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

Jun 10, 2017

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

#### Explanation:

$\text{Oxidation reaction}$

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

$\text{Reduction reaction}$

$A {l}^{3 +} + 3 {e}^{-} \rightarrow A l$ $\left(i i\right)$

And we cross multiply the individual redox equations so that electrons are absent from the final redox equation, i.e. $3 \times \left(i\right) + 2 \times \left(i i\right) = \ldots \ldots \ldots . .$

$3 M g + 2 A {l}^{3 +} + \cancel{6 {e}^{-}} \rightarrow 2 A l + 3 M {g}^{2 +} + \cancel{6 {e}^{-}}$

And finally.............

$3 M g + 2 A {l}^{3 +} \rightarrow 2 A l + 3 M {g}^{2 +}$

Jun 10, 2017

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

#### Explanation:

The Aluminum is reduced as it charge goes from + 3 to zero, as the Al gains electrons. That means the charge or oxidation number of aluminum goes down. The Aluminum acts as an oxidizing agent in this reaction.

The Magnesium is oxidized as it charge goes from zero to +2 . The magnesium has lost electrons. This causes the charge or oxidation number magnesium to become more positive. The Magnesium acts as a reducing agent in this reaction.

Electrons are transferred from the Magnesium solid to the Aluminum ion.