Double Displacement: How can Iron (III) Chloride and Sodium Hydroxide react if Fe is below Na in the Reactivity Series?

Question

I just wanted to check if I am thinking of this the right way.

My understanding of using the reactivity chart is to check if the metal in the first reactant (on the left of the + sign) is higher than the metal in the second reactant (on the right of the + sign).

So for Iron (III) Chloride I would see if Fe is higher than Na on the Reactivity Series and if it isn't, there is no reaction.

However, when I set up the equation for Iron (III) Chloride and Sodium Hydroxide, the reactants are a solid and an aqueous solution which means the reaction works. If Fe and Na can't switch places and the reaction can't work, shouldn't it produce two aqueous products?

I hope this makes sense. Help would be very appreciated!

Answer 1

Reactivity series is helpful when determining reactions between metals and water, metals and acids, and single displacement reactions. Reactivity series does not help with double displacement series because reactivity series focuses on the relative activity of the metal which is not the key factor of determining if a double displacement reaction will occur.

For the reaction in question, you have two aqueous substances meaning for both species the substances are soluble and present in solution as their aqueous ions:

Fe³⁺ (aq) + 3 Cl^- (aq) + Na⁺(aq) + OH^- (aq)

If the combination of displacement ions forms an insoluble compound (based on solubility rules and meeting ion concentrations available), the reaction will occur and a precipitate (reddish-brown solid) will form. In this case iron III and hydroxide form an insoluble compound; therefore, the reaction is as follows:

FeCl₃ (aq) + 3 NaOH (aq) --> Fe(OH)₃ (s) + 3 NaCl (aq)

If no insoluble compounds are formed, there is no reaction and all the ions are in solution as ions.