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Sacrificial metals and displacement of metals

  1. Feb 21, 2012 #1
    When I want to prevent iron from rusting, I use a more reactive metal such as magnesium and connect it to the iron. This protects iron from being rusted as the magnesium will transfer electrons into the iron when it rusts, thus reducing Fe2O3 into Fe metal. However, Mg will be oxidised to Mg2+ but how does it get oxidised? It's electrons are transferred to the Fe2O3 already, so how does a layer of MgO form around it? Since ionic compounds form when the electrons transfer from the metal to the non metal (layman terms) so hoe does the magnesium receive its oxygen ion?

    Also, I thought a lot of heat is required for Mg to displace Fe2O3? So why just a wire is needed for it to "cure" the metal. And why can't we extract iron ore Fe2O3 from this method too?

    Lastly, if I place a wire onto a zinc block and dip the other end of the wire into a solution of copper sulfate, will displacement of the metal occur to form copper metal in the solution while the since forms a block of Zn2+ ions?

    Thanks for the help!
  2. jcsd
  3. Feb 22, 2012 #2
    I guess that the water present in Fe2O3 , like yes : Fe2O3.xH2O ionizes

    Fe2O3 <--------------> 2Fe3+ + 3O2-
    Now Mg is oxidized because of loss of electrons as well as addition of oxygen.

    3Mg ----------> 3Mg2+ +6e-

    2Fe3+ + 6e- ---------------> 2Fe

    Hence we see that Mg reduces Fe2+ to Fe with itself being oxidized without the requirement of heat energy !

    Now 3Mg + 3O2- ---------> 3MgO

    Overall reaction : Fe2O3 + Mg ------> 3MgO + 2Fe

    And we can not extract Fe from haematite ore by this method because its not efficient as lot of impurities remains. We prefer blast furnace method.

    And what wire are you talking about in last paragraph ?

    Note : I am not sure if I am cent percent correct.
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