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6. Electrochemistry Candidates should be able to:

6. Electrochemistry Candidates should be able to: Describe and explain redox processes in terms of electron transfer and/or of changes in oxidation number (oxidation state) Explain, including the electrode reactions, the industrial processes of:

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6. Electrochemistry Candidates should be able to:

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  1. 6. Electrochemistry • Candidates should be able to: • Describe and explain redox processes in terms of electron transfer and/or of changes in oxidation number (oxidation state) • Explain, including the electrode reactions, the industrial processes of: • (i) the electrolysis of brine, using a diaphragm cell • (ii) the extraction of aluminium from molten aluminium oxide/cryolite • (iii) the electrolytic purification of copper

  2. How to balance Redox Equations • For this example, let's consider a redox reaction between KMnO4 • and HI in an acidic solution: • MnO4- + I- I2 + Mn2+ • Separate the two half reactions: • I- I2 • MnO4- Mn2+

  3. Balancing Redox Reactions – 2. Balance the Atoms To balance the atoms of each half-reaction, first balance all of the atoms except H and O. For an acidic solution, next add H2O to balance The O atoms and H+ to balance the H atoms. In a basic solution, we would use OH- and H2O to balance the O and H. Balance the iodine atoms: 2I- I2 The Mn in the permanganate reaction is already balanced, so let's balance the oxygen: MnO4- Mn2+ + 4H2O Add H+ to balance the 4 waters molecules: MnO4- + 8H+ Mn2+ + 4H2O The two half-reactions are now balanced for atoms: MnO4- + 8H+ Mn2+ + 4H2O

  4. Balancing Redox Reactions – 3. Balance the Charge Next, balance the charges in each half-reaction so that the reduction half-reaction consumes the same number of electrons as the oxidation half-reaction supplies. This is accomplished by adding electrons to the reactions: 2I- I2 + 2e- 5e- + 8H+ + MnO4- Mn2+ + 4H2O Now multiply the oxidations numbers so that the two half-reactions will have the same number of electrons and can cancel each other out: 5(2I- I2 + 2e-) 2(5e- + 8H+ + MnO4- Mn2+ + 4H2O)

  5. Balancing Redox Reactions – 4. Add the Half-Reactions Now add the two half-reactions: 10I- 5I2 + 10e- 16H+ + 2MnO4- + 10e- 2Mn2+ + 8H2O This yields the following final equation: 10I- + 10e- + 16H+ + 2MnO4- 5I2 + 2Mn2+ + 10e- + 8H2O Get the overall equation by canceling out the electrons and H2O, H+, and OH- that may appear on both sides of the equation: 10I- + 16H+ + 2MnO4- 5I2 + 2Mn2+ + 8H2O

  6. Brine Electrolysis The solution changes to alkali (sodium hydroxide…NaOH) At NEGATIVE electrode…. 2H+ + 2e = H2 (g) Na + At POSITIVE electrode….. 2Cl- = Cl2 (g)+ 2e Cl- H+ Bubbles of chlorine at the +electrode Bubbles of hydrogen at the - electrode OH-

  7. Brine Electrolysis [NaCl(aq)] An Anode (+): 2Cl-(aq) Cl2(g) + 2e- At Cathode (-): 2H2O(l) + 2e- 2OH-(aq) + H2(g) Overall: 2NaCl(aq) + 2H2O(l) Cl2(g) + H2(g) + 2NaOH(aq)

  8. Extraction of Aluminium At the cathode (-): 4Al3+ + 12e- 4Al(s) At the anode (+): 6O2- 3O2(g) + 12e-

  9. Purification of Copper impure copper anode pure copper cathode The anode (+) is impure copper. At this electrode, the copper dissolves and copper ions (Cu2+) move into the solution. Copper ions are attracted to the cathode (-) to form copper atoms. Impurities fall to the bottom.

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