# Why same metal salt is required in electrolytic refining?

Why we use same metal salt in electrolytic refining? Ex. For $$\ce{Cu^2+}$$ if we take suppose $$\ce{FeSO4}$$ so there are 2 ions ready to react on cathode ie $$\ce{Fe^2+}$$ and $$\ce{Cu^2+}$$.

But from a series (who will react first) we know $$\ce{Cu^2+}$$ is more likely to react with cathode instead of $$\ce{Fe^2+}$$.

Then why we need only same metal salt we can take any metal salt that will not react with cathode first according to series?

• Answers on CH SE site are figuratively paid by the user's own effort. When you ask, you are supposed to provide explicit compact summary of partial answers or at least ideas you have got until then. Effort not shown may be considered as effort not done. May 31 '21 at 10:28
• Let me suggest you look up the meaning of «refine». This time, it need not be a reference related to crystallography, nor chemistry, because its meaning is so old (in terms of etymology) that a good English dictionary plus some thought may rapidly clarify the situation. May 31 '21 at 11:06
• I have literally no idea what you are asking here. In which processes is the same metal salt "required"? There are a lot of different processes. May 31 '21 at 11:38

You see, in electrolytic refining, the goal is to get pure metal on the anode. Let us take a specific case of electrolytic refining of $$\ce{Cu}$$. Normally we would use $$\ce{CuSO4}$$ solution, so let us take that case.
Impure copper is set as the anode, $$\ce{Cu^{2+}}$$ ions are released. But you see, the $$\ce{Cu^{2+}}$$ ions from the anode don't directly go to the cathode.
It is a sort of chain reaction if you will, the $$\ce{Cu^{2+}}$$ ions make the solution positively charged (i.e. there are more $$\ce{Cu^{2+}}$$ ions than $$\ce{SO^{2-}4}$$ ions), so the positive ions near the cathode will try and neutralize themselves to maintain the solution's neutrality. Now, because we have taken $$\ce{CuSO4}$$ solution, there will be only $$\ce{Cu^{2+}}$$ ions near the cathode, so they will proceed to the cathode become $$\ce{Cu}$$ atoms.
But if we take $$\ce{FeSO4}$$ solution, there will be some deposition of iron too on the cathode, implying impure metal which is the opposite of what we want.