Question: [![enter image description here](https://i.stack.imgur.com/r18K6.jpg)](https://i.stack.imgur.com/r18K6.jpg)

This is an open circuit. Now, a chemical reaction happens in the cell which makes the current to flow in the wire. Then, the current has no other place to go, so it just gets grounded. It is similar to if I was holding the end of this circuit & put it on my finger. I would experience the current. So, the current does flow right. I do agree the circuit will not complete.

Why is it said that $I = 0$ in open circuit because of potential difference.

The charge inside the cell will move from the negative terminal to the end of circuit through chemical reaction inside the cell.

Answer: 
*Why is it said that I = 0 in open circuit . . .*

This is a steady state condition assuming that the resistance of the open circuit is infinite (very, very high).

The processes occurring in a cell are very complex.  

There is an electro-chemical reaction which results in charges being migrating from one electrode to the other.  

The imbalance in charge between the two electrodes leads to a potential difference across the electrodes which in turn results in an electric field between the electrodes.  

The electric field opposes the migration of changes between the electrodes and eventually, if the electrodes are not connected to the outside world, the migration of charges stops and a steady potential difference is set up between the terminals (the emf/voltage of the cell).

If a conducting circuit is connected between the electrodes of the cell an electric current flows whilst the electro-chemical reaction maintains a potential difference across the cell.  

This is the state when your circuit above has a complete conducting path.

Now suppose a break is made in the circuit.  

The current cannot stop instantaneously as the circuit has an inductance, but rather reaches the final steady state zero value over a period of time which in this instance will be very short.

In the final state the two ends of the open circuit can be thought of as a charged capacitor with the potential difference across it being equal to the emf of the cell and no current is flowing in the circuit.
A current is created due to the potential difference in a circuit (which means when one side of the circuit has more + charges than the other side (and vice-versa for the other side))), this causes a charge to flow from one side
to the other(current). When its open due you think it will ever reach the other side? And even consider that potential drop only exists in the circuit (meaning the battery and wires). If they are not connected (closed), how will a potential drop exist?

And therefore consider, if there will be a flow of charge without a potential drop?

To your statement: *"I = 0 in open circuit because of potential difference"* - I do not think thats true.

Consider the simple Ohms law, that relates potential drop and current: $$V=IR$$where $R > 0$ (otherwise you have a short circuit). If $V = 0$, then $I = 0$, as $R > 0$. If there is a potential drop without a current, then either the resistance is $\infty$, or the system is malfunctioning or simply your circuit is non-ohmic.

*"Now, a chemical reaction happens in the cell which makes the current to flow in the wire"* - "**chemical reaction**" - this is not what causes current to flow in the circuit. Remember its the potential drop (imbalance of charges on either ends of the closed circuit (or cell)). You cannot force the flow of charge (electrons) just because of a chemical reaction...