For a capacitor, the rate of change of the voltage is equal to the current divided by the capacitance:

** dv i**

---- = ---

dt C

For a constant current, the voltage changes at a constant rate.

Two 150F capacitors in series will have a capacitance of 75F. For a current of, say, 30 mA, the rate of change of the voltage will be 0.03/75, or 0.4 mV / s. If the series combination of the capacitors is charged to its limit, 5.4V, and the critical voltage is, say, 4.5V, it'll hit that voltage in (5.4 - 4.5) / 0.0004 = 2250 s, or 37 1/2 minutes.

Taking the positive terminal as the terminal that's, well, more positive, and the direction of current as out of the capacitor, the general formula for the time it takes a capacitor to discharge from an initial voltage

**V**_{i} to a final voltage

**V**_{f} is:

** i * t**

( V_{i} - V_{f} ) = --------

C

and that resolves to:

** ( Vi - Vf ) * C**

------------------ = t

i

The same formula applies for charging the capacitors -

**Vf** is bigger than

**Vi**, making the voltage term negative, and the current goes into the capacitor, making

**i** negative, so the quotient is positive. Using the formula, it takes 75 amp-seconds for each volt of charging. At 1/2 amp, it'll take thirteen and a half minutes to charge the capacitors from a fully discharged state to 5.4V. It'll take nearly two and a half minutes to get it from 4.5V to 5.4V at 1/2 amp.

Directly answering the question:

But how long is 2 150F cap's in Series giving a little over 5v going to power a standard Arduino Uno?

Translating "a little more than 5V" to, "exactly 5V," and adding the notion that the capacitors start at 5.4V, the answer comes to

( 30 / ( Arduino supply current in amps ) ) seconds.

I don't find any really credible estimates for the supply current of the UNO, but I'll guess that it's about 30 mA, making the time come to a neat 1000 seconds, nearly 17 minutes.