Sacrificial metals and displacement of metals

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Using magnesium as a sacrificial metal effectively prevents iron from rusting by transferring electrons to iron, reducing Fe2O3 to Fe while magnesium itself is oxidized to Mg2+. The oxidation of magnesium occurs through the loss of electrons and the addition of oxygen, forming magnesium oxide (MgO) in the process. The method does not require significant heat energy, which is why a simple wire connection suffices for the reaction. Extracting iron from Fe2O3 using this method is inefficient due to impurities, making the blast furnace method preferable. The discussion also raises a question about the displacement reaction involving zinc and copper sulfate, which suggests further exploration of metal displacement principles.
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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!
 
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sgstudent said:
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 how 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!

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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