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$$\ce{MnS + 2HCl -> MnCl2 + H2S}$$

If oxidation and reduction is taken to be the loss of hydrogen and the gain of hydrogen respectively, then I can see how the aforementioned reaction is redox — $\ce{S}$ gains hydrogen and is therefore reduced while $\ce{Cl}$ loses it, and hence is oxidized. But how can this be explained in terms of electron transfer? The oxidation state of every atom seems to be unchanged, which should mean that no reduction or oxidation has taken place:

$$\ce{\overset{\color{blue}{(2+)}}{Mn}\overset{\color{red}{(2-)}}{S} + \overset{\color{orange}{(+)}}{2H}\overset{(-)}{Cl} -> \overset{\color{blue}{(2+)}}{Mn}\overset{(-)}{Cl_2} + \overset{\color{orange}{(+)}}{H_2}\overset{\color{red}{(2-)}}{S}}$$

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    $\begingroup$ It isn't. End of story. $\endgroup$ Jan 23, 2022 at 10:46
  • $\begingroup$ The reason for my asking here was because a reputable online learning platform claimed it was a redox reaction when it didn't quite fully seem to be one $\endgroup$
    – Shane
    Jan 23, 2022 at 11:17
  • $\begingroup$ Please let us know the source. It looks a little less reputable now, maybe? $\endgroup$ Jan 23, 2022 at 11:20
  • $\begingroup$ It's from Khan Academy, but I won't be able to post a link to the quiz containing the problem because the questions don't seem to be static — they change for each viewer. $\endgroup$
    – Shane
    Jan 23, 2022 at 11:26
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    $\begingroup$ Well, at least now you know that Khan Academy is not a "reputable online learning platform". $\endgroup$
    – andselisk
    Jan 23, 2022 at 11:48

2 Answers 2

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If oxidation and reduction is taken to be the loss of hydrogen and the gain of hydrogen respectively...

The loss of hydrogen is oxidation in cases of homolytic breaking the bond with hydrogen when the electron goes away with the proton. Heterolytic losing hydrogen ion(=hydrated proton), when the electron stays, is not oxidation, but acid dissociation.

The former case leads to change in oxidation number/state, the latter does not.

E.g. $\ce{CH4 + O2 -> C + 2H2O}$ is oxidation, as carbon is oxidized from the oxidation state -IV to 0.

Exchanging an acidic hydrogen and a ion of an electropositive metal between some acid and salt (i.e. the metal keeps its oxidation state) in cases like above:

$$\ce{MX_n + n HA -> n HX + MA_n}$$

does not change oxidation states of elements in involved compounds.

Therefore conversion $$\ce{MnS + 2HCl -> MnCl2 + H2S}$$

is not a redox reaction, as Mn keeps the oxidation state +II, sulphur -II hydrogen +I and chlorine -I.

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Oxidation state involves a counting of valence electrons, so you have to look at whether and how electrons are transferred between atoms, not how the atoms themselves are transferred. Follow the definitions given by the IUPAC Gold Book for all species:

Oxidation:

  1. The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').

  2. An increase in the oxidation number of any atom within any substrate

  3. Gain of oxygen and/or loss of hydrogen of an organic substrate.

(Hydrogen and ixygen are understood to be neutral species, not hydrogen or oxide ions, in the organic chemistry definition.)

Reduction:

The complete transfer of one or more electrons to a molecular entity (also called 'electronation'), and, more generally, the reverse of the processes described under oxidation (2) and (3).

It is true that the transfer of neutral hydrogen atoms (implied in definition 3 for oxidation given above) provides a way to transfer electrons because each hydrogen atom picks up an electron, or perhaps leaves one behind if it is initially bound to an electropositive element, as it gets transferred. But here that mechanism does not apply because the hydrogen is transferred as ions, and the ions do not pick up or drop off any electrons as they move from $\ce{HCl}$ (or its dissociated solvated hydrogen ions) into $\ce{H2S}$.

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