This preview has intentionally blurred parts. Sign up to view the full document

View Full Document

Unformatted Document Excerpt

When is Oxidation the loss of electrons by a chemical species." sodium (Na) forms a compound, Na+ is formed, so sodium is oxidized." Reduction Chloride is the gain of electrons by a chemical species." ions (Cl-) are formed when Cl2 reacts, so Cl is reduced." An oxidation-reduction reaction, or redox reaction, is one in which electrons are transferred. " In a redox reaction, at least one species is oxidized and at least one species is reduced." 2Na(s) + Cl2(g) 2NaCl(s) is a redox reaction, because Na is oxidized and Cl is reduced. " Oxidizing The agent is the reactant that gains electrons, causing an oxidation to occur." oxidizing agent is reduced in the reaction." Reducing The agent is the reactant that loses electrons, causing reduction to occur." reducing agent is oxidized in the reaction." In the reaction of sodium with chlorine, Na is the reducing agent, and Cl2 is the oxidizing agent." In a half-reaction either the oxidation or reduction portion of a redox reaction is given, showing the electrons explicitly." Half-reactions emphasize the transfer of electrons in a redox reaction." + - Na Na + 1e "oxidation half-reaction" "Cl2 + 2e 2Cl "reduction half-reaction" Oxidation state is the charge on a monatomic ion, or the charge on an atom when the shared electrons in covalent bonds are assigned to the more electronegative atom." Electron pairs shared by atoms of the same element are divided equally." Calcium in CaCl2, an ionic compound, has an oxidation state of +2. Hydrogen in H2O, has an oxidation state of +1." Determine the oxidation state of H H each atom in formaldehyde" O is more electronegative than C, C is more electronegative than H" O O C is assigned 8 electrons; oxidation state = - 2 " C is assigned 4 electrons; oxidation state = 0" H is assigned no electrons; oxidation state = +1" Draw the Lewis structure and determine the oxidation states of each atom in SO2" Draw the Lewis structure and determine the oxidation states of each atom in SO2 " "S = +4, O = -2" Answer: 1.Oxidation states for atoms in uncombined " elements is 0." 2.The oxidation state of a monatomic ion is " equal to the charge of the ion. Group IA is +1, group IIA is +2." 3.In compounds, F is always -1. Other halogens are -1 unless combined with a more electronegative element (oxygen or a halogen above it in the periodic table)." 4.In compounds, H is +1 except in metal " hydrides where it is -1." 5.In compounds, oxygen is always -2 except " for peroxides (O22-) or combined with F" 6.The sum of the oxidation states must equal " zero in a neutral compound, or the charge of the ion for polyatomic ions." Assign oxidation states for each atom in K2CrO4" Assign oxidation states to each atom in a) CO b) NH4Cl" Assign oxidation states to each atom in a) CO b) NH4Cl "a) C = +2, O = -2" "b) N = -3, H = +1, Cl = -1" Answer: " Balance the equation" "Cu + NO3- Cu2+ + NO + H2O" Determine oxidation states to identify species being oxidized and reduced." Write 2 skeleton half-reactions" Balance each half-reaction separately" all elements except H and O" Balance O by adding H2O as needed" + Balance H by adding H " Add electrons to balance charge" Multiply one or both half reactions by a small integer so the number of electrons lost is equal to the number of electrons gained." Add the two half reactions, canceling out the electrons and any species which appear on both sides of the equation." Balance Balance the equation" "Cu + NO3- Cu2+ + NO + H2O" Cu is oxidized from zero to +2" Write a skeleton equation for Cu" 2+ " " "Cu Cu Balance charge with electrons" 2+ + 2 e-" " "Cu Cu N is reduced from +5 to +2 " Write a skeleton equation for N" " "NO3- NO" Balance O by adding water" " "NO3- NO + 2H2O" + Balance H by adding H " - + 4H+ NO + 2H O" " "NO3 2 Balance charge by adding electrons - + 4H+ + 3e- NO + 2H O" "NO3 2 Make number of electrons lost equal number of electrons gained" 2+ + 2e-)" "3(Cu Cu - + 4H+ + 3e- NO + 2H O)" "2(NO3 2 Add the two half-reactions" + "3Cu + 2NO3 + 8H 3Cu2+ + 2NO + 4H2O" In basic solutions, balance as above and add H+ + OHH 2O " "or H 2O H+ + OH- to eliminate H+ from the equation." Balance the redox equation (basic solution)" " "Ag + 2CN + O2 Ag(CN)2 + OH-" Balance the redox equation (basic solution)" " "Ag + 2CN + O2 Ag(CN)2 + OH-" Answer:" O2 + 4Ag + 8CN- + 2 H2O >>> " " "4 Ag(CN)2- + 4 OH-" Balance the equation (acid solution)" 23+ Cr2O7 + C2H5OH Cr + CO2 + H2O" Balance the equation (acid solution)" 23+ Cr2O7 + C2H5OH Cr + CO2 + H2O" Answer: 2Cr2O7 + C2H5OH + 16H "4Cr3+ + 2CO2 + 11H2O" 2- + Balance " the equation (in basic solution)" 2-" "Zn + ClO Zn(OH)4 + Cl Balance " the equation (in basic solution)" 2"Zn + ClO Zn(OH)4 + Cl " Answer:" Zn + ClO + H2O + 2OH Zn(OH)4 + Cl - - 2- -" Voltaic cell (or galvanic cell): an apparatus that converts chemical energy directly into electrical energy" 2+ 2+ "Zn(s) + Cu (aq) Zn (aq) + Cu(s) is oxidized, Cu2+ is reduced" Zn Electrons move toward the positive electrode where the reduction takes place. Zn2+(aq) + 2e-" oxidation half-cell" 2+ Cu (aq) + 2e Cu(s) reduction half-cell" Physically separate the half cells so that the electrons must travel through a wire" Zn(s) Assume that reduction occurs in the right halfcell and oxidation in the left one" "Cu2+(aq) + 2eCu(s) " Reduction, Right Zn(s) Zn2+(aq) + 2e" oxidation, left Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)" Attach the positive lead (red) of the voltmeter to the right electrode, and the negative lead (black) to the left one." If the measured voltage is positive, the reaction is spontaneous as written; if it is negative, the reaction is spontaneous in the reverse direction." For the cell on the next slide" Write the half-reaction for each half-cell" Write the equation for the cell reaction" Determine if the reaction is spontaneous" Identify the electrode that is the source of electrons?" Determine the direction in which sodium ions migrate through the salt bridge?" Cu2+(aq) + 2e- (oxidation half-cell)" + Ag (aq) + 1e Ag(s) (reduction halfcell)" + 2+ Cu(s) + 2 Ag (aq) Cu (aq) + 2Ag(s)" The reaction is spontaneous (V is positive)" The Cu electrode is the electron source" + Na moves to the right in the salt bridge" Cu(s) Redox couples not involving metals can be used in half-cells with an inert electrode, such as gold, platinum, or graphite providing electrical contact." Two soluble ions " "Fe3+(aq) + eFe2+(aq)" Gas-ion couple " "Cl2(g) + 2e2Cl-(aq)" Metal-slightly soluble salt couple " "AgCl(s) + eAg(s) + Cl-" Sn and Sn4+ are both in contact with an inert electrode" 2+ Hydrogen gas is bubbled over an inert platinum electrode" Electromotive force or emf (E): the electrical driving force which pushes electrons from the oxidation half-cell to the reduction half-cell" Cell potential: the potential energy difference between the cell electrodes, measured in volts (V)" 1 V = 1 Joule/Coulomb" The potential of a cell depends on the concentration of species in the reaction." Standard Potential, Ecell: the cell potential when each reactant and product is in its standard state" solids, liquids, and gases in pure state at 1 atm pressure" solutes at 1 M concentration" Zn(s) + Cu2+(1 M) Cu(s) + Zn2+(1 M)" Ecell = +1.10 V (measured A voltage)" positive cell potential means reaction is spontaneous in forward direction" Cell potentials are additive " + 2+ " " E ocell Cu(s) + 2Ag (1 M) Zn(s) + Cu2+(1 M) Zn(s) + 2Ag+(1 M) 2Ag(s) + Cu (1 M) +0.46 V Zn2+(1 M) + Cu(s) +1.10 V 2Ag(s) + Zn2+(1 M) Ecell =+1.10 V + 0.46 V = +1.56 V" The standard reduction potential of a halfreaction is the potential of the reduction relative to the standard hydrogen electrode. " F2(g) + 2e2F-(1 M) " +2.87 V is the potential of the cell with the reaction F2(g) + H2(g) 2H+(1 M) + 2F-(1 M) Zn2+(1 M) + 2eZn(s) -0.76 V " is the potential of the cell with the reaction Zn2+(1 M) + H2(g) 2H+(1 M) + Zn(s)" The half-reaction is written as a reduction (electrons are reactants)" The standard reduction potential for" "2H+(1 M) + 2eH2(g, 1 atm) is exactly 0.0 V" The oxidation potential of any half-reaction is the negative of the reduction potential." "Ag+(1 M) + eAg(s)Eored = +0.80 V " Ag(s) Ag +(1 M) + e"Eoox = -0.80 V" A voltaic cell is made up of an Al electrode in a 1 M solution of Al(NO3)3 and a Cu electrode in a 1 M solution of Cu(NO3)2 . What is the cell emf, which electrode is positive, which electrode is negative, what is the oxidizing agent, and what is the reducing agent?" The half reactions involved are:" 3+ "Al (aq) + 3e Al(s) "Ered = -1.66 V" "Cu2+(aq) + 2e- Cu(s) "E red = +0.34 V" The more positive half-reaction is the reduction half-reaction" The more negative half-reaction is the oxidation half-reaction" "Al(s) Al3+(aq) + 3e- "Eox = +1.66 V" 2+ "Cu (aq) + 2e Cu(s) "E red = +0.34 V" 2+ 3+ "2Al(s) + 3Cu (aq) 2Al (aq) +3Cu(s)" Ecell = +1.66 V +0.34 V = 2.00 V" The Al electrode is the negative electrode. Al is the reducing agent." The Cu electrode is the positive 2+ electrode. Cu is the oxidizing agent." Write the cell reaction and calculate " Ecell for the voltaic cell made up of a standard Mg/Mg2+ half-cell and a standard Sn2+/Sn4+ half-cell." Mg+2 + 2e- >>>>Mg(s) E(V) = -2.37" Sn+4 + 2e- >>>> Sn+2 E(V) = +0.16" Write the cell reaction and calculate E cell for the voltaic cell made up of a 2+ standard Mg/Mg half-cell and a standard Sn2+/Sn4+ half-cell." Answer: " Mg(s) + Sn4+(aq) Mg2+(aq) +Sn2+(aq)" Ecell = +2.53 V" A reaction is spontaneous in the forward direction if E is positive." Determine the direction in which the following reaction is spontaneous:" Zn(s) + 2H2O(l) Zn(OH)2(aq) +H2(g)" The half-reactions are:" 2+ Zn(s) Zn (aq) +2e "E = +0.76 V" 2H2O(l) + 2e- 2OH-(aq) +H2(g) E = -0.83 V" Ecell = +0.76 V -0.83 V = -0.07 V" The reaction is spontaneous in the reverse direction." Determine the direction in which the following reaction is spontaneous:" " "Pb(s) + Br2(l) PbBr2(aq)" E(V) = -0.126" E(V) = +1.06" Pb+2 (aq) + 2e- >>> Pb(s) Br2(l) + 2e- >>> 2Br- (aq) Determine the direction in which the following reaction is spontaneous:" " "Pb(s) + Br2(l) PbBr2(aq)" Ecell = +1.19 V" "The reaction is spontaneous in the forward direction." Answer: Half-reactions listed in order of reduction potentials are also in order of chemical reactivity." When two half-reactions are combined, the one higher on the list proceeds spontaneously as a reduction; the one lower on the list proceeds spontaneously as an oxidation." What happens when Fe is added to a solution of 1 M Zn(NO3)2 and 1 M Co (NO3)2?" Co2+(aq) + 2e- Co(s) E = -0.28 V" 2+ "Fe (aq) + 2e Fe(s) "E = -0.44 V" "Zn2+(aq) + 2e- Zn(s) "E = -0.76 V" 2+ 2+ Fe reacts with Co but not with Zn " 2+ 2+ "Fe(s) + Co (aq) Fe (aq) + Co(s) " G = -nFE" "where"G = "free energy change" "(negative for spontaneous reactions)" "n = number of moles of electrons transferred" " "F = " faraday, 96,485 C/mole e " " "E = "cell potential (volt = J/C)" Under standard state conditions," " " G = -nFE" Spontaneous reactions (negative G) have positive cell potentials." Calculate G for the reaction" + "AgCl(s) Ag (aq) + Cl (aq)" AgCl(s) + e- Ag(s) + Cl-(aq) "E = 0.222 V" + Ag(s) Ag (aq) + e "E = -0.80 V" " "Ecell = -0.58 V" G = -nFE = -(1)(96,485 C/mol)(-0.58 V)" G = +56 kJ (remember V = J/C)" G = -RT ln Keq" G = -nFE" -nFE = -RT ln Keq" RT" 2.303 RT" E = ln Keq = " log Keq" nF" nF" 0.0591" at 298 K, E = " log Keq" n" Determine Keq for the reaction at 25C" Fe(s) + Pb2+(aq) Pb(s) + Fe2+(aq)" Ecell = +0.44 V -0.13 V = +0.31 V" 0.0591" E = " log Keq" n" 0.0591" +0.31 V = log Keq" 2" 10" log Keq = 10.49 Keq = 3.09 x 10 The Nernst equation is used to calculate cell potentials under non-standard conditions." "G = G + RT lnQ" "-nFEcell = -nFEcell + RT ln Q" 2.303 RT" "Ecell = Ecell log Q" nF" at 298 K" 0.0591" "Ecell = Ecell log Q" n" What is the cell potential for a cell made up of Zn, 0.500 M Zn(NO3)2, Cu, and 1.500 M Cu(NO3)2?" 2+ - Zn (aq) + 2e Zn(s) "Ered = -0.76 V" Cu2+(aq) + 2e- Cu(s) "Ered = +0.34 V" 2+ 2+ Zn(s) + Cu (aq) Zn (aq) + Cu(s) " " "Ecell = 1.10 V " 0.0591" Ecell = Ecell log Q" n" [Zn ]" 0.0591" Ecell = 1.10 log" 2+ 2" [Cu ]" (0.500)" Ecell = 1.10 - 0.0295 log " (1.500)" Ecell = 1.11 V" 2+ Oxidation half-reaction:" Pb(s) + SO42-(aq) PbSO4(s) + 2e-"" Reduction half-reaction:" PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- PbSO4(s) + 2H2O(l) " Overall reaction:" + 2Pb(s) + PbO2(s) + 4H (aq) + 2SO4 (aq) 2PbSO4(s) + 2H2O(l)" No salt bridge is needed, because the oxidizing and reducing agents are solids, and cannot come in direct contact with each other. They "sit in the same beaker"." The battery is rechargeable as long as the PbSO4 formed adheres to the electrode surfaces." A Zn can is the anode, in contact with moist paste of MnO2, NH4Cl, and ZnCl2 in starch. A carbon rod cathode is in contact with the metal cap on top of the cell." Oxidation half-reaction:" Zn(s) Zn2+(aq) + 2e"" Reduction half-reaction:" + 2NH4 (aq) + MnO2(s) + 2e " " "Mn2O3(s) + 2NH3(aq) + H2O(l) " Overall reaction:" + Zn(s) + 2NH4 (aq) + MnO2(s) " 2+ " Zn (aq) + Mn2O3(s) + 2NH3(aq) + H2O(l)" Alkaline Battery: NH4Cl is replaced with KOH or NaOH. The Zn anode corrodes less under alkaline conditions and the cell voltage is more constant because the net reaction contains all condensedphase species." Oxidation half-reaction:" Zn(s) + 2OH (aq) ZnO(s) + H2O(l) + 2e Reduction half-reaction:" 2MnO2(s) + H2O(l) + 2e " " Mn2O3(s) + 2OH (aq) " Overall reaction:" Zn(s) + MnO2(s) ZnO(s) + Mn2O3(s) " Fuel cell: a voltaic cell in which the reactants are supplied continuously. Usually, the products are removed as they are formed." The H2/O2 fuel cell uses the reaction" " "2H2(g) + O2(g) 2H2O(l)" "to produce electricity and water on space missions." States" Balancing Redox Equations" Oxidation Ion electron method" Acidic and basic solutions" Voltaic Half Cells" cells" Electron transfer " Ion transfer" Potential of Voltaic Cells" of a reaction" Spontaneity Standard additive" Potentials for Half-Reactions" Cell Potentials, G, Keq" G = -nFE" E=(0.0591)/n log Keq" Dependence The of Voltage on Concentration" Nernst Equation" E = E- (0.0591)/n log Q" Batteries" ... View Full Document

End of Preview

Sign up now to access the rest of the document