Concentration cell based on the Ni2 -Ni cell reaction. (a) Concentrations of Ni2 (aq) in the two half-cells are unequal, and the cell generates an electrical current and a voltage.
(b) The cell operates until [Ni2 ] is the same in the two half-cells, at which point the cell has reached equilibrium and the emf goes to zero.

Explanation:

The missing diagram contained in this question is first attached in the image below.

The objective of this question is to determine how ions migrate when the cells are operating by assuming the solutions are composed of Ni(NO3)2.

From the information provided:

In this instance, the ions tend to move first from cathode to anode in terms of raising the concentration of Ni(2+) at the anode, resulting in the development of a dead cell. The initial concentration of [Ni(2+)] in the anode solution is 1.00 × 10⁻³ M, which gradually increases to 0.5 M, during which both the cathode and the anode possess the same concentration at the same point.

This causes Q(equilibrium constant) to equal  1 as well as log(Q) to equal 0, indicating that the cell is dead.

As a result, the cell will cease to operate, and nothing will migrate from the left to the right side.

acceleration = 18/3 = 6 m/s per second.

hope i

-lavira: )