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ELECTROLYSIS

ELECTROLYSIS. Electrolysis: electricity to break apart substance. "electro"=electricity "lysis"=break Electrolytic cell Process of using electricity to force a nonspontaneous reaction .

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ELECTROLYSIS

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  1. ELECTROLYSIS

  2. Electrolysis: electricity to break apartsubstance. • "electro"=electricity "lysis"=break • Electrolytic cell • Process of using electricity to force a nonspontaneousreaction. • A direct current source (battery) is attached to inert electrodes - carbon or platinum. e-still flow from the anode to the cathode, but they are forced.

  3. 1. Molten Electrolytic Cells • Some metals only occur naturally as salts • (NaCl, CuCl2 ) • Or contain oxygen impurities - Cu2O, TiO2 • Forcing the non-spontaneous redox to occur is a way of obtaining pure metals. • Accomplished by melting salts (> 800°C).

  4. NaCl(s) → Na+(l) + Cl–(l) 2 Cl-(l) → Cl2(g) + 2e- 2 Na+(l) + 2e-→ 2 Na(s) E°R = -2.71 V E°O = -1.36 V • > 4.07 V needed to operate this cell. • Called a Down's Cell.

  5. Note: • Inert electrodes are charged to attract the ions. • Cations migrate toward (-) cathode to be reduced. • Anions migrate to (+) anode and are oxidized. • Electrodes are oppositely • charged versus voltaic cells.

  6. 2. Electroplating: Electric current is used to plate a layer of metal onto another surface by reducing itsions. • Object (to be plated) acts as the cathode. • Metal – anode –immersed in a solution containing ions of the metal to be plated. • Notice: Only one ion.

  7. Electrolysis of water: • Water is bothreduced at the cathode and oxidized at the anode. 2 H2O(l) → 2 H2(g)+ O2(g) Oxidation: 2 H2O(l) → 4 H+(aq) + O2(g) + 4e– Reduction: 4 H2O(l) + 4e– → 4 OH–(aq) + 2 H2(g) Net: 6 H2O(l) → 2 H2(g)+ O2(g) + 4 H+(aq) + 4 OH–(aq)

  8. 4. Aqueous Electrolytic Cells: • Three possible reactantsthat will compete for • electrons: the cations, anions and water. • Must predict which substances will be most easily oxidized and which reduced. • Compare oxidation and reduction potentials.

  9. What are the products formed at each electrode during the electrolysis of aqueous KI? K+ I– H2O K+ will not lose more electrons - only I– and H2O can be oxidized. 2 I–(aq) → I2(s) + 2e– E°o = – 0.54V H2O(l) → 2 H+(aq) + ½ O2(g) + 2e– E°o = –1.23V Iodide ion has most positive oxid. potential. Solid iodine is formed at the anode.

  10. I- will not gain more electrons - only K+ and H2O could be reduced. K+(aq) + 1e– → K(s)E°R = – 2.93V 2 H2O(l) + 2e– → 2 OH-(aq) + H2(g)E°R = –0.83V Water has most positive reduction potential. Anode (O): 2 I–(aq) → I2(s) + 2e– Cathode (R): 2 H2O(l) + 2e– → 2 OH-(aq) + H2(g) Note: If solution was evaporated, KOH, would be left.

  11. Corrosion: Process of the oxidation of metals. Metals oxidize easily because they have reduction potentials less than oxygen (except gold). Presence of oxygen - oxidation is spontaneous.

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