2. Electrochemistry Chapter 17

3. “Leo” the lion says “ger” §LEO - Lose Electrons = Oxidation GER - Gain Electrons = Reduction §What are the half reactions of l Mg(s) + 2H + (aq)  Mg 2+ (aq) + H 2 (g) ? § Mg(s)  Mg 2+ (aq) + 2e - l 2H + (aq) + 2e -  H 2 (g)

4. Galvanic Cell §Which is the oxidation and which is the reduction? Mg(s)  Mg 2+ (aq) + 2e - 2H + (aq) + 2e -  H 2 (g) leOgeR

5. Galvanic Cell §Electrons flow from the anode to the cathodes Mg(s)  Mg 2+ (aq) + 2e - 2H + (aq) + 2e -  H 2 (g) cathodeanode What’s this?

6. Electromotive Force ( E º) §…is the combination (sum) of the potentials of two half reactions. §Reactions are listed on tables (17.1 pg 833) as reductions.  Galvanic cells require E º cell > 0. §One of the tabulated red. potentials will have to be reversed. §Sum of the redox half reactions must be > 0.

7. Electromotive Force (cont)  When you reverse the reaction E º gets the opposite sign.  When you multiply a reaction by a coefficient E º is NOT changed.

8. Practice §Which of these is galvanic? §. Mg(s) + 2H + (aq)  Mg 2+ (aq) + H 2 (g) Cu 2+ (aq) + 2Ag(s)  2Ag + (aq) + Cu(s) 2Zn 2+ (aq) + 4OH - (aq)  2 H 2 O(l) + O 2 (g) + 2Zn(s) §+2.37V, yes §-0.46V, no §-1.16V,no The more positive the E value, the more likely the species is to be reduced. (It’s a stronger oxidizing agent).