Also available in python:

  import heapq
  n = 2 
  heapq.nlargest(n, vals)[n-1]

Edit: and out of morbid curiosity, here is the same implemented by a scan. It could be arranged as a single statement.

  swapIfGt = lambda xs, q: xs if xs[0] >= q else sorted([q]+xs[1:])
  n = 2
  reduce(swapIfGt, vals, [None]*n)[0]

Edit2: rewrote swapIfGt

The pedant in me squirms at seeing a quadratic-time solution to something so linear. Use a heap or a scan, sir!