Will a Schottky diode save my LEDs against reversed voltage?
I do not think that there will be any issue as the reverse current of the Schottky diode is very likely far to small to cause any damage to the LEDs. Due to the small leakage current not much power can dissipate in the LEDs so I do not expect that any damage can occur.

However, the reverse current of a Schottky diode is very temperature dependent, at higher temperatures it could become in the order of 10 mA! (Source: datasheet of IN5817, a 1N4001 silicon diode has a reverse current of only 50 uA at 100C) Even that might not be enough to damage the LEDs though.

Fortunately there are simple measures we can take to prevent all issues. Just add another diode in parallel with the LEDs such that this diode conducts the reverse current of the other diode:







simulate this circuit &ndash; Schematic created using CircuitLab

Note that I copied the Schottky diode, D9 does not need to be a Schottky diode, a Silicon diode (like 1N4007) will do the job just as well. The same is true for D8, there is little need to make that a Schottky diode, you can use a 1N4001 as well as long as the current is less than 1 A.

D9 will be in forward mode when the voltage from V1 is negative. That means that the reverse voltage across the LEDs is limited to less than 1 Volt so nothing will happen to the LEDs.

In your circuit however I do miss a resistor to limit the current through the LEDs. Using LEDs without a current limiting resistor is not advised as the current through a LED is heavily dependent on the voltage, type, aging, temperature. So in my circuit I added R1.