Hi, can someone please explain what is happening if the electrolyte is NaCl?
If the Metal is more reactive than Sodium, then the metal will go into solution (the metal electrode is the anode) and the Sodium will deposit onto the cathode which is the graphite electrode.
What is the redox reaction happening?
So the anode reaction would be: Metal(s) ---> Metal+(aq) + e-
I'm assuming that for something to be more reactive than sodium (such as Potassium, Rubidium, Cesium), it will only release one electron (electro-negativity rules).
The cathode reaction: Na+(aq) +e- ---> Na(s).
So overall: Na+(aq) + Metal(s) ---> Metal+(aq) + Na(s)
In more detail: NaCl(aq) + Metal(s) ---> MetalCl(aq) + Na(s).
How come there is a potential difference?
There is a potential difference because there is a movement of charge. Potential Difference refers to the movement of charge in a circuit in simple terms.
And in a normal galvanic cell with two metal electrolyte, say Zn and Cu and two electrolyte, what will happen if I use something (e.g. NaCl) other than ZnSO4 and CuSO4?
Out of the scope of the syllabus I think. The anode will be Zn, the Cathode Cu. For this reaction to occur, you need two separate electrodes.
As the anode is Zn, it will lose electrons and dissolve into solution. However, an excess of Cl- ions are required.
As Cu is the Cathode, the Na+ ions will gain electrons, which means that Sodium metal will deposit on the Cu cathode.
As a result, the excess Cl- ions will migrate across a salt bridge (preferably KNO3, as these ions are soluble in all solutions) and will form a ZnCl2(aq) solution in the anode beaker.
This whole section I put in italics is too complicated though for HSC standards, so I may be completely wrong and no reaction may occur. You probably don't need to worry about it.