# 6. Electrochemistry Candidates should be able to: (a)Describe and explain redox processes in terms of electron transfer and/or of changes in oxidation.

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6. Electrochemistry Candidates should be able to: (a)Describe and explain redox processes in terms of electron transfer and/or of changes in oxidation number (oxidation state) (b)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

How to balance Redox Equations For this example, let's consider a redox reaction between KMnO4 and HI in an acidic solution: MnO 4 - + I - I 2 + Mn 2+ 1.Separate the two half reactions: I - I 2 MnO 4 - Mn 2+

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 H 2 O to balance The O atoms and H + to balance the H atoms. In a basic solution, we would use OH - and H 2 O to balance the O and H. Balance the iodine atoms: 2I - I 2 The Mn in the permanganate reaction is already balanced, so let's balance the oxygen: MnO 4 - Mn 2+ + 4H 2 O Add H + to balance the 4 waters molecules: MnO 4 - + 8H + Mn 2+ + 4H 2 O The two half-reactions are now balanced for atoms: MnO 4 - + 8H + Mn 2+ + 4H 2 O

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 - I 2 + 2e - 5e - + 8H + + MnO 4 - Mn 2+ + 4H 2 O 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 - I 2 + 2e - ) 2(5e - + 8H + + MnO 4 - Mn 2+ + 4H 2 O)

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

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

An Anode (+): 2Cl - (aq) Cl 2 (g) + 2e - At Cathode (-): 2H 2 O(l) + 2e - 2OH - (aq) + H 2 (g) Overall: 2NaCl(aq) + 2H 2 O(l) Cl 2 (g) + H 2 (g) + 2NaOH(aq) Brine Electrolysis [NaCl(aq)]

Extraction of Aluminium At the cathode (-): 4Al 3+ + 12e - 4Al(s) At the anode (+): 6O 2- 3O 2 (g) + 12e -

Purification of Copper impure copper anode pure copper cathode The anode (+) is impure copper. At this electrode, the copper dissolves and copper ions (Cu 2+ ) move into the solution. Copper ions are attracted to the cathode (-) to form copper atoms. Impurities fall to the bottom.

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