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?

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    $\begingroup$ 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. $\endgroup$
    – Poutnik
    May 31 '21 at 10:28
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    $\begingroup$ 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. $\endgroup$
    – Buttonwood
    May 31 '21 at 11:06
  • $\begingroup$ 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. $\endgroup$
    – matt_black
    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.

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    $\begingroup$ Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. $\endgroup$
    – Buttonwood
    May 31 '21 at 11:13
  • $\begingroup$ Yah that i wanna to ask why there is some iron deposits, because there is a series like reactivity series that tell if there r 2 or more ion than which would react on amode and cathode. And according to that series copper should react first iron is late in series. Or can little iron could also react???? $\endgroup$ May 31 '21 at 11:15
  • $\begingroup$ Iron deposits are there because Fe2+ ions also neutralise or reduce to give Fe, iron atom. Yeah iron can also react, the reactivity series is supposed to be basically how much can a metal react (in very basic terms). It tells that copper will react faster than iron to give copper atom. But it doesn't say, copper will only react and iron won't, but the copper reaction will be faster. Also, another thing to consider is that at first, there will be a lot of Fe+2 ions. So, the Fe+2 ions are more prone to react that the lower amount of Cu+2 ions. $\endgroup$
    – Nova Azure
    May 31 '21 at 11:20

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