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

2. Consider a solution of Zn(NO 3 ) 2 (aq) and Cu(NO 3 ) 2 (aq) with electrodes as shown… e–e– e–e– Cu cathode Zn anode Cu 2+ NO 3 – Zn 2+ NO 3 – salt bridge containing electrolyte (e.g., NaNO 3 ) in a porous gel Zn 2+ Cu 2+ NO 3 – Na + WHY do the e – go the way they do? (needed to neutralize both solutions) -- Zn anode dissolves into sol’n -- Cu 2+ plates out as Cu on the cathode

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

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

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

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

7. E o red, cath For Cr(s) + Cu 2+ (aq)  Cr 2+ (aq) + Cu(s), E o cell is measured to be 1.25 V. Given that E o red for Cr 2+ to Cr is –0.91 V, find E o red for the reduction of Cu 2+ to Cu. E o cell = E o red,cath – E o red,an 1.25 V =–0.91 V (Cu 2+  Cu) E o = Cu 2+  Cu +0.34 V – Cu 2+ solutionCu metalCr metalCr 2+ solution

8. E o cell = A galvanic cell has half-rxns: (a) Al 3+ (aq) + 3 e –  Al(s)E o red = –1.66 V (b) Ba 2+ (aq) + 2 e –  Ba(s)E o red = –2.90 V Calculate E o cell and write the balanced equation. Al 3+ (aq) + Ba(s)  Al(s) + Ba 2+ (aq) E o cell = E o red,cath – E o red,an For a galvanic cell, E o must be > 0. Thus, (a) represents the cathode and (b) represents the anode. –1.66 V – –2.90 V= 1.24 V 2233