If the battery is not connected to anything, the chemical force is pulling on the ions, trying to draw them across the electrolyte to complete the reaction, but this is balanced by the electrostatic force-- the voltage between the electrodes.
In your battery example, there is no return current path so no current will flow. There is obviously a more deep physics reason for why this works but as the question asked for a simple answer I'll skip the math, google Maxwell's Equations and how they are used in the derivation of Kirchhoff's voltage law.
When you add a wire between the ends of the batteries, electrons can pass through the wire, driven by the voltage. This reduces the electrostatic force, so ions can pass through the electrolyte. As the battery is discharged, ions move from one electrode to the other, and the chemical reaction proceeds until one of the electrodes is used up.
The result would be that the lamp would not light. In either the open or short case, the light does not function but there is an important difference: for the short circuit, excessive current will flow out of the battery because there is little to resist the flow of current. Thus, the battery will be drained very quickly.
(a) No short-circuit occurs when the batteries are properly connected in series. (b) Attempted series-connection of two grounded batteries would result in a short-circuit as the current could flow through the ground connection as indicated by the red arrow. BAT3 is short-circuited while BAT4 is not.
When the battery is being drained by a circuit, the voltage is a little lower, allowing reactions to proceed. The higher the drain, the lower the voltage, the faster the chemical reactions, and the greater the current. That effect creates what we can measure as the battery's internal resistance.
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