patino_chm_2046_chapter18

patino_chm_2046_chapter18 - Chapter 18 Electrochemistry...

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Unformatted text preview: Chapter 18 Electrochemistry GOALS Balancing redox reactions Voltaic cells Electrochemical potentials Electrolysis …& the calculations!! Review: oxidation states oxidation/reduction oxidizing/reducing agent 2 ch. 17 Why Study Electrochemistry? Why Study Electrochemistry? • • • Batteries Corrosion Industrial production Industrial of chemicals such as of Cl2, NaOH, F2 and Al • Biological redox Biological reactions reactions The heme group Electron Transfer Reactions Electron Transfer Reactions • Electron transfer reactions are oxidation-reduction or redox reactions (i.e. changes in oxidation states). • Redox reactions can result in the generation of an electric current (battery), or, may be caused by applying an electric current (electroplating). • Therefore, this field of chemistry is often called ELECTROCHEMISTRY. ELECTRON TRANSFER REACTIONS ELECTRON TRANSFER REACTIONS 0 0 1+ 1+ 1- 2Na(s) + Cl2(g) → 2NaCl(s) (g) 1+ 0 2+ 0 2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g) 0 0 Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) Cu(s) (aq) 0 0 3+ 3+ 2- 2Fe(s) + xH2O(l) + 1½O2(g) → Fe2O3.xH2O(s) (g) O(s) Review of Terminology for Redox Reactions •• OXIDATION—loss of electron(s) by a species; OXIDATION—loss of electron(s) by a species; iincrease in oxidation number ;;e-- to the right of arrow.. ncrease in oxidation number e to the right of arrow Na → Na++ + e-Na → Na + e •• REDUCTION—gain of electron(s); decrease in REDUCTION—gain of electron(s); decrease in oxidation number; e-- to left of arrow.. oxidation number; e to left of arrow ½Cl22(g) + e- → Cll½Cl (g) + e- → C •• OXIDIZING AGENT—electron acceptor; it is OXIDIZING AGENT—electron acceptor; it is reduced: ½ Cll2(g) + e- → Cll(g) + e- → C reduced: ½ C 2 •• REDUCING AGENT—electron donor; it is oxidized REDUCING AGENT—electron donor; it is oxidized Na → Na++ + e-Na → Na + e Electrochemical Cells Electrochemical Cells • Apparatus for generating an Apparatus electric current through the use of a product favored reaction (spontaneous): voltaic or galvanic cell. • An electrolytic cell is used to carry out electrolysis (an electric current is used to bring about a nonspontaneous chemical reaction). chemical Batteries are voltaic Batteries cells cells Electrochemistry Electrochemistry Alessandro Volta, Alessandro 1745-1827, Italian scientist and inventor. scientist Luigi Galvani, 1737-1798, Luigi Italian scientist and inventor. Italian Balancing Equations for Redox Reactions Balancing Equations for Redox Reactions Some redox reactions have equations that Some must be balanced by special techniques. must MnO4-(aq) + 5 Fe2+(aq) + 8 H+(aq) (aq) Fe = +2 Mn = +7 → Mn2+ (aq) + 5 Fe3+(aq) + 4 H2O(liq) Mn2+ (aq) Mn = +2 Fe = +3 Rules for Assigning Oxidation States • rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) 1. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = -1 in NaCl 1. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0 10 Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl2 11 Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below (grp # - 8) nonmetals higher on the table take priority Nonmetal Oxidation State Example F -1 CF4 H +1 CH4 O -2 CO2 Group 7A -1 CCl4 Group 6A -2 CS2 Group 5A -3 NH3 12 Balancing Equations Balancing Cu + Ag+ −give→ Cu2+ + Ag Balancing Equations Balancing Equations Step 1: Divide the reaction into halfreactions, one for oxidation and the reactions, other for reduction. other Ox Cu → Cu2+ Cu Red Ag+ → Ag Step 2: Balance each for mass. Already Balance done in this case. done Step 3: Balance each half-reaction for Balance charge by adding electrons. charge Ox Cu → Cu2+ + 2e2eRed Ag+ + e- → Ag Balancing Equations Balancing Equations Step 4: Multiply each half-reaction by a factor Multiply so that the reducing agent supplies as many electrons as the oxidizing agent requires. electrons Reducing agent Cu → Cu2+ + 2eOxidizing agent 2 Ag+ + 2 e- → 2 Ag Step 5: Add half-reactions to give the overall Add equation. equation. Cu + 2 Ag+ → Cu2+ + 2Ag The equation is now balanced for both The charge and mass (the 2e- of the left are charge cancelled out with those on the right). cancelled Reduction of VO2+ with Zn Reduction Balancing Equations Balancing Equations Balance the following in acid solution— Balance acid 2+ VO2+ + Zn → VO2+ + Zn2+ VO Zn Step 1: Ox Red Write the half-reactions Zn → Zn2+ Zn VO2+ → VO2+ Step 2: Balance each half-reaction for Balance mass. mass. Ox Zn → Zn2+ + 2eZn 2e Red VO2+ + e- → VO2+ excess of 2+ on the right VO eRed Zn lost two electrons. They are written on the right. Zn V is 5+ on the left and 4+ on the right. It gained one is electron. That is written on the left side. electron. Balancing Equations Balancing Equations Step 3: Balance half-reactions for charge Reaction is acidic, then we can use H+. Reaction Ox Zn → Zn2+ + 2eZn Red e- + 2 H+ + VO2+ → VO2+ + H2O eStep 4: Multiply by an appropriate factor. Ox Zn → Zn2+ + 2eZn 2eRed 2e- + 4 H+ + 2 VO2+ → 2 VO2+ + 2 H2O Step 5: Add balanced half-reactions. Add balanced Zn + 4 H+ + 2 VO2+ → Zn2+ + 2 VO2+ + 2 H2O Tips on Balancing Equations Tips on Balancing Equations • Never add O2, O atoms, or O2to balance oxygen. to Balance O with OH- or H2O. • Never add H2 or H atoms to balance hydrogen. balance Balance H with H+/H2O in acid or OH-/H2O iin n base. base. Tips on Balancing Equations {Equations that include oxoanions like SO42-, NO3-, ClO- , CrO42-, and MnO4-, also fall into this category}. Be sure to write the correct charges on all the Be ions. ions. •Check your work at the end to make sure mass Check and charge are balanced. and •PRACTICE!!!!!!!!!!! More Practice - Balance the equations below! I−(aq) + MnO4−(aq) → I2(aq) + MnO2(s) in basic solution An alkaline (basic) solution of hypochlorite ions reacts with solid chromium(III) hydroxide to produce chromate and chloride ions. ClO3- + Cl- → Cl2 (in acid) Cr2O72- + I- → IO3- + Cr3+ (in acid) MnO4- + H2SO3 → SO42- + Mn2+ (in acid) Cr(OH)4- + H2O2 → CrO42- + H2O (in basic soln) Zn + NO3- → Zn(OH)4- + NH3 (in basic soln) 21 VOLTAIC CELLS VOLTAIC - use a chemical rxn to produce an electric current. wire wire e le c t r o ns e le c t r o ns Zn Zn Zn 2+ ions Zn 2+ ions salt salt bridge bridge Cu Cu Cu 2+ ions Cu 2+ ions The Zn|Zn2+ and Cu|Cu2+ Cell Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (aq) A CHEMICAL CHANGE PRODUCES AN ELECTRIC CURRENT Zn metal With time, Cu plates out onto Zn metal With time, Cu plates out onto Zn metal sstrip,and Zn strip “disappears.” trip, and Zn strip “disappears.” Cu2+ ions Electrons are transferred from Zn to Electrons are transferred from Zn to Cu2+, ,but there is no useful electric Cu2+ but there is no useful electric ccurrent. urrent. Oxidation: Zn(s) → Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e- → Cu(s) -------------------------------------------------------Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) (aq) A CHEMICAL CHANGE PRODUCES AN ELECTRIC CURRENT •To obtain a useful current, To we separate the oxidizing and reducing agents so that electron transfer occurs through an external wire. Zn(s) → Zn2+(aq) + 2eZn(s) (aq) 2ewire wire eeee c t r o ns l l c t r o ns Zn Zn salt salt bridge bridge Zn 2+ ions Zn 2+ ions Cu Cu Cu 2+ ions Cu 2+ ions This is accomplished in a GALVANIC or VOLTAIC This GALVANIC VOLTAIC cell. cell. A group of such cells is called a battery. group battery Zn → Zn2+ + 2eZn Oxidation Oxidation Oxidation Anode Anode Negative Negative wire wire e llec ttrro ns e e c o ns Zn Zn salt salt bridge bridge Cu2+ + 2e- → Cu Cu Cu ← Anions Anions Cations → Cations Zn 2+ iions Zn 2+ ons Reduction Reduction Cathode Cathode Positive Positive Cu 2+ iions Cu 2+ ons ••Electrons travel through external wire. Electrons travel through external wire. ••Salt bridge allows anions and cations to move Salt bridge allows anions and cations to move Salt Salt between electrode compartments. between electrode compartments. between between CELL POTENTIAL, E CELL POTENTIAL, E wire 1.10 V wire e lle c t r o ns e e c t r o ns Zn Zn Zn and Zn2+, anode Zn2+ iions Zn2+ ons 1.0 M salt salt bridge bridge Cu Cu Cu2+ iions Cu2+ ons 1.0 M Cu and Cu2+, cathode • Electrons are “driven” from anode to cathode Electrons • by an electromotive force or emf. electromotive emf For Zn/Cu cell, this is indicated by a voltage of For 1.10 V at 25 ˚C and when [Zn2+] = [Cu2+] = 1.0 M. 1.10 Need Calculate Cell Voltage Need Calculate Cell Voltage • Balanced half-reactions can be added together to get the overall, balanced equation. Zn(s) → Zn2+(aq) + 2eZn(s) → Zn2+(aq) + 2eZn(s) Cu2+((aq) + 2e- → Cu(s) aq) Cu2+(aq) + 2e- → Cu(s) (aq) --------------------------------------------------------------------------------------Cu2+((aq) + Zn(s) → Zn2+(aq) + Cu(s) aq) Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) (aq) If we know Eo for each half-reaction, we add them to get Eo for overall reaction. them -need Eo for Zn & Cu halfcells Standard Reduction Potential • We can measure ALL other half-cell potentials relative to another half-reaction. • We select as a standard half-reaction, the reduction of H+ to H2 under standard conditions (1atm, 1 M @ 25 oC) and which we assign a potential difference = 0 V Standard Hydrogen Electrode, SHE 28 Zn/Zn2+ half-cell hooked up to a SHE. Zn/Zn2+ half-cell hooked up to a SHE. Eoo ffor the cell = +0.76 V E or the cell = +0.76 V Negative Negative electrode electrode Volts Volts Zn Zn Supplier Supplier of electrons electrons -- Salt Bridge Salt Bridge Zn2+ Zn2+ Zn n →Zn2++ 2eZ Zn2+ + 2eZn Zn2+ + eZn OXIDATION Oxidation Oxidation OXIDATION ANODE ANODE Anode H+ H+ Positive Positive + electrode + electrode H2 H2 Acceptor Acceptor of electrons electrons H + 2e22H++ + 2e- → HH2 2 2 H+ + 2eH2 REDUCTION Reduction REDUCTION CATHODE CATHODE Cathode Zn is a better reducing agent than H2 Zn better Cu/Cu2+ half-cell hooked up to a SHE. Cu/Cu2+ half-cell hooked up to a SHE. Eoo ffor the cell = +0.34 V E or the cell = +0.34 V Volts Volts Positive Cu Cu Acceptor of electrons + + -Salt Bridge Salt Bridge Cu2+ Cu2+ 2+ Cu2++ 2e- → Cu 2+ + 2eCu Cu Reduction Cu + 2eCu REDUCTION REDUCTION Cathode CATHODE CATHODE H+ H+ Negative H2 H2 Supplier Supplier of electrons electrons H2 → 2 H+ + 2eH2 Oxidation 2e2 H+ + 2eH2 2 H+ + OXIDATION OXIDATION Anode ANODE ANODE H2 is now a better reducing agent than Cu! better Zn/Cu Electrochemical Cell + wire wire e le c ttrr o ns e le c o n s Anode; negative; source of electrons Zn Zn Zn 2+ ions Zn 2+ ions Cu Cu salt salt bridge bridge Cu 2+ ions Cu 2+ ions Cathode; positive; sink for electrons oxid: Zn(s) → Zn2+(aq) + 2eEo = +0.76 V red: Cu2+(aq) + 2e- → Cu(s) Eo = +0.34 V red: (aq) --------------------------------------------------------------Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) Cu (aq) (aq) Eo (calc’d) = +1.10 V Uses of Eo Values Organize half-reactions by relative ability to act as oxidizing/reducing agents. Half-rxns are written as reduction rxns!! Cu2+(aq) + 2e- → Cu(s) (aq) Zn2+(aq) + 2e- → Zn(s) (aq) wire wire eeee c t r o ns l l c t r o ns Zn Zn Zn 2+ ions Zn 2+ ions Cu Cu salt salt bridge bridge Cu 2+ ions Cu 2+ ions Eo = +0.34 V Eo = –0.76 V When a reaction is reversed, the sign of E˚ is reversed! oxidizing agents reducing agents 34 Using Standard Potentials, Eo Using Standard Potentials, E Which is the best oxidizing agent: O2 (1.23 V); H2O2 (1.77 V) or Cl2 (1.36 V)? H2O2 (1.77 V) Which is the best reducing agent: Hg (+0.79 V), Al (-1.66 V), or Sn (-0.14 V)? Al (-1.66 V) Using Standard Potentials, Eo Using Standard Potentials, E Which substance is the best oxidizing agent? Cr2O72- + 6e- + 14H+ → 2Cr3+ + 7H2O (+1.33 V) O2 + 4e- + 4H+ → 2H2O (+1.23 V) Fe3+ + e- → Fe2+ (+0.77 V) Cr2O72Which element/ion is the best reducing agent? Fe3+ + e- → Fe2+ (+0.77 V) I2 + 2e- → 2I(+0.54 V) Sn4+ + 2e- → Sn2+ (+0.15 V) Sn2+ Standard Redox Potentials, Eo Standard Redox Potentials, E Cu2+ + 2e- Cu +0.34 Any substance on the right will reduce any substance HIGHER than it on the HIGHER 2 H+ + 2e- H2 0.00 LEFT. LEFT. Zn2+ + 2e- Zn -0.76 oxidizing ability of ion Eo (V) reducing ability of element • Zn can reduce H+ and Cu2+. • H2 can reduce Cu2+ but not Zn2+ Zn • Cu cannot reduce H+ or Zn2+. Zn Standard Redox Potentials, Eo Standard Redox Potentials, E Ox. agent Cu2+ + 2e- --> Cu +0.34 2 H+ + 2e- --> H2 0.00 Zn2+ + 2e- --> Zn -0.76 Red. agent Any substance on the right will reduce any substance Any higher than it on the left. higher Northwest-southeast rule: product-favored reactions occur between • reducing agent at southeast corner (ANODE) & • oxidizing agent at northwest corner (CATHODE) oxidizing Standard Redox Potentials, Eo Standard Redox Potentials, E Ox. agent Cu2+ + 2e- --> Cu +0.34 V Ni 2++ 2e- --> Ni -0.25 V Zn2+ + 2e- --> Zn -0.76 V Red. agent Zn will reduce Ni2+, Cu2+; Ni will reduce Cu2+. Northwest-southeast rule: product-favored reactions occur between • reducing agent at southeast corner (ANODE) & • oxidizing agent at northwest corner (CATHODE) oxidizing Using Standard Potentials, Eo Using Standard Potentials, E • In which direction do the following reactions go? • Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) +0.46 V (aq) (aq) +0.46 • Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) +1.10 V (aq) (aq) +1.10 Go to the right as written • Fe2+(aq) + Cd(s) → Fe(s) + Cd2+(aq) -0.04 V Goes LEFT, opposite to direction written • What is Eonet for this reverse reaction? +0.04 V Eo for a Voltaic Cell Fe(s) + Cd2+(aq) → Cd(s) + Fe2+(aq) Volts Volts Cd Cd Fe Fe Salt Bridge Salt Bridge Cd2+ Cd2+ Cd → Cd2+ + 2eCd or Cd2+ + 2e- → Cd (-0.40 V) (-0.40 Fe2+ Fe2+ Fe → Fe2+ + 2eFe or Fe2+ + 2e- → Fe (-0.44 V) (-0.44 Which way does the reaction proceed? In which direction is it spontaneous? Eo for a Voltaic Cell From the table, table Fe Fe • Fe is a better reducing agent Fe Volts Volts Cd Cd S altBridge alt Bridge S Cd2+ Cd2+ Fe2+ Fe2+ than Cd (-0.44 V; anode) than • Cd2+ is a better oxidizing agent than Fe2+ (-0.40 V; agent cathode) cathode) Eo = E˚cathode - E˚anode (reverse the smaller, more anode negative, then add) negative cathode: Cd2+(aq) + 2e- → Cd(s) -0.40 V (red) - anode: Fe(s) → Fe2+(aq) + 2e- +0.44 V (oxid) Overall : Fe(s) + Cd2+(aq) → Cd(s) + Fe2+(aq) +0.04 V More 0n Cell Voltage When two half-rxns (written as reduction) are joined in an electrochemical cell, the one with the larger half-cell potential occurs in the forward direction, and the one with the smaller potential occurs in the reverse direction. Cd2+ + 2e- → Cd (-0.40 V) (-0.40 Fe2+ + 2e- → Fe (-0.44 V) larger smaller (reverse this rxn) Cd2+ + 2e- → Cd (-0.40 V) larger → Fe2+ + 2e- (+0.44 V) overall: Cd2+ + Fe → Cd + Fe2+ (+0.04 V) Cd Fe More 0n Cell Voltage When two half-rxns (written as reduction) are joined in an electrochemical cell, the one with the larger half-cell potential occurs in the forward direction, and the one with the smaller potential occurs in the reverse direction. Ni2+ + 2e- → Ni (-0.23 V) larger (-0.23 Mn2+ + 2e- → Mn (-1.18 V) smaller (reverse this rxn) Ni2+ + 2e- → Ni (+1.18 V) Mn → Mn2+ + 2e- (-0.23 V) overall: Ni2+ + Mn → Ni + Mn2+ (+0.95 V) Ni More 0n Cell Voltage Assume I- ion can reduce water. 2 H22O + 2e- → H22 + 2 OH-Cathode 2 H O + 2e- → H + 2 OH Cathode 2 II- → II2 + 2eAnode 2 → 2 + 2eAnode ------------------------------------------------------------------------------------------------2 II- + 2 H22O → II2 + 2 OH-- + H22 2 + 2 H O → 2 + 2 OH + H Assuming reaction occurs as written, Assuming E˚net = E˚cathode - E˚anode (from values in table) E˚ = (-0.828 V) - (+0.535 V) = -1.363 V (-0.828 -1.363 Minus Enet˚ means net rxn. occurs in the opposite means direction (favors I- + H2O). direction I2 + 2 OH- + H2 → 2I- + H2O (+1.363 V)!!! OH Calculate E°cell for the reaction at 25°C Al(s) + NO3−(aq) + 4 H+(aq) → Al3+(aq) + NO(g) + 2 H2O(l) (This is the reaction of Al with nitric acid) Separate the reaction into the oxidation and reduction half-reactions Find the E° for each halfreaction and sum to get E°cell ox: Al(s) → Al3+(aq) + 3 e− E°ox = −E°red = +1.66 V red: NO3−(aq) + 4 H+(aq) + 3 e− → NO(g) + 2 H2O(l) E°red = +0.96 V E°ox = −(E°red) = +1.66 V E°red = +0.96 V E°cell = (+1.66 V) + (+0.96 V) = +2.62 V 46 For the reaction Al(s) + NO3−(aq) + 4 H+(aq) → Al3+(aq) + NO(g) + 2 H2O(l) ox: Al(s) → Al3+(aq) + 3 e− E°ox = −E°red = +1.66 V red: NO3−(aq) + 4 H+(aq) + 3 e− → NO(g) + 2 H2O(l) E°red = +0.96 V E°cell = (+1.66 V) + (+0.96 V) = +2.62 V The symbol of the cell is Al(s)| Al3+(aq) || NO3−(aq), 4 H+(aq) , NO(g) |Pt This is the symbol for the salt bridge This is the symbol for the electrode-solxn contact 47 E at Nonstandard Conditions E at Nonstandard Conditions 0.0257 V [products] E=E − ln n [reactants] o RT E =E − ln Q • The NERNST EQUATION nF • E = potential under nonstandard conditions o • • • • • n = no. of electrons exchanged ln = “natural log” If [P] and [R] = 1 mol/L, then E = E˚ If [R] > [P], then E is LARGER than E˚ If [R] < [P], then E is smaller than E˚ Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Zn is anode (-0.76 V) in 0.40 M Zn2+(aq) and Cu is cathode (0.34 V) in 4.8 × 10-3 M Cu2+(aq) . Calculate the cell potential. Solution: (need standard cell potential, Eocell) Eºcell = Eºcathode– Eºanode = (0.34 V) – (–0.76 V) Substituting E = Ecell = 0.34 + 0.76 = +1.10 V o 2+ 0.0257 V [Zn ] − ln 2+ n [Cu ] Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Zn is anode (-0.76 V) and Cu is cathode (0.34 V) n = 2 (Cu2+ + 2e- = Cu0) Eocell = 1.10 V Solution: 0.0257 V [0.40] E = 1.10 V − ln −3 2 [4.8 × 10 ] E = 1.10 V – (0.01285)(4.42) = 1.04 V 2Fe3+(aq) + 3Mg(s) → 2Fe(s) + 3Mg2+(aq) Fe3+ = 1.0 × 10-3 M ; Mg2+ = 2.5 M Calculate the cell potential. Solution: (need standard cell potential, Eocell) Eºcell = Eºcathode– Eºanode = (-0.036 V) – (–2.37 V) = -0.036 + 2.37 = +2.33 V Substituting E = Ecell What is n??? o 2+ 3 0.0257 V [ Mg ] − ln 3+ 2 n [ Fe ] 2Fe3+(aq) + 3Mg(s) → 2Fe(s) + 3Mg2+(aq) Fe3+ = 1.0 × 10-3 M ; Mg2+ = 2.5 M Calculate the cell potential. Solution: (need standard cell potential, Eocell) Eºcell = +2.33 V; n = 6 Substituting 0.0257 V [2.5]3 E = 2.33 V − ln −3 2 6 [1.0 ×10 ] 0.0257 V E = 2.33 V − × 16.564 6 E = 2.33 V − 0.071 V Ans = 2.26 V BATTERIES BATTERIES Primary, Secondary, and Fuel Cells Dry Cell Battery Dry Cell Battery Primary battery — uses redox reactions that cannot be restored by recharge. restored Anode (-) Zn → Zn2+ + 2eZn Cathode (+) 2 NH4+ + 2e- → 2 NH3 + H2 NH Alkaline Battery Nearly same reactions as in common dry cell, but under basic conditions. cell, Anode (-): Zn + 2 OH- → ZnO + H2O + 2eAnode Zn Cathode (+): 2 MnO2 + H2O + 2e- → Cathode MnO 2eMn2O3 + 2 OHMn Lead Storage Battery Lead Storage Battery • Secondary battery • Uses redox reactions Uses • that can be reversed. that Can be restored by Can recharging recharging Lead Storage Battery Lead Storage Battery Anode (-) Eo = +0.36 V Anode Pb + HSO4- → PbSO4 + H+ + 2eCathode (+) Eo = +1.68 V PbO2 + HSO4- + 3 H+ + 2e→ PbSO4 + 2 H2O Ni­Cad Battery Ni­Cad Battery Anode (-) Cd + 2 OH- → Cd(OH)2 + 2eCathode (+) Cathode NiO(OH) + H2O + e- → Ni(OH)2 + OHNiO(OH) e- Fuel Cells: H2 as a Fuel Fuel Cells: H •Fuel cell - reactants are supplied continuously from an external source. an •Cars can use electricity Cars generated by H2/O2 fuel generated cells. cells. •H2 carried in tanks or generated from hydrocarbons. hydrocarbons. Storing H2 as a Fuel Storing H One way to store H2 is to adsorb the gas onto a metal or metal alloy. Hydrogen—Air (O2) Fuel Cell Hydrogen—Air Anode: 2H2(g) → 4H+(aq) + 4eAnode: (g) (aq) Cathode: O2(g) + 2H2O(liq) + 4e- → 4OH- (aq) O(liq) ------------------------------------------------------------------Net: O2(g) + 2H2(g) → 2H2O(liq) (g) O(liq) Electrolysis Electrolysis Using electrical energy to produce chemical change. Sn2+(aq) + 2 Cl-(aq) → Sn(s) + Cl2(g) Electrolysis of water; electroplating; refining metals; production of chemicals. Electrolysis Electrolysis Electric Energy → Chemical Change • Electrolysis of molten NaCl. molten • Here a battery “pumps” electrons from Cl- to Na+. from • NOTE: Polarity of NOTE: electrodes is reversed from batteries. batteries. electrons electrons BATTERY BATTERY + + Anode Anode Cathode Cathode Cl- Na+ Cl- Na+ Electrolysis of Molten NaCl Electrolysis of Molten NaCl electrons electrons Anode (+) BATTERY BATTERY 2Cl-(l) → Cl2(g) + 2e2Cl (g) ++ Anode Anode Cathode Cathode Cl- - Na++ Cl Na (-1.36 V) (-1.36 Cathode (-) Na+(l) + e- → Na Na e- (-2.71 V) (-2.71 Eo for cell (in melted NaCl) = E˚c + E˚a = - 2.71 V + (-1.36 V) 2.71 = - 4.07 V (in melted NaCl) rxn is nonspontaneous rxn External electrical energy needed because Eo is (-). Electrolysis of Aqueous NaOH Electrolysis of Aqueous NaOH NaOH + H2O → Na+(aq) + OH-(aq) Electric Energy → Chemical Change Electric Anode Anode (+) Anode Cathode E° = -0.40 V E° 4 OH- → O2(g) + 2 H2O + 4e- Cathode (-) 4 H2O + 4e- → 4e2 H 2 + 4 OH Eo for cell = -1.23 V H2O is more easily reduced is than Na+!! (Eo -2.71 V) than !! (E -2.71 Electrolysis of Aqueous NaCl Electrolysis of Aqueous NaCl NaCl + H2O → Na+(aq) + Cl-(aq) electrons electrons Anode (+) Anode 2 Cl- → Cl BATTERY BATTERY Cl2(g) + 2e E° =-1.36 V Cl + + Cathode (-) Cathode Anode Cathode Anode Cathode 2 H2O + 2e- → 2eH 2 + 2 OH Cl-- Na+ Cl Na+ o E for cell = -2.19 V H2O H2O Note that H2O is more is Easily reduced than Na+(E° = -2.71 V) Easily Also, Cl-- is oxidized in preference to H2O Also, 2H2O(l) → 2H O(l) 2 because O2(g) + 4H+ + 4e- because of kinetics (overvoltage) E°red < -1.23 V, may be down to ~ -2.00 V down red Eo and Thermodynamics he • Eo is related to ∆Go, tthe free energy • change for the reaction. change ∆G˚ proportional to –nE˚ ∆Go = ­nFEo where F = Faraday constant where Faraday = 9.6485 x 104 J/V•mol of e9.6485 (or 9.6485 × 104 coulombs/mol) (or and n is the number of moles of electrons and transferred. transferred. Electrolysis of Aqueous CuCl2 Electrolysis of Aqueous CuCl CuCl2 + H2O → Cu2+(aq) + 2Cl-(aq) electrons electrons Anode (+) Anode 2 Cll → Cl2(g) + 2eC BATTERY BATTERY - Cathode (-) Cathode + + Anode Anode Cu2+ + 2e- → Cu Cu Eo for cell = -1.02 V Note that Cu is more easily Note reduced than either H2O or reduced or Na+ (check redox potentials). Na Cathode Cathode Cl-- Cu2+ Cl Cu2+ H2O H2O Calculate ∆ Go for the reaction, Zn2+(aq) + Ni(s) → Zn(s) + Ni2+(aq) (aq) (aq) Solution: use ∆ Go = -nFE° no. of electrons, n = 2 F = 9.6485 × 104 C 9.6485 need Eocell Zn2+(aq) + 2e- → Zn(s), -0.763 V (aq) Ni(s) → Ni2+(aq) + 2e-, +0.25 V Ni(s) cathode anode Eocell = Ecathode – (Eanode) Eocell = -0.763 – (-0.25 V) = -0.51 V ∆ Go = (-2 × 96485 J/V × -0.51 V) × 1 kJ/1000 J kJ/1000 ≈ 98 kJ ∆Go > 0, reaction is nonspontaneous Reagents are favored E and ∆G o ∆Go = - n F Eo For a product-favored reaction product-favored Reactants → Products ∆Go < 0 and so Eo > 0 Eo is positive For a reactant-favored reaction For reactant-favored Reactants ← Products ∆Go > 0 and so Eo < 0 Eo is negative o E°cell, G° and K!! • for a spontaneous reaction ∆ G° < 0 (negative) E° > 0 (positive) K > 1 (large) E =E 0 cell RT − ln Q nF When Ecell = 0 (no net rxn), reactants and products are at equilibrium……and Q = K o 0.0257 nE cell 0 Ecell = ln K = ln K 72 n 0.0257 Quantitative Aspects of Electrochemistry Quantitative Consider electrolysis of aqueous silver ion. Ag+ (aq) + e- → Ag(s) 1 mol e→ 1 mol Ag If we could measure the moles of e-, we could If know the quantity of Ag formed. know But do we measure moles of e-? charge passing Current = time cu ms o lo b I (am s = p) scns eod Quantitative Aspects of Electrochemistry cu ms o lo b charge passing I (a p ) = ms Current = scns eod time But , how is charge related to moles of electrons? But how = 96,500 C/mol e= 1 Faraday Faraday 96,500 C 1 mol e − or 1 mol e − 96,500 C Michael Faraday 1791-1867 Quantitative Aspects of Electrochemistry I (amps) = coulombs seconds 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? What Solution (a) Calc. charge Charge (C) = current (A) x time (t) Charge = (1.5 amps)(15.0 min)(60 s/min) = 1350 C (1.5 Q ua ntita tive As p e c ts o f Ele c tro c h e m is try coulombs I (amps) = seconds 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is (aq) deposited? Ag+ + e- → Ag(s) Ag Solution (a) Charge = 1350 C (b) Calculate moles of e- used 1 mol e 1350 C • = 0.0140 mol e96, 500 C (c) Calc. quantity of Ag 1 mol Ag 0.0140 mol e - • = 0.0140 mol Ag or 1.51 g Ag 1 mol e - Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e(aq) If a battery delivers 1.50 amp, and there is 454 g of Pb, how long will If the battery last? the Solution Solution a) 454 g Pb = 2.19 mol Pb a) b) Calculate moles of eeach Pb atom is loosing 2e− each 2 mol e 2.19 mol Pb • = 4 .38 mol e1 mol Pb c) Calculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C 4.38 Q u a ntita tive As p e c ts o f Ele c tro c h e m is try The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e(aq) If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the If battery last? battery Solution a) 454 g Pb = 2.19 mol Pb b) Mol of e- = 4.38 mol c) Charge = 423,000 C d) Calculate time Charge (C) Time (s) = I (amps) 423,000 C Time (s) = = 282,000 s About 78 hours 1.50 amp ...
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This note was uploaded on 11/07/2011 for the course CHM 2045 taught by Professor Geiger during the Fall '08 term at University of Central Florida.

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