How do you write an ionic equation for the following reaction? A student repeats the experiment with copper instead of magnesium. State whether a reaction would still occur. Why or why not?

A student reacts magnesium with an aqueous solution of iron chloride to produce magnesium chloride and iron. #Mg(s) + FeCL_2(aq) -> MgCl_2(aq) + Fe(s)#

1 Answer
Aug 8, 2018

Answer:

  • #"Mg"(s) + "Fe"^(2+)(aq) to "Mg"^(2+)(aq) + "Fe"(s)#
  • No reaction if #"Cu"(s)# is used in place of #"Mg"(s)# .

Explanation:

Steps for rewriting this chemical equation in its ionic form:

Rewrite all strong electrolytes- soluble salts and strong acids/bases- in their fully disassociated ionic form. For example, #"FeCl"_2(aq)# is soluble in water and ionizes through the following process:

#"FeCl"_2(aq) to "Fe"^(2+)(aq) + 2 color(white)(l) "Cl"^(-) (aq)#

Each "#"FeCl"_2(aq)#" found in the chemical equation shall thus be represented as "#"Fe"^(2+)(aq) + 2 color(white)(l) "Cl"^(-) (aq)#".
Similarly, each #"MgCl"_2(aq)# found in the equation would be represented as "#"Mg"^(2+)(aq) + 2 color(white)(l) "Cl"^(-)(aq)#".

#"Mg"(s) + "Fe"^(2+)(aq) + 2 color(white)(l) "Cl"^(-) (aq) to "Mg"^(2+)(aq) + 2 color(white)(l) "Cl"^(-) (aq)+ "Fe"(s)#

Eliminate spectator ions: spectator ions do not participate in the overall reaction and present on both the reactant side and the product side. They are to be canceled out in the net ionic equation the way one would cancel terms in an addition/subtraction algebra operation.

#"Mg"(s) + "Fe"^(2+)(aq) + color(red)(cancel(color(black)(2 color(white)(l) "Cl"^(-)(aq)))) to "Mg"^(2+)(aq) + color(red)(cancel(color(black)(2 color(white)(l) "Cl"^(-)(aq)))) + "Fe"(s)#

#"Mg"(s) + "Fe"^(2+)(aq) to "Mg"^(2+)(aq) + "Fe"(s)#

Recall the metal reactivity series:

#"Mg" >"Fe" > "Cu"#

Metals the series are capable of displacing less active ones that ensue in the series but not in the opposite way. As a result,

  • #"Mg"(s)# would be capable of displacing #"Fe"^(2+)(aq)# from its solution;
  • #"Fe"(s)# would be capable of displacing #"Cu"^(2+)(aq)# from its solution
  • #"Cu"(s)# is more stable than #"Fe"# and is thus unable to displace #"Fe"^(2+)# from its solution the way #"Mg"(s)# does. No reaction would take place when #"Cu"(s)# is added to #"Fe"^(2+)#.