# Why does Mg react vigorously in NaCl solution and less so in water?

When you put $\ce{Mg}$ into water a few $\ce{H2}$ bubbles appear. But when you put $\ce{Mg}$ into a $\ce{NaCl}$ solution there is a vigorous release of $\ce{H2}$, why is this and what reactions are taking place?

• My first thought is that the salt is providing nucleation sites for $\ce{H_{2}}$, but I wasn't able to find definitive support with modest searching. I could be totally wrong. Do you have a reference or two that you can share, or is it your observation you are sharing? – Todd Minehardt Aug 6 '15 at 16:33
• It might be wrong but my hypothesis is different from the actual answer. When magnesium hydroxide reacts it forms a protective layer above pure magnesium, as in passived aluminum. When NaCl is added it increases hydroxide solubility (because of activities and so on); as a consequence, makes the reaction faster... – user43021 Oct 13 '18 at 0:57

\begin{align} \ce{Mg (s) &-> Mg^2+ + 2 e-} & E^\circ&=\pu{2.36 V}\\ \ce{2 e- + 2H2O (l) &-> H2 (g) + 2OH-} & E^\circ&=\pu{-0.828 V} \times 2 \end{align}
The combined reaction would be spontaneous ($$E > 0$$), but $$\ce{Mg(OH)2 (s)}$$ is highly insoluble, \begin{align} \ce{Mg(OH)2 (s) &<=> Mg^2+ (aq) + 2OH- (aq)}, & K_\mathrm{sp}=7.1 \times 10^{-12}, \end{align} leaving the solution insufficient ions to conduct the electrons necessary to fast complete the reaction (despite of the escaping hydrogen gas). $$\ce{Mg (s) + 2H2O (l) -> H2 (g) + Mg(OH)2 (s)}$$
In contrast, calcium solid, having comparable standard reduction potential, will undergo a similar redox reaction in water and needs no help for additional ions because $$\ce{Ca(OH)2 (s)}$$ is about a million times more soluble ($$K_\mathrm{sp}=6.5 \times 10^{-6}$$) than magnesium hydroxide, providing sufficient free ions in the solution to drive the vigorous release of hydrogen gas. (There are a couple of YouTube video demos.)