I'm not certain about this but it did occur to me that it could involve the equation V = IR, where I is current and R is resistance in a circuit. Obviously with a higher concentration of ions, it's going to be possible for the electricity to bridge the electrodes more easily, unfortunately, not only am I uncertain as to whether that's the correct path to go down, I'm also uncertain as to what would happen further down the path. Shooting from the hip, I think that while the voltage is lower, the current is correspondingly higher. But, you could also argue that the higher concentration of circuit-bridging ions amounts to a reduction in resistance. There again, the galvanic cell is playing the role of the electricity provider in this circuit, and the value of R should describe the overall resistance on the other part of the circuit, ie the wires connecting the electrodes and the voltmeter. So my money's on low voltage but high current. I could be talking completely nonsense however.