Final answer:
The water potential of the cell is -3 bar, and the water potential of the solution is -5 bar. Water will move out of the cell and into the solution due to osmosis, from the higher (less negative) water potential within the cell to the lower (more negative) water potential of the beaker solution.
Step-by-step explanation:
A cell has a solute potential of -3 bar and it's in an open beaker solution with a water potential of -5 bar. To calculate the water potential of the cell, you would add the solute potential and pressure potential. In this case, we can assume the pressure potential of the cell is zero (since it's not specified and the cell is in an open beaker). Hence, the water potential of the cell is also -3 bar.
Water moves from an area of higher water potential to an area of lower water potential, thus in this case, water will move out of the cell towards the beaker solution. This is described by the equation Ψsystem = Ψtotal = Ψs + Ψp + Ψg + Ψm, where Ψs is the solute potential, Ψp is the pressure potential, Ψg is gravity potential, and Ψm is matric potential. Gravity and matric potentials are negligible in this scenario. The osmotic potential, in other words, the solute potential, greatly influences the movement of water through osmosis.