A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains
1 answer:
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Answer:
Hydride ion
Explanation:
You often see the reduction with NADH written as
NADH ⟶ NAD⁺ + H⁺ + 2e⁻
If you think about it, H⁺ + 2e⁻ is equivalent to H:⁻, so we could write the reaction as
NADH ⟶ NAD⁺ + H:⁻
In terms of a mechanism, the dihydropyridine ring of NADH transfers a hydrogen atom with its pair of electrons (a hydride ion) to the substrate and becomes the more stable, aromatic pyridinium ion in NAD⁺.
