if Cu is the anode and Fe is the cathode Eocell = Eored − Eoox Eocell = −0.44 − (0.337 V) Eocell = − 0.78 V Eocell = Eored − Eoox Eocell = 0.337 V − (−0.44 V) Eocell = 0.78 V Fe(s) + Cu2+ (aq) ⟶ Fe2+(aq) + Cu(s) Nernst Equation (nonstandard condition) The equation that describe cell potentials under nonstandard conditions. o F – Faraday constant, 96, 485 J V-1 mol-1 or 96, 485 C/mol o n – number of electron transferred in the redox reaction o T –temperature in Kelvin o R – gas constant, 8.314 J mol-1K-1 o Eo – standard cell potential Nernst Equation Sample Problem 7: Assume that you have a cell that has an iron(II) concentration of 0.015 M and an H+ concentration of 0.0010 M. The cell temperature is 38∘C, and the concentration of hydrogen gas is 0.04 M. What would be the cell potential under these nonstandard condition? Battery Battery a cell or series of cells that generates an electric current. Composed of many different materials that harness electrical work of a galvanic cell. Types of Batteries 1. Primary cells or primary batteries 2. Secondary cells or secondary batteries Battery Primary Cells or Primary Batteries Single-use batteries that cannot be recharged Example: alkaline battery or dry cell Battery Primary Cells or Primary Batteries Single-use batteries that cannot be recharged Example: mercury battery – long lasting battery, anode is zinc, while cathode is mercury(II) oxide Battery Secondary Cells or Secondary Batteries Single-use batteries that can be recharged (rechargeable batteries) Example: Nickel-Cadmium Batteries (anode: cadmium; cathode: nickel) Battery Secondary Cells or Secondary Batteries Single-use batteries that can be recharged (rechargeable batteries) Example: Nickel-Metal Hydride Batteries (anode: metal alloy; cathode: nickel) Battery Secondary Cells or Secondary Batteries Single-use batteries that can be recharged (rechargeable batteries) Example: Lead-Acid Storage Batteries (anode: Lead; cathode: Lead oxide) Fuel Cells Fuel Cells A voltaic cell in which the reactants can be supplied continuously and the products of the reaction are continuously removed. Like batteries, fuel cells use a chemical reaction to produce electrical energy. But unlike batteries, it can be refueled on an ongoing basis. Example: reaction of hydrogen and oxygen gases to produce water. Ø Hydrogen gas flows into the anode compartment, and oxygen gas flows into the cathode compartment. Ø Oxygen is reduced at the cathode, while hydrogen is oxidized at the anode END