How do we use the method of half-equations, and redox formalism to represent the oxidation of magnesium metal by hydrochloric acid?

Jul 12, 2017

The $\text{oxidation state}$ of magnesium metal is a big fat $\text{ZERO}$........ And the metal oxidation state in $M g C {l}_{2}$ is $+ I I$.

Explanation:

You have performed a redox reaction, the which we could formally separate into oxidation/reduction half equations.......

$M g \left(s\right) \rightarrow M {g}^{2 +} + 2 {e}^{-}$ $\left(i\right)$, i.e. the electron rich metal formally LOSES 2 valence electrons

And a reduction half-equation.......

$H C l \left(a q\right) + {e}^{-} \rightarrow \frac{1}{2} {H}_{2} \left(g\right) \uparrow + C {l}^{-}$ $\left(i i\right)$

A oxidizing protium ion formally gains one electron........

We add $\left(i\right)$ and $\left(i i\right)$ such that electrons DO NOT APPEAR in the final redox equation, i.e. $\left(i\right) + 2 \times \left(i i\right)$ gives.........

$M g \left(s\right) + 2 H C l \left(a q\right) \rightarrow M g C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right) \uparrow$

........alternatively.......

$M g \left(s\right) + 2 H C l \left(a q\right) \rightarrow M {g}^{2 +} + 2 C {l}^{-} + {H}_{2} \left(g\right) \uparrow$

Charge and mass are CONSERVED, as is absolutely required for ANY chemical reaction.......

Capisce?