Why is it not equal to zero since a reverse biases diode is essentially an open circuit?
You may be confusing open and short circuits. An open component is like a component which is not there. The voltage across a non-conducting diode between the points in the circuits where it is connected is the same as what the voltage would be if we removed the diode.
A voltage is a potential difference between two points that are at different places in an electric field. If we move a charge through this field from one point to the other, we have put in work (or obtain work, if we go the other way).
A potential difference does not require a conducting path, since electric fields can exist even in a vacuum. An electron and a proton in a vacuum have a potential difference (i.e.) voltage between them. Current does not have to flow for voltage to be present. That's why it is a "potential": it represents stored energy that can potentially be used to do work, if it is released.
When a conducting path is provided between points at a different potential, that is what in fact erodes potential differences. A conductor, such as a piece of copper wire, can have a voltage between its two ends, but that means that current is present. (If no current is present, it means there must not be any voltage). If the source of voltage has a limited capacity (such as a battery or capacitor), then the conductor will eventually drain the potential difference down to zero. Charges will flow from the region of higher potential to the region of lower potential, until the two are at the same potential.
If multiple components are connected in series, and a voltage is applied this series arrangement, they each have a share of the voltage, such that their individual shares add up to the applied voltage.
Suppose that the components all have some nonzero resistance, but one of them is open. In that case, the open one will have the full voltage across it, and the others have zero. Since the circuit is broken, no current flows. According to Ohm's law (V = IR), since I is zero, V must be zero for each one of the R's. However, this formula doesn't apply to the open component because its R is infinite. We just know that the voltage is the same as the total voltage, since all the other components have zero voltage.
Now suppose we have nearly the same arrangement, but the open component is replaced by a short. In that case, the short has nearly zero voltage across it. This is again from V = IR. Resistance is nearly zero, and I is some reasonable value limited by the other R's, so V is nearly zero. The other components which have a nonzero resistance now pick up the entire voltage and divide it among themselves in proportion to their resistances.
So, as a rule of thumb, an open or nonconducting component has full voltage across it, and a short has nearly zero voltage (if that short is in series with some resistances which limit current).
If the voltage was zero then what has happened to the voltage source?
For example take a coin cell battery, it has an open circuit voltage of ~3V, not 0V. If you connect a wire between the terminals (briefly) then the voltage will be 0V (note that this is NOT recommended with any voltage source capable of sourcing more than a few mA of current)
The above assumes real world non-ideal voltage sources with finite internal resistance.
Let's go back to basic definitions. Think of a switch - a switch is either an open circuit or a closed circuit (short circuit).
Now, think of the definitions of open and short circuit.
For an (ideal) open circuit, the current must be zero while the voltage can be any value.
For a (ideal) short circuit, the voltage must be zero while the current can be any value.
In light of these definitions, let's consider your question:
The answer is that the current must be zero for an (ideal) open circuit while the voltage can be any value (consistent with the reverse bias constraint of the ideal diode).