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# Ch21_4 - CELL POTENTIAL E CELL POTENTIAL E 1 2 3 CELL...

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Unformatted text preview: CELL POTENTIAL, E CELL POTENTIAL, E 1 2 3 CELL POTENTIAL, E CELL POTENTIAL, E Zn and Zn2+, anode Cu and Cu2+, cathode Calculating Cell Voltage Calculating Cell Voltage • Balanced half-reactions can be added together to get overall, balanced equation. Zn(s) ---> Zn 2+((aq) + 2eZn(s) ---> Zn 2+ aq) + 2eaq) Zn(s) Cu2+((aq) + 2e- ---> Cu(s) aq) Cu2+ aq) + 2e- ---> Cu(s) --------------------------------------------------------------------------------------Cu2+((aq) + Zn(s) ---> Zn 2+((aq) + Cu(s) aq) aq) Cu2+ aq) + Zn(s) ---> Zn 2+ aq) + Cu(s) Zn(s) If we know Eo for each half-reaction, we for could get Eo for net reaction. for • For Zn/Cu cell, voltage is 1.10 V at 25 ˚C and Zn/Cu when [Zn2+] and [Cu2+] = 1.0 M. • This is the STANDARD CELL STANDARD • Electrons are “driven” from anode to cathode by an electromotive force or emf. emf POTENTIAL, Eo • For Zn/Cu cell, this is indicated by a voltage Zn/Cu of 1.10 V at 25 ˚C and when [Zn 2+] and [Cu2+] = 1.0 M. • —a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C. 4 5 6 Volts Zn + Salt Bridge H2 Zn2+ Zn Zn2+ + 2eOXIDATION ANODE H+ 2 H+ + 2eH2 REDUCTION CATHODE CELL POTENTIALS, Eo CELL POTENTIALS, Eo Can’t measure 1/2 reaction E o directly. Therefore, measure it relative to a STANDARD HYDROGEN CELL, SHE. Zn/Zn2+ half-cell hooked to a SHE. Zn/Zn2+ half-cell hooked to a SHE. Eo ffor the cell = +0.76 V Eo or the cell = +0.76 V Negative electrode Zn Volts + Salt Bridge H2 Positive electrode 2 H+((aq, 1 M) + 2e- <----> H 22(g, 1 atm) aq, 2 H+ aq, 1 M) + 2e- <----> H (g, 1 atm) atm) Eo = 0.0 V 0.0 Supplier of electrons Zn2+ H+ Acceptor of electrons Overall reaction is reduction of H + by Zn metal. by Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g) Zn(s) aq) Eo = +0.76 V +0.76 Zn --> 2+ + 2e2eZn Zn + 2eOxidation OXIDATION Anode ANODE Zn2+ 2 + 2e- --> H2 2e2 H+ + 2eH Reduction 2 REDUCTION Cathode CATHODE H+ Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is +0.76 V for aq) Zn is a (better) (poorer) reducing agent than H 2. (better) reducing Page 1 Cu/Cu2+ and H2/H+ Cell Cu/Cu2+ and H2/H+ Cell and Cell Eo = +0.34 V +0.34 Volts 7 Cu/Cu2+ and H2/H+ Cell Cu/Cu2+ and H2/H+ Cell and Cell Volts Cu + Salt Bridge H2 8 Zn/Cu Electrochemical Cell Zn/Cu Electrochemical Cell Zn/Cu wire 9 electrons + Cu Cathode, positive, sink for electrons Zn Anode, negative, source of electrons salt bridge Positive Cu + Salt Bridge Negative H2 Cu2+ Cu2+ + 2eCu REDUCTION CATHODE H+ H2 2 H+ + 2eOXIDATION ANODE Acceptor of electrons Cu2+ Zn2+ ions Cu2+ ions Cu2+ H+ Supplier of electrons H+ + 2e- --> Cu 2eCu2+ + 2eCu Reduction REDUCTION Cathode CATHODE H2 --> 2 + 2e--> 2eH2 Oxidation + 2e2 H+ OXIDATION Anode ANODE Overall reaction is reduction of by H2 gas. by gas. Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H +(aq) aq) aq) Measured Eo = +0.34 V +0.34 Therefore, Eo for Cu2+ + 2e- ---> Cu is for 2e- Cu 2+ +0.34 V Zn(s) ---> Zn 2+(aq) + 2eEo = +0.76 V Zn(s) aq) +0.76 Cu2+(aq) + 2e- ---> Cu(s) Eo = +0.34 V aq) +0.34 --------------------------------------------------------------2+(aq) + Zn(s) ---> Zn 2+(aq) + Cu(s) Cu aq) Zn(s) aq) Eo (calc’d) = +1.10 V calc’d) Uses of Eo Values Eo Values • Organize halfreactions by relative ability to act as oxidizing agents • Table 21.1 • Use this to predict cell potentials and direction of redox reactions. wire 10 TABLE OF STANDARD TABLE OF STANDARD REDUCTION POTENTIALS REDUCTION POTENTIALS oxidizing ability of ion Eo (V) Cu H2 Zn +0.34 0.00 -0.76 11 12 Standard Redox Potentials, Eoo Standard Redox Potentials, E oxidizing ability of ion Cu2+ + 2e2 H+ + 2eZn2+ + 2eCu H2 Zn Eo (V) +0.34 0.00 -0.76 electrons Any substance on the right will reduce any substance higher than it on the left. • Zn can reduce H+ and and Cu2+. • H2 can reduce Cu 2+ but can but not Zn2+ • Cu cannot reduce H + or or Zn2+. Zn salt bridge Cu Cu2+ + 2e2 H+ + 2eZn2+ + 2e- Zn2+ ions Cu2+ ions reducing ability of element reducing ability of element Page 2 Using Standard Potentials, Eo Table 21.1 13 Eo for a Voltaic Cell Eo for a Voltaic Cell for Volts Cd Salt Bridge Fe 14 Eo for a Voltaic Cell Eo for a Voltaic Cell for From the table, you see • Fe is a better reducing agent than Cd • Cd2+ is a better is oxidizing agent than 2+ Fe Overall reaction Fe + Cd2+ ---> Cd + Fe2+ ---> Eo = +0.04 V +0.04 15 • Which is the best oxidizing agent: O2, H2O2, or Cl2? _________________ • Which is the best reducing agent: Hg, Al, or Sn? ____________________ • In which direction does the following reaction go? Cu(s) + 2 Ag +(aq) ---> Cu 2+(aq) + 2 Ag(s) Cd2+ Volts Cd Salt Bridge Fe Fe2+ Cd --> Cd2+ + 2e2eor Cd2+ + 2e- --> Cd 2e- Fe --> Fe2+ + 2e2eor Fe2+ + 2e- --> Fe 2e- Cd2+ Fe2+ 16 Calculating Cell Voltage Calculating Cell Voltage • Balanced half-reactions can be added together to get overall, balanced equation. 2 II- ---> II2 + 2e2 ---> 2 + 2e- E˚ and G? YES! Michael Faraday Michael Faraday 1791-1867 1791-1867 Originated the terms anode, cathode, anion, cation, cation, electrode. Discoverer of • electrolysis • magnetic props. of matter • electromagnetic induction • benzene and other organic chemicals Was a popular lecturer. 18 2 H22O + 2e- ---> 2 OH-- + H22 2 H O + 2e- ---> 2 OH + H ------------------------------------------------------------------------------------------------2 II- + 2 H22O --> II2 + 2 OH-- + H22 2 + 2 H O --> 2 + 2 OH + H If we know Eo for each half-reaction, we for could get Eo for net reaction. for Page 3 19 Eo and G o Eo and G o Eo is related to Go, the free energy change for the reaction. Eo and G o Eo and G o Go = - n F E o For a product-favored reaction Reactants ----> Products Go < 0 and so Eo > 0 and Eo is positive is For a reactant-favored reaction Reactants <---- Products Go > 0 and so Eo < 0 and Eo is negative is 20 G o = - n F Eo where F = Faraday constant = 9.6485 x 10 4 J/V•mol J/V•mol and n is the number of moles of Michael Faraday electrons transferred 1791-1867 Page 4 ...
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