# Given the reaction below, which is the reduced substance?

## $M g + C {l}_{2} \to M {g}^{2 +} + 2 C {l}^{-}$

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........

$C {l}_{2} + 2 {e}^{-} \rightarrow 2 C {l}^{-}$ $\left(i\right)$

Zerovalent chlorine gas, $\stackrel{0}{C} {l}_{2}$ is reduced to $\stackrel{- I}{C} {l}^{-}$

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............

$M g \rightarrow M {g}^{2 +} + 2 {e}^{-}$ $\left(i i\right)$

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 $\left(i\right) + \left(i i\right)$, which give us.......

$M g + C {l}_{2} + \cancel{2 {e}^{-}} \rightarrow M {g}^{2 +} + 2 C {l}^{-} + \cancel{2 {e}^{-}}$

i.e. $M g + C {l}_{2} \rightarrow M {g}^{2 +} + 2 C {l}^{-}$