[cliverlong]

Hi.

I can find

Lithium Aluminium Hydride  LiAlH₄
Sodium Hydride NaH
SodiumBoroHydride NaBH₄

as "good" reducing agents.

This must be because their E values (standard electrode potential) are very negative in comparison to most other chemicals

However, I can't find actual E values, nor half-cell equations, for these chemicals to show this. I have googled for tables of redox potentials and these three aren't in any that I have found -even though they are widely quoted as "standard" reducing agents. Can anyone point me at a more comprehensive list that shows the half-cells and E values for these three?

Now, in the redox lists I have found I see

Li⁺ + e⁻ → Li(s)  E=-3.05 V
K⁺ + e^- → K(s)    E=-2.93V

(both K and Li lose electrons easily - so the other chemical that GAINS those electrons is REDUCED. Therefore Li and K are reducing agents)

Why aren't K and Li used as reducing agents in preference to the rather more complex compounds I have listed above?


Ta

Clive

[cliverlong]

I read

(I think from here, but I'm not sure now) http://pages.towson.edu/ladon/orgrxs/reagent/reducers.htm

that all such reducing agents

Lithium Aluminium Hydride,
Sodium BoroHydride
Sodium Hydride

should be considered equivalent to the hydride on its own

H₂ + 2e⁻ ⇌ 2H−; E^o = −2.25 V

[DevaDevil]

I read

(I think from here, but I'm not sure now) http://pages.towson.edu/ladon/orgrxs/reagent/reducers.htm

that all such reducing agents

Lithium Aluminium Hydride,
Sodium BoroHydride
Sodium Hydride

should be considered equivalent to the hydride on its own

H₂ + 2e⁻ ⇌ 2H−; E^o = −2.25 V

correct

and the reason K as a metal (and Li and Na etc) is rather not used is because it is more complicated and dangerous to work with, and far too strong for most reactions