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anorexic ox

anode. cathode. Voltaic (or Galvanic) Cells. In a voltaic (or galvanic ) cell , e – transfer occurs via an external pathway that links the reactants. a battery. e.g.,. electrodes : the two solid metals in a voltaic/galvanic cell. anorexic ox. fat, red cat.

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anorexic ox

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  1. anode cathode Voltaic (or Galvanic) Cells In a voltaic (or galvanic) cell, e– transfer occurs via an external pathway that links the reactants. a battery e.g., electrodes: the two solid metals in a voltaic/galvanic cell anorexic ox fat, red cat -- oxidation occurs here -- reduction occurs here -- written w/a (–) sign -- written w/a (+) sign

  2. e– e– Zn anode Cu cathode NO3– Na+ Zn2+ Cu2+ Cu2+ Zn2+ NO3– NO3– Consider a solution of Zn(NO3)2(aq) and Cu(NO3)2(aq) with electrodes as shown… salt bridge containing electrolyte (e.g., NaNO3) in a porous gel (needed to neutralize both solutions) -- Zn anode dissolves into sol’n -- Cu2+ plates out as Cu on the cathode WHY do the e– go the way they do?

  3. Difference in is important; PE charge Cell EMF > PE of anode’s e– PE of cathode’s e– Thus… anode’s e–s spontaneously flow towards cathode, if given a path. # of charges is NOT. Voltage (or “potential”) difference is also called the electromotive force (emf).

  4. (Look it up!) For a particular cell, (i.e., a particular anode and cathode), the cell’s emf is written Ecell and is called the cell potential. it is + for spontaneous cell reactions -- it depends on materials, conc., and temp. -- -- standard emfs occur at 25oC -- To calculate Ecell, look up tabulated standard reduction potentials for each half-cell… e.g., Ag+(aq) + e–Ag(s) Eored = +0.80 V …and then use the equation: Eocell = Eored,cath – Eored,an

  5. (Eored = –0.76 V) The reference point for reduction potentials is the standard hydrogen electrode (SHE): 2 H+(aq, 1 M) + 2 e–H2(g, 1 atm) Eored = 0 V This is analogous to the arbitrary ref. line when calc. PEg in physics. J Multiples of coefficients don’t affect Eored. V = C e.g., Zn2+(aq, 1 M) + 2 e– Zn(s) 2 Zn2+(aq, 1 M) + 4 e– 2 Zn(s) Because the same chemicals are used for each, AAA-, AA-, C-, and D-batteries all have an emf of 1.5 V.

  6. A 9-volt battery is six AAA batteries wired together in series.

  7. Eo = Cu2+ Cu Cu2+ solution Cr metal Cu metal Cr2+ solution For Cr(s) + Cu2+(aq)  Cr2+(aq) + Cu(s), Eocell is measured to be 1.25 V. Given that Eored for Cr2+ to Cr is –0.91 V, find Eored for the reduction of Cu2+ to Cu. Eocell = Eored,cath – Eored,an Eored, cath – 1.25 V = –0.91 V –0.91 V (Cu2+ Cu) +0.34 V

  8. A galvanic cell has half-rxns: (a) Al3+(aq) + 3 e–Al(s) Eored = –1.66 V (b) Ba2+(aq) + 2 e– Ba(s) Eored = –2.90 V Calculate Eocell and write the balanced equation. For a galvanic cell, Eo must be > 0. Thus, (a) represents the cathode and (b) represents the anode. Eocell = Eored,cath – Eored,an Eocell = –1.66 V – –2.90 V = 1.24 V 2 3 2 3 Al3+(aq) + Ba(s) Al(s) + Ba2+(aq)

  9. For a half-reaction, the more (+) the Eored value, the greater the tendency for that reaction to “go” in that direction (i.e., reduction). Strongest oxidizer is… F2 F2(g) + 2 e– 2 F–(aq) Eored = +2.87 V Other strong oxidizers are… the other halogens (Cl2, Br2, I2) and oxyanions in which the central atom has a large ___ charge. (+) e.g., MnO4–, Cr2O72–, ClO3–, etc. Strong oxidizers LOVE their own e–s (and they want everybody else’s, too).

  10. (–) sign indicates poor tendency to “go” in this direction, but large magnitude (i.e., 3.05 V) shows strong tendency to “go” in other direction (i.e., oxidation). Lithium batteries take advantage of lithium’s strong tendency to BE oxidized (i.e., to REDUCE other stuff.) Poorest oxidizer is… Li. Li+(aq) + e– Li(s) Eored = –3.05 V Poor oxidizers HATE their own e–s (and have no interest in accepting anyone else’s e– s, either.)

  11. (+) V 0 V (–) V In comparing the reduction potentials of two half-reactions, consider the scale shown. The “higher-up” reaction is the reduction half-cell; the “lower-down” reaction is the oxidation half-cell. A– Red. A– Ox. C– Red. B– Red. B– Ox. C– Ox.

  12. standard conditions nonstandard conditions “e– haters” Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H2 Sb Bi Cu Hg Ag Pt Au Spontaneity of Redox Reactions Eocell = Eored,cath – Eored,an (same equation as before) Activity Series for Metals If Eo (or E, or emf) is +…spontaneous. –…nonspontaneous. -- The Activity Series is based on standard reduction potentials. “e– lovers”

  13. ( ) C 96,500 mol e– J V . mol e– Michael Faraday Josiah Gibbs (1791–1867) (1839–1903) Relationship between E and DG… DG = –nFE DGo = –nFEo In standard states… n = # of mol of transferred e– F = Faraday’s constant =

  14. J V . mol e– ( ) = –5 mol e– (0.74 V) 96,500 For… 5 Fe2+ + MnO4– + 8 H+  5 Fe3+ + Mn2+ + 4 H2O (b) Find DGo. (a) What is n? (a) n = 5 Fe3+ + e– Fe2+Eored = 0.77 V (b) MnO4– + 8 H+ + 5 e– Mn2+ + 4 H2O Eored = 1.51 V Eocell = Eored,cath – Eored,an Eocell = 1.51 V – 0.77 V = 0.74 V DGo = –nFEo (SPONTANEOUS) = –357 kJ

  15. Walther Nernst (1864–1941) Effect of Concentration on Cell EMF Cell emf drops gradually due to changing concentrations of reactants and products. When emf = 0 V, cell is “dead.” Nernst equation: At 25oC (298 K):

  16. nonspont. spont. Fe(s) + Cd2+(aq)  Cd(s) + Fe2+(aq) Find emf at 25oC when [Cd2+] = 0.030 M and [Fe2+] = 2.0 M. Eo = +0.04 V Fe2+(aq) + 2 e– Fe(s) Eored = –0.44 V Cd2+(aq) + 2 e– Cd(s) Eored = –0.40 V = –0.01 V If [Cd2+] = 2.0 M and [Fe2+] = 0.030 M… = 0.09 V

  17. 0 K 0 0 K At equilibrium, DG = ___, E = ___ and Q = ___. The Nernst equation can be rearranged to give the relationship between K and Eo. – = log K = At 25oC (298 K): –DGo log K = 16.912 n Eo = 5706

  18. galvanized electrical conduit Corrosion -- spontaneous redox reactions in which a metal reacts with some substance in its environment to form an unwanted compound -- For some metals (e.g., Al and Mg)… a protective oxide coating (Al2O3, MgO) prevents further corrosion of the underlying substrate. -- Galvanized iron is coated with a protective layer of ____. zinc

  19. Mg is used in the cathodic protection of underground Fe pipe. The Mg has to be replaced every so often. Fe e– Mg Offshore oil rigs are often cathodically protected. Mg2+ -- cathodic protection: protecting a metal by making it the cathode in an electrochemical cell sacrificial anode  oxidized metal is called the ______________ EX.

  20. Ag-plated brass trumpet Cr-plated Fe pipe Cu-plated Pb shot Ni-plated steel rotor using an outside source of electrical energy to cause nonspontaneous redox reactions to “go” Electrolysis: -- Electrolysis occurs in electrolytic cells, which consist of two electrodes in a molten salt or a solution. • reduction at cathode; oxidation at anode

  21. Consider plating chromium onto an iron pipe: Fe2+/Fe  Eored = –0.44 V Cr3+/Cr  Eored = –0.74 V For Cr to plate out on the Fe pipe, the equation is: 2 Cr3+ + 3 Fe  3 Fe2+ + 2 Cr and Eo (for the galvanic cell) would be: Cathode – Anode = “Cr3+/Cr”– “Fe2+/Fe” = –0.74 V – –0.44 V = –0.30 V The rxn is nonspontaneous in the forward direction. A galvanic cell would run opposite the way we want. We need to put an external “oomph” into the rxn to make it go…which is the definition of an electrolytic cell.

  22. Ni anode Vext Fe cathode to be plated with Ni (Faraday’s constant) Ni2+(aq) electroplating: using electrolysis to deposit a thin layer of one metal onto another For electrolysis, current in amperes (1 A = 1 C/s) is passed through the liquid. One mole of transferred e– carries with it 96,500 C of charge. 1 e– = 1.60 x 10–19 C ~ 6.02 x 1023 e– = 96,500 C

  23. Charles Martin Hall (1863–1914) Paul Heroult (1863–1914) Hall-Heroult process for purifying Al from Al2O3 and Na3AlF6 (1886) For how long must a 50.0 A current be passed through molten BaBr2 in order to produce 500. g of barium? **To solve these problems, use unit cancellation. 500. g Ba = 14,100 s = 3.90 h

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