I have two possible chemical equations for a reaction between $\ce{SnCl2}$ and $\ce{H2O}$. The first one is:

$$\ce{2SnCl2 + 2H2O +O2 -> 2SnO2 + 4HCl}$$

where the reaction draws on oxygen from the air. The second possibility is:

$$\ce{SnCl2 + H2O -> SnO + 2HCl}$$

I know that in the first reaction, tin is oxidized and oxygen is reduced, but I can't figure out if that reaction is favorable, or if it is possible that it's using oxygen from the air. The second equation is the only other possible reaction that I can find, but it doesn't form the product that we want.

  • $\begingroup$ SnCl2 does react with O2 from air, but slowly. $\endgroup$
    – Mithoron
    Jul 1, 2016 at 16:01

2 Answers 2


The actual reaction of tin (II) chloride with water in the presence of air is given here.

$$\ce{6SnCl2 + O2 + 2H2O → 2SnCl4 + 4Sn(OH)Cl}$$

Tin(II) chloride react with oxygen and water to produce tin(IV) chloride and hydroxotin(II) chloride. Interaction with oxygen and air humidity. This reaction takes place slowly.

As Mithoron noted, the reaction takes place very, very slowly and thus get hydrolysed partially to form the basic salt, $\ce{Sn(OH)Cl}$ instead of forming the hydroxide or the oxide.

The basic salt will also form when tin (II) chloride reacts with water in the absence of air, thanks to Gannex.

$$\ce{SnCl2 + H2O ⇄ Sn(OH)Cl + HCl}$$

Tin(II) chloride react with water to produce hydroxotin(II) chloride and hydrogen chloride. This reaction is proceeds by diluting the solution tin(II) chloride. (source)

As noted earlier, the reaction is very, very slow and maintains an equilibrium between the reactants and the products.

However when tin (II) chloride is present in a concentrated, acidified solution, it will form a complex.(source)

$$\ce{SnCl2 + H2O → [Sn(H2O)Cl2]}$$

Tin(II) chloride react with water to produce dichloridoaquatin. Tin(II) chloride - concentrated solution. pH < 7.


Your tin will not spontaneously change oxidation state just from adding water. On dilution, some of the tin (II) chloride may be hydrolyzed, precipitating the basic salt:

$\ce{SnCl2 + H2O -> Sn(OH)Cl + HCl}$

  • $\begingroup$ We did initially mix the SnCl2 with water and the HCl...I don't know if that makes a difference. $\endgroup$
    – Meguna
    Jul 1, 2016 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.