(e) ElectrochemistryTEXTBOOK READING: BLB-10, Chapter 20.3-9 Practice PROBLEMS: (Ch 20) 24, 32, 34, 39, 40, 47, 52, 58, 60, 62, 85, 86, 91, 94 . (i) Spontaneous vs. Nonspontaneous Redox Reactions: In spontaneous redox reactions, electrons flow from a position of high potential energy to low potential energy. Consider the following four chemical situations: 1. Placing a Zn strip in HCl(aq): Bubbles observed 2. Placing a Cu strip in HCl(aq): Nothing happens 3. Placing a Cu strip in Zn2+(aq): Nothing happens 4. Placing a Zn strip in Cu2+(aq): Black solid forms on Zn strip, blue color of the solution gradually fades. What can we conclude from these examples? 1. Zn spontaneously reacts with H+(aq) to give H2(g). 2. Cu does not spontaneously react with H+(aq) to give H2(g). 3. Zn spontaneously reacts with Cu2+(aq) to give Cu(s) (the black solid is CuO, which forms since the finely divided Cu metal is exposed to O2dissolved in water). 4. Cu does not spontaneously react with Zn2+(aq). Therefore, we can write the following chemical reactions: 1. Zn(s) + 2 H+(aq) ⎯→Zn2+(aq) + H2(g); ∆G< 0 2. Cu(s) + 2 H+(aq) ⎯×→Cu2+(aq) + H2(g); ∆G> 0 3. Zn(s) + Cu2+(aq) ⎯→Zn2+(aq) + Cu(s); ∆G< 0 4. Cu(s) + Zn2+(aq) ⎯×→Cu2+(aq) + Zn(s); ∆G> 0 Therefore, Zn(s) has a higher potential energy for electrons than H2(g), and H2(g) has a higher potential energy for electrons than Cu(s). OR, the strength as a reducingagent increases from Cu(s) to H2(g) to Zn(s) (Zn is most easily oxidized; Cu is least easily oxidized). The potential energy of an electron is proportional to its charge (in units of coulombs). The charge of one electron is −1.61 ×10−19coulombs. The charge of one moleof electrons is (−1.61 ×10−19coulomb/electron) ×(6.02 ×1023electrons/mole) = −96,500 coulombs/mole. The Faradayis defined as the number of coulombs in one mole of electrons 1 F= 96,500 coulombs/mole During a spontaneous electrochemical reaction, electrons flow from high potential energy to low potential energy. This change in potential energy is just the free energy change for the reaction, ∆G. Therefore, the electrochemical potential∆Eis: ∆G= −nF∆E
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