# Why can't electrons be transferred using hydrogen ions?

I have a homework question that asks:

Recognizing how oxidation and reduction occurs is essential for analyzing and understanding biochemical reactions. Which of the following methods would NOT be employed in transferring electrons in biochemical reactions?

a) As hydrogen ions (correct)

b) Through direct combination with oxygen

c) Directly as electrons

d)As hydride ions

My problem is that it doesn't say why hydrogen ions cannot transfer electrons, so I think I'm not understanding what it means by hydrogen ions. What if you have a hydrogen ion with 2 or 3 electrons? Why can't that be used to transfer electrons?

• While H$^{-}$ exists, H$^{2-}$ would autoionize very very quickly... And, normally I would read 'hydrogen ion' as meaning the positive ion unless it was specifically called out as a negative ion. – Jon Custer Mar 3 '16 at 21:02
• H- is hydride, your D option... De/protonation isn't redox reaction. – Mithoron Mar 3 '16 at 22:03

Firstly, $\ce{H+}$ would be called a hydrogen ion, and $\ce{H-}$ would be called a hydride ion. It would take a lot of energy to get three electrons onto a hydrogen atom, and considering that we are talking about a biological system, this just won't happen.
$\ce{H+}$ would not be employed in the transfer of electrons because it is a lone proton, and has no electrons to give. Whatever electron the hydrogen atom originally had was left behind, and the proton now exists on its own. This, I believe, should explain your homework question.
To answer your question, $\ce{H-}$ ions are common reducing agents in non-biological organic reactions, in the forms of substances like $\ce{NaBH4}$ and $\ce{LiAlH4}$, but these are extremely unstable in aqueous solution.
Instead, cells uses NAD+ to facilitate the transfer of electrons from biomolecules, like those used in cellular respiration, in the form of $\ce{H-}$. During the transfer, one electron from the $\ce{H-}$ is transferred to a nitrogen, and the other electron along with the proton forms a bond to a carbon directly across the ring from the nitrogen, converting NAD+ to NADH, a reaction that is readily reversible.