We have a guess at the chipset given the iPad 2 has very similar battery performance with the same battery as the iPad 1.
However, voltage is not "consumed." Whatever voltage is applied on the chip, it will all drop through the circuitry because you have to reach ground potential at the end. Thus, the question is does the higher potential applied lower, increase, or keep current the same?
Normally you increase voltage when you're in a bad timing situation, but the A5 is in the iphone is likely to be a slower clock in the iPad 2, just as it was in the iphone 4 versus the iPad. Thus, the higher voltage won't gain you anything there. So all you would be doing is forcing higher leakage (because now you've got a higher a higher potential on the gate and more voltage difference between transistors and bulk).
AFAIK, the voltage for the chipset in the iPad 2 didn't change from the iPad 1, so it seems likely that the battery is reporting a different voltage than what is actually being delivered to the chipset.