playing with vertices of a cube

This is more of a logic puzzle.

Its not possible. Because:

The numbers that can appear on the edges range from 3 to 15, thats 13 numbers. So if all the numbers are to be distinct we have to drop a number.

We can, in fact, find that number. No matter how the vertices are numbered, the sum of the numbers on the edges is always 108 (why?). Since 3+4+5+...+15 = 117, we have to drop 9.

What this means is that 3,4,5 and 6 all have to appear on the edges. This means that 1 is connected to 2,3 and 4 and further 2 and 4 lie on different edges. But this means the number 6 cannot appear as we cannot pair 1 and 5 too as all three 'valencies' of 1 are taken up and (1,5) and (2,4) are the only pairings that can produce a 6