Given the balanced ionic equation representing a reaction below. In this reaction, where are electrons transferred from?

Question

Cameron Andrews · Accepted Answer

This is a redox reaction in which electrons are formally transferred....

#"Oxidation reaction"# #Mg rarr Mg^(2+) + 2e^(-)# #(i)# #"Reduction reaction"# #Al^(3+) + 3e^(-) rarr Al# #(ii)# And we cross multiply the individual redox equations so that electrons are absent from the final redox equation, i.e. #3xx(i) + 2xx(ii)=...........# #3Mg+2Al^(3+) + cancel(6e^(-)) rarr 2Al + 3Mg^(2+) + cancel(6e^(-))# And lastly... #3Mg+2Al^(3+) rarr 2Al + 3Mg^(2+)#

Henry Antrim · Answer

The electrons are transferred from the magnesium metal solid to the Aluminum ions

Aluminum acts as an oxidizing agent in this reaction and is reduced as its charge decreases from + 3 to zero as Al gains electrons. This decrease in charge or oxidation number of aluminum results from this process. Magnesium acts as a reducing agent in this reaction and is oxidized as its charge increases from zero to +2. This is because it has lost electrons, making its charge or oxidation number more positive. The aluminum ion receives electrons from the magnesium solid.

Jeremiah Adams · Answer

undefined The copper ions receive electrons from the zinc atoms.