If a battery is open on one end but connected on the other, does it raise the potential on the connected end? If a battery has one end open and other end connected to a circuit will the battery increase the potential at the point of connection?
 A: Within a battery a chemical reaction is responsible for moving mobile electrons from one terminal to the other.
The terminal from which the mobile electrons come from is called the positive terminal (deficit of mobile electrons) of the battery and the one that they go to is called the negative terminal (surplus of negative electrons).
If there is no external conducting circuit then the build-up of charges on the terminals leads to an electric field within the battery which opposes the motion of mobile electrons within the battery produced by the chemical reaction.
Eventually the electric field produced by the surplus/deficit of charges on the terminals is strong enough to stop to movement of the mobile electrons within the battery and the potential difference across the terminals of the battery stays constant.
This potential difference is equal to the emf of the battery.
You can think of the two terminals of the battery as a capacitor with air in between and that capacitor of extremely small value has become fully charged.
Now consider the circuit in your question and let the circuit be connected to the negative terminal of the battery but not the positive terminal.
Since there is no complete conducting circuit there is no current flowing in the conducting part of your circuit.
This means that there can be no potential difference between the negative terminal of your battery and the open end of your circuit but there is a potential difference between the positive terminal of the battery and the negative terminal of the battery..
So there is no potential difference across your circuit but there is a potential difference between the positive (deficit of mobile electrons) terminal of the battery and the open end of your circuit (surplus of electrons) which is in a sense has become the new negative terminal of your battery.
In terms of mobile electron movement in your circuit this is what happens.
Suppose your circuit was just a piece of wire neither end of which is connected to the battery.
As you bring one end of the wire towards the negative (surplus of electrons) terminal of the battery a minute current flows in the wire (your circuit).
Those charges flow within the wire under the influence of the electric field produced outside the battery by the terminals of the battery.
Charges within the wire redistribute themselves and in electrostatics you called those charges induced charges with induced positive charges (deficit of mobile electrons) at the end of the wire closest to the negative terminal of the battery and induced negative (surplus of electrons) at the end which is remote from the negative terminal of the battery.  
As you bring the wire closer to the negative terminal more charges are induced and this continues until the wire touches the negative terminal.
The deficit of electrons in the wire is then equal to the surplus of electrons on the terminal and they "neutralise" themselves.
So you end up with no potential difference across the wire but there is a potential difference between the positive terminal of the battery and the open end of your wire – the new negative terminal of the battery.
