[kenny1999]

about simple chemical cells, several concepts questions

as far as I understand, simple chemical cell requires two different metals connected by
metal wire immersed in an electrolyte to make it function.

Take Mg/Cu as an example, we should use Copper(II) Sulphate to work as electrolyte.

I have some questions concerning this.

1. Will the displacement reaction happen when the chemical cell do not work or open circuit? (because no electron flow through the circuit)

2. What happen if I use Magnesium Sulphate to replace Copper(II) sulphate ?

3. What happen if I use NaCl (and what happen if I use distilled or tap water?)

Thank you!

[kenny1999]

doesn't anyone know?

[DevaDevil]

I presume everyone (like me) is waiting for you to reason first before they answer outright.

in reactions with electron transfer, look at the table of standard reduction potentials first to see what would have any chance of happening. Then reason if it could in your given situation.

Show us what you THINK will happen in Mg|CuSO4|Cu

[kenny1999]

I presume everyone (like me) is waiting for you to reason first before they answer outright.

in reactions with electron transfer, look at the table of standard reduction potentials first to see what would have any chance of happening. Then reason if it could in your given situation.

Show us what you THINK will happen in Mg|CuSO4|Cu

I really don't understand what will happen if I use Mg and Cu as electrodes with NaCl as electrolyte@.@. I don't need to concern about electrode potential because it is out of syllabus... I don't think Sodium ion will reduce to become Na(s), but I just can't think of anything else which could reduce. However, the fact is that NaCl can be used as electrolyte in this chemical cells, so I think there must be something to accept the electrons from the external wire. I really can't figure out what it is and what would happen if Magnesium Sulphate is used in Mg/Cu cell.

For your question (Mg/Cu in copper(II) sulphate electrolyte), I am certain that copper(II) ion from the solution will reduce to Copper at cathode while Mg electrode gives up Mg ion to the solution. However, I don't really know what would happen if I use Magnesium Sulphate instead. I think there should be some reaction because it fits the basic criteria of a chemical cell --> different metals, electrolyte. I am thinking about that as Mg electrode release electrons to form Mg ion, then the Mg ion will flow to cathode to accept the electron to become Mg again, forming on another electrode. This is my thought, however, I don't feel comfortable about this. I think there must be something wrong.

Thanks

[DevaDevil]

Well, what you propose (Mg-deposition on Cu electrode) would be an equilibrium, right?
(aka, you would have exactly the same reaction, just reversed on both electrodes; with the potential of the whole cell = 0)
This is not an ongoing reaction of course.

Mg²⁺ + 2e^- <-->  Mg (s); E⁰ = -2.372 V
this means a lot of energy comes free in oxidising magnesium.

The next step is to look through the table and to see which reduction reaction with the molecules/ions present has the highest potential. Do not forget that you also have water and air present!

possibilities:
O₂(g) + 2 H⁺ + 2 e⁻ <-->  H₂O₂(aq)     E⁰ = +0.70 V
O₂(g) + 2 H₂O + 4 e⁻ <--> 4 OH⁻(aq)   E⁰ = 0.40 V
2 H⁺ + 2 e⁻ <--> H₂(g)       E⁰ = 0 V

Likely both the second and the third will happen (unless you evacuate the oxygen first)