Sort of. the midpoint charges up (or down) by reverse leakage when the "wrong" capacitor is reverse biased, once it's charged you have a bipolar device.
a high resistance to a suitable supply could be added to ensure the charge remains otherwise the pair will go non-linear near the absolute peaks in whatever current signal the capacitors pass .
same as any other two capacitors in series.
probably not enough to charge the forward one either, include an 1meg (etc. - scale aprropriate to supply voltage and leakage current) pull up (or down) on the midpoint to ensure they are correctly biased.
only the first time, the second time it's got a charge on it and is working like a capacitor.
it's fairly simple to model, IIRC it was done in SEB some time last year.
C1 1F C2 1F ||+ +|| 1V - --- --- --- -o--||--o--||----. / \\ / \\ / \\ a || b || | 0V - + + + + + ---+-- \\ / \\ / ////// -1V - --- --- 0 1 2 3 4 5 6 7 0) start with both caps discharged, a at 0V b at 0V
1) ramp the voltage on a upto 1V , C1 is reverse biased and leaks current , c2 charges up with 1C b is at 1V a is at 1V 2) reduce the voltage on a to 0V again and 0.5C charge flows out of C2 into C1, a is a 0V b is at 0.5V , 3) now take A negative, the other 0.5C flows out of C2 and into C1, b is at 0v a is at -1 points 4,5,6,7 are left as an exercose for the reader should he need more convincing,
for a swing of 2V 1C flows... capacitance is 0.5F a real reverse biased electrolytic is going to be some non-zero voltage drop, so you may need to use more than 1V for the signal :)