Given the reaction below, which is the reduced substance?

#Mg + Cl_2 -> Mg^(2+) + 2Cl^-#

1 Answer
anor277
Jun 10, 2017

The chlorine molecule is REDUCED........

Explanation:

And we may represent this by the reduction reaction in which electrons appear as stoichiometric reagents........

#Cl_2+2e^(-) rarr2Cl^(-)# #(i)#

Zerovalent chlorine gas, #stackrel(0)Cl_2# is reduced to #stackrel(-I)Cl^(-)#

And of course, for every reduction, for every electron GAIN, there must be a corresponding oxidation, an electron loss. And metals, especially alkali and alkaline earth metals, are typically oxidized............

#MgrarrMg^(2+) + 2e^(-)# #(ii)#

And for the final reaction we SUM the individual oxidation and reduction reactions in such a way that electrons DO NOT appear in the final reaction. Here it is simply #(i) + (ii)#, which give us.......

#Mg+Cl_2+cancel(2e^(-))rarrMg^(2+) + 2Cl^(-) + cancel(2e^(-))#

i.e. #Mg+Cl_2rarrMg^(2+) + 2Cl^(-)#

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