Redox of NAD+ and NADH

[altitude]

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I know that NADH --> NAD+ is an oxidation, but I don't see how there is an increase in oxidation number/state if a hydrogen atom is lost- wouldn't this reduce the oxidation state because H and an oxidation number of +1?

 

[Arctangent]

What is lost is hydrogen, or more specifically hydride. Hydride has 2 electrons and 1 proton, so it is net negative. As NADH is neutral, and NAD+ is positive, this suggests that what is lost should correspondingly have a -1 charge.

 

[altitude]

What is lost is hydrogen, or more specifically hydride. Hydride has 2 electrons and 1 proton, so it is net negative. As NADH is neutral, and NAD+ is positive, this suggests that what is lost should correspondingly have a -1 charge.

Thank you for the explanation.

How do you know it is H- and not H+?

Edit: Because it is NAD+ rather than NAD-?

 

[Arctangent]

Yeah, basically from a neutral molecule (NADH) you get NAD+ and H- for charge balance. If theoretically NADH did turn into NAD-, that would suggest an H+ is lost. NADPH/NADP+ behaves likewise.

 

[altitude]

Yeah, basically from a neutral molecule (NADH) you get NAD+ and H- for charge balance. If theoretically NADH did turn into NAD-, that would suggest an H+ is lost. NADPH/NADP+ behaves likewise.

Thank you