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Reduction is gain of hydrogen (Source)

Reduction is gain of electrons (Source)

Is it because a hydrogen has an electron so gaining hydrogen is technically gaining electrons? But that doesn't seem right as oxygens also have electrons and gaining oxygens is oxidation.

What is confusing me is when we usually talk about hydrogens, its a hydrogen ion. And hydrogen ions and electrons are opposite. I realise this situation is not a hydrogen ion but a full hydrogen, but it's still doing my head in.

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  • $\begingroup$ Indeed, we're talking about "full" hydrogen. Gain of hydrogen ions would have pretty much nothing to do with reduction. $\endgroup$ – Ivan Neretin Nov 30 '16 at 9:47
  • $\begingroup$ Perhaps its easier to understand reduction and oxidation by defining what happens as (a) an oxidising agent is an electron acceptor and (b) a reducing agent is an electron donor. $\endgroup$ – porphyrin Nov 30 '16 at 9:48
  • $\begingroup$ @porphyrin Yeah, until we move on to study the Krebs cycle, where you'd literally die of old age before you assign all the oxidation states and figure out where the electrons are going, yet everybody is talking about redox processes, and that quite confidently. $\endgroup$ – Ivan Neretin Nov 30 '16 at 9:51
  • $\begingroup$ @Ivan Neretin ? don't understand what the Krebs cycle has to do with at, its even a pain just to look at :) I've worked on electron transfer extensively and thinking of electron donors and acceptors is by far the easiest way of dealing with these reactions, also you get $\Delta G$ directly as donor-acceptor redox. $\endgroup$ – porphyrin Nov 30 '16 at 10:00
  • $\begingroup$ @porphyrin Krebs cycle is just an example of redox system where thinking of electron donors and acceptors is probably not the easiest way of dealing with it. $\endgroup$ – Ivan Neretin Nov 30 '16 at 10:16
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As a blanket statement the gain of a hydrogen atom cannot be considered a reduction. The IUPAC gold book defines it as follows:

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).

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.

All oxidations meet criteria 1 and 2, and many meet criterion 3, but this is not always easy to demonstrate. Alternatively, an oxidation can be described as a transformation of an organic substrate that can be rationally dissected into steps or primitive changes. The latter consist in removal of one or several electrons from the substrate followed or preceded by gain or loss of water and/or hydrons or hydroxide ions, or by nucleophilic substitution by water or its reverse and/or by an intramolecular molecular rearrangement. This formal definition allows the original idea of oxidation (combination with oxygen), together with its extension to removal of hydrogen, as well as processes closely akin to this type of transformation (and generally regarded in current usage of the term in organic chemistry to be oxidations and to be effected by 'oxidizing agents') to be descriptively related to definition 1. For example the oxidation of methane to chloromethane may be considered as follows:
example for oxidation

As you can see, the reverse of (3) is your first statement with one significant addition: organic substrate.

To understand this one only has to look at the electronegativities. With an electronegativity of 2.20 (Pauling) for hydrogen, it is less electronegative than most other non-metals, or in general, elements that organic substrates are made out of.[1] If you follow the electronegativity scheme of assigning oxidation states, then adding a hydrogen atom (one proton and one electron) results in an decrease of the oxidation number of the element that the hydrogen atom was added to.[2]

On the other hand, adding oxygen would increase the oxidation number of the element (except fluorine) it was added to, since its electronegativity is the second highest, i.e. 3.44 (Pauling).[3]

For example, adding dihydrogen to ethene, the carbons are reduced, while hydrogen is oxidised. $$\ce{\overset{\color{orange}{-2}}{C}_2\overset{+1}{H}_4 + \overset{0}{H}_2 -> \overset{\color{orange}{-3}}{C}_2\overset{+1}{H}_6}$$ Adding dioxygen to ethane to form ethane-1,2-diol, the carbons are oxidised, while oxygen is reduced. $$\ce{\overset{\color{orange}{-3}}{C}_2\overset{+1}{H}_6 + \overset{0}{O}_2 -> \overset{\color{orange}{-1}}{C}_2\overset{+1}{H}_6\overset{-2}{O}_2}$$ You can go as far as looking at the addition of water to ethane to form ethanol. One carbon will be oxidised and one will be reduced. $$\ce{\overset{\color{orange}{-2}}{C}_2\overset{+1}{H}_4 + \overset{+1}{H}_2\overset{-2}{O} -> (\overset{+1}{H}\overset{-2}{O})H2\overset{\color{orange}{-1}}{C}-\overset{\color{orange}{-3}}{C}\overset{+1}{H}_3}$$

Just keep in mind that your first statement is only true when hydrogen is added to to more electronegative elements. The reversal is the case when adding it to a less electronegative element, like a metal.
A second point to keep in mind is that oxidation states are bookkeeping tools only. The world of bonding is not strictly ionic or covalent; it is often somewhere in between. Therefore electrons are almost never completely transferred and the actual charge distribution might be quite different.

References:

  1. Hydrogen: electronegativity. WebElements [http://www.webelements.com/]. [Online]; Dated 30th Nov. 2016; https://www.webelements.com/hydrogen/electronegativity.html
  2. Electronegativity Considerations in Assigning Oxidation States
  3. Oxygen: electronegativity. WebElements [http://www.webelements.com/]. [Online]; Dated 30th Nov. 2016; https://www.webelements.com/oxygen/electronegativity.html
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The hydrogen definition is a simplified form which works well with organic compounds or other electronegative non-metals. It fails for metals. Specifically, if you react a metal with hydrogen, the metal will be oxidised in spite of additional hydrogen being in the compound.

Or in other words, $\ce{NaH}$ is the oxidised form, $\ce{Na}$ the reduced one.

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