Sunday, July 9, 2017

1.59C: Write Ionic Half - Equations Representing the Reactions at the Electrodes during Electrolysis and Understand Why These Reactions are Classified as Oxidation or Reduction



ELECTROLYSIS: Decomposition of an ionic substance by passing an electric current through it
s
REACTION AT ELECTRODES:
s
SOLUTION
PRODUCT AT POSITIVE ELECTRODE
PRODUCT AT NEGATIVE ELECTRODE
LEAD (II) BROMIDE ( PbBr2 )

Bromine   -   Br2
2Br -     -     2e-     →     Br2
Lead   -   Pb
Pb2+     +     2e-     →     Pb
SODIUM CHLORIDE ( NaCl )

Chlorine - Cl
2Cl-  -  2e-  →  Cl2
Hydrogen - H2
2H2O  +  2e-  →  H2  +  2OH-
DILUTE SULFURIC ACID ( H2SO4 )

Oxygen - O2
2H2O  -  4e-  →  O2  +  4H+
Hydrogen - H2
2H+  +  2e-  →  H2
COPPER (II) SULFATE ( CuSO4 )

Oxygen - O2
2H2O  -  4e-  →  O2  +  4H+
Copper - Cu
Cu2+  +  2e-  →  Cu

3 comments:

  1. LEAD (II) BROMIDE ( PbBr2 ) the PRODUCT AT NEGATIVE ELECTRODE can't be Lead because it in a solution and Hydrogen is less reactive then Lead. So the product is Hydrogen.

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    Replies
    1. Lead (II) Bromide in this case is molten, check the previous syllabus statement. It's the only one of the few not in an aqueous solution. If it was you'd be correct, though.

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    2. yeah thats right

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