Let me try yet one more type of explanation, which I will confine to the PN junction diode (covers virtually all diodes used in modern circuits).
The diode consists of a p-doped region (p-type) slapped up against an n-doped region (n-type). In the p-type, the electron (e-) flow is largely accomplished by electrons moving from hole to hole. This is, electrically, exactly analogous (and is often visualized) as holes moving in a direction opposite to e- flow (although there is no physical movement of postivie charge) In the n-type, there are loosely bound e- which can be donated (moved).
At the PN junction of the diode, loosely bound e- in the n-type fall into the holes of the adjacent p-type. What you then have is an abundance of e- in a thin layer of the p-type layer at the junction, and a depletion of them (creating a net positive charge) in a thin layer of the n-type. This sets up a voltage field of positive in the n-type relative to negative in the p-type. This pushes any free e- in the n-type further away from the junction. The result is a thin PN layer which has no free holes and no free e-. The layer becomes an insulator.
Now, if you apply a positive voltage to the p-type and a negative at the n-type, e- in the p-type are removed, making free holes. Simultaneously, the positive voltage is conteracting the reverse voltage which had been set up in the PN junction, and e- in the n-type are force closer to the p-type, where they can cross over and fill up the new holes. Current flows.
If, however, you apply positive voltage to the n-type, and negative to the p-type ("reverse-biasing" the diode) you simply reinforce the voltage gradient which was already naturally set up in the PN junction. The e- are forced even farther away from the PN junction, and the insulative boundary (depletion region) thickens. No current flows.
To get more in-depth than that might take a good portion of a graduate course in materials science. I hope what I have written suffices.