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Diluted hydrochloric acid oxidizes tin to $\ce{Sn^2+}$ ion:

$$\ce{Sn + HCl(aq) -> SnCl2(aq) + H2(g)}$$

Why is oxidation not complete? $\ce{Sn^4+}$ has closed shell and is more stable.

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    $\begingroup$ Neither Fe being dissolved in HCl form Fe^3+. Additionally, SnCl4 is a liquid with covalent bonds, there would not be Sn^4+. There is no shortcut for studying chemistry of elements. $\endgroup$
    – Poutnik
    Commented Sep 29, 2022 at 6:36
  • $\begingroup$ @ Poutnik. Anhydrous $\ce{SnCl4}$ is a liquid, and it is decomposed in water. But if produced by a redox reaction in aqueous solution, $\ce{Sn^{4+}}$ is hydrated, and its evaporation may produce hydrated compounds like $\ce{SnCl4·4H2O}$. $\endgroup$
    – Maurice
    Commented Sep 29, 2022 at 6:52
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    $\begingroup$ @Maurice There is no real M^4+(aq) ion in water, the field is too strong for that. Neither the famour Ce^4+ ( cerimetry ) is true Ce^4+(aq), being a dimer with bridges, not handy if oxo- or hydroxo- ones. // BTW, I have not said SnCl4(aq) remains in the same form as SnCl4(l). $\endgroup$
    – Poutnik
    Commented Sep 29, 2022 at 7:07

2 Answers 2

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Look at the standard electrode potentials here

The std electrode potential for Sn/Sn(II) is -0.14, the std electrode potential for 2H+/H2 is 0 (by definition) so the reaction of Sn + 2H+ to Sn(II) + H2 is energetically favourable.

The std electrode potential of Sn(II)/Sn(IV) is +0.15 so H+ going to H2 is not a strong enough oxidant to oxidise Sn(II) to Sn(IV).

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The reaction $\ce{Sn + HCl}$ produces $\ce{SnCl2}$ and $\ce{H2}$. If by chance or by accident some $\ce{SnCl4}$ or $\ce{Sn^{4+}}$ is present in solution, it will be reduced by the reaction : $\ce{Sn^{4+} + H2 -> Sn^{2+} + 2 H+}$.

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  • $\begingroup$ With such small differences in potentials there possibly are measurable amounts of oxidation to Sn(IV) but a see-saw reaction? $\endgroup$
    – jimchmst
    Commented Sep 30, 2022 at 17:12
  • $\begingroup$ If both concentrations of $\ce{Sn^{2+}}$ and $\ce{H+}$ are $1$ M, Nernst law states that the equilibrium concentration of $\ce{Sn^{4+}}$ is about $0.003$ M. It is small, but it may be considered as measurable. $\endgroup$
    – Maurice
    Commented Sep 30, 2022 at 19:43

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