Question

iron nail and copper (II) sulfate the compound contains Fe2+

molecular equation

ionic equation

net ionic equation

spectator ions

Answer 1

You could represent this single replacement reaction as....
Fe(s) + CuSO4(aq) --> FeSO4(aq) + Cu(s)
In a single replacement reaction an element reacts with a compound to form a compound and an element.

In reality, you have the following ....
Fe(s) + Cu2+ + SO4^2- --> Fe2+ + SO4^2- + Cu(s)
The sulfate ion really does nothing in this reaction. It is what is called a "spectator ion."

The net ionic equation is ....
Fe(s) + Cu2+ --> Fe2+ + Cu(s)
This represents the actual reaction between iron metal and copper(II) ions. Iron metal is capable of reducing Cu2+ ions to copper metal, while being oxidized to Fe2+. Note the positions of iron and copper in the activity series. Iron is higher than copper, which allows us to predict that Fe metal will reduce Cu2+ ions. It also tells us that Cu metal will do nothing with Fe2+ ions.

========== Follow up ============

This equation fails to have the correct symbol for oxygen, O.
CuSo4 + Fe --> FeSo4 + Cu

BluVulpix goes on to say something about "sulfuric acid." There is no sulfuric acid, H2SO4, in this reaction. Sulfate ion is not sulfuric acid, nor does it turn into sulfuric acid.

The iron didn't separate the copper and sulfate. That happened when solid copper(II) sulfate dissolved in water, and had nothing to do with iron. Instead, it had everything to do with water since CuSO4 is soluble in water and dissolves to form copper(II) ions and sulfate ions.

The only reaction is between iron metal and Cu2+ ions.