What causes the voltage potential in a chemistry battery

Consider a simple chemistry battery: Zn and Cu with H2SO4 as the electrolyte. The electron will travel from Zn to Cu. And at the Cu side, H ion will get electron and turns to H2. If the circuit is open, Zn itself will react with the acid, and H2 will be generated at the Zn side.
Why there is voltage potential when the circuit is closed? The voltage force the electron to travel around the circuit.




In your setup, the metal zinc is transformed into $\ce{Zn^{2+}}$ ions plus electrons. The electrons can and will react with $\ce{H^+}$ ions in contact with the zinc plate, and this takes place when the circuit is open and when it is closed. It produces bubbles of $\ce{H2}$ gas. But this reaction is slow, because the reaction produces a positive charge ($\ce{Zn^{2+}}$) which repels the ions $\ce{H+}$, and prevent them from touching the zinc plate.


On the contrary, if the circuit is closed, the $\ce{H^+}$ ions are in contact of the copper plate without being repelled, and the reaction can proceed with a higher rate, as nothing prevents the electrons from reacting with the $\ce{H+}$ ions present in the solution. So Hydrogen is produced with a higher rate on the copper plate, even if it is far away from the zinc atoms which have produced them.


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