Question

Given the reaction below, which is the reduced substance?

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

Answer 1

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^(-)`