If the circuit isn’t complete the cell can’t gain and lose electrons in equal measure, it’s either accepting them *or* donating them.
This causes a buildup of charge, the donor becomes positively charged and the acceptor becomes negatively charged.
Since electrons are negative, they will not flow from a positive towards a negative for very long.
You may get a very brief trickle of current while the system charges up like a capacitor, but this will quickly go to zero.
There’s no complete path
Current would try to flow from the second cell’s positive terminal, through the wire, into the first negative, through the cell, out the positive, and then try to cross the insanely resistive air to get back to the negative terminal
Some tiny tiny current flows but like Pico amps not anything meaningful
Electrons at any point in a circuit need to be conserved. If current is leaving the positive terminal then it needs to flow up in the battery from the negative terminal. If there’s no connection to the negative terminal for current to flow into then there’s nothing to replace the charges and it’s like putting your thumb on the end of a straw, the water can’t leave because it forms a vacuum behind it
Are you saying you have these cells in a loop, or three of them in series ?
Assuming series, a small amount of electrons may flow for a very short while from one battery terminal to the next, but the charges will soon equalise.
There is nowhere for the electrons at either end of the cell-chain to go, so they will just stop. What you have is then the same situation as a single unconnected cell; there is nothing at the ends to discharge into.
Current is the amount of charge carriers moving per time. I=dQ/dt. As there is nothing moving, there is no current Maybe plenty of voltage, but no path so no current
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