The choice of electrodes, as well as the solutes involved in each solution, are important. The relative ease of discharging of each cation depends on the relative reactivity of the cation involved as compared to other cations, and likewise, there is something similar to be said for anions.

Let's look at each option. Option A first...

Concentrated sodium chloride solution between carbon electrodes. For most circumstances, carbon electrodes are considered to be inert (and thus only facilitate in keeping the circuit closed).

Because hydrogen is far less reactive than sodium, it will get discharged first, even though "concentrated" suggests a sufficiently significant quantity of sodium ions present. Chloride ions are not far above hydroxide ions in the relative ease of discharging, and thus, the numerous chloride ions will far outweigh the lower quantity of hydroxide ions present. Thus, chloride ions are discharged first.

All in all, hydrogen ions and chloride ions will get discharged. This will result in two conclusions. One, some water of dissolution is lost (because the hydrogen ion component of water is gone). Two, some of the hydroxide ions will bind with sodium ions to form sodium hydroxide. The amount of particles dissolved remains somewhat unchanged (since some sodium chloride particles are effectively replaced by sodium hydroxide particles), but the amount of water decreases.

Thus, the concentration of the solution increases.

Let's look at Option B...

Unlike earlier, copper electrodes do react to some degree. Copper (II) sulfate solution contains copper, sulfate, hydrogen and hydroxide ions.

Remember that the electrode attached to the positive terminal of the battery is the anode and oxidation occurs here, while the electrode attached to the negative terminal is the cathode and reduction occurs here.

Because copper is not fully inert, it will prefer to lose electrons to form copper (II) ions which dissolve into the solution. No gas or metal will be deposited at the anode, and instead, the copper anode itself dissolves into the solution.

On the other end, electrons are received. Of course, the copper cathode is not going to dissolve into the solution because the cathode is gaining electrons and not losing electrons. Cations are absorbed here.

Because copper (II) ions are less reactive that hydrogen ions, copper (II) ions prefer to be discharged, so copper (II) ions are lost from the solution.

The same electrons are involved in the entire process, so the rate of loss of mass of the anode is the same as the rate of gain of mass of the cathode, so there is no net change in the concentration of the copper (II) sulfate solution.

Let's look at option C...

This is same as option B but with inert platinum electrodes. No copper anode here so nothing is dissolved.

Copper (II) ions continue to be discharged at the cathode due to the inferiority of its reactivity compared to hydrogen ions.

At the anode, hydroxide ions are more preferentially discharged than sulfate ions, so hydroxide ions are lost.

Similar in argument to option A, hydroxide ions and copper (II) ions are lost. The loss of hydroxide ions causes the water of dissolution to decrease, leading to a change in concentration of the particles.

Let's look at option D...


Without a doubt, we know by now that platinum electrodes are inert, and clearly hydrogen ions are preferentially discharged to sodium ions at the cathode and oxide ions are preferentially discharged to chloride ions at the anode. This indicates the loss of water (since both hydrogen ions and hydroxide ions are lost).

The concentration will also change here.