Home >
Community >
NaCl electrolysis with copper electrodes
Upvote
VOTE
Downvote
+ Salt
+ Ph
+ Chemistry
+ Electrochemistry
Posted by
Momina Hussain
NaCl electrolysis with copper electrodes
Electrolysis of sodium chloride solution produces results at each electrode. At the cathode, water is reduced to $\ce{OH-}$ and makes the solution more alkaline. Copper hydroxide is a blue, bulky precipitate, feebly acidic, and soluble in concentrated alkali hydroxide solutions. So if you got copper ions into solution at the anode, you could form some cupric hydroxide which is reported to form blue polymeric species in solution [1].
Crystals from concentration of this solution will likely be $\ce{NaCl}$ contaminated with some $\ce{CuCl2}$. Examination with a microscope could show whether $\ce{CuCl2}$ has formed, as the blue-green rhombic dihydrate (because, after all, it is in water), or, if you actually did enough electrolysis to do more than color the water, whether you precipitated the yellow-green hexagonal basic salt $\ce{CuCl2.Cu(OH)2}$ or the hydroxide or some mixture [2].
Although cuprous chloride is highly insoluble in water, it does form complexes in concentrated chloride solutions: the main species in $1\rm{M}$$\ce{KCl}$ is
$\ce{CuCl2^-}$ [1]. So you really have a complex situation with anode/cathode, $\ce{OH-}$/$\ce{Cl2}$, $\ce{Cu^0}$/$\ce{Cu+}$/$\ce{Cu^{+2}}$, $\ce{CuCl2}\cdot2\ce{H2O}$/$\ce{Cu(OH)2}$, and polymeric complexes with $\ce{OH-}$ and $\ce{Cl-}$, and various combinations of products depending on how long you do the electrolysis and, of course, whether you emit $\ce{O2}$ or $\ce{Cl2}$.
Of course, that's what makes chemistry so much fun. There are many variables, some of which you can control.
[1]: Advanced Inorganic Chemistry, Cotton and Wilkinson, 1966
Electrolysis of sodium chloride solution produces results at each electrode. At the cathode, water is reduced to $\ce{OH-}$ and makes the solution more alkaline. Copper hydroxide is a blue, bulky precipitate, feebly acidic, and soluble in concentrated alkali hydroxide solutions. So if you got copper ions into solution at the anode, you could form some cupric hydroxide which is reported to form blue polymeric species in solution [1].
Crystals from concentration of this solution will likely be $\ce{NaCl}$ contaminated with some $\ce{CuCl2}$. Examination with a microscope could show whether $\ce{CuCl2}$ has formed, as the blue-green rhombic dihydrate (because, after all, it is in water), or, if you actually did enough electrolysis to do more than color the water, whether you precipitated the yellow-green hexagonal basic salt $\ce{CuCl2.Cu(OH)2}$ or the hydroxide or some mixture [2].
Although cuprous chloride is highly insoluble in water, it does form complexes in concentrated chloride solutions: the main species in $1\rm{M}$$\ce{KCl}$ is$\ce{CuCl2^-}$ [1]. So you really have a complex situation with anode/cathode, $\ce{OH-}$/$\ce{Cl2}$, $\ce{Cu^0}$/$\ce{Cu+}$/$\ce{Cu^{+2}}$, $\ce{CuCl2}\cdot2\ce{H2O}$/$\ce{Cu(OH)2}$, and polymeric complexes with $\ce{OH-}$ and $\ce{Cl-}$, and various combinations of products depending on how long you do the electrolysis and, of course, whether you emit $\ce{O2}$ or $\ce{Cl2}$.
Of course, that's what makes chemistry so much fun. There are many variables, some of which you can control.
[1]: Advanced Inorganic Chemistry, Cotton and Wilkinson, 1966
Electrolysis of sodium chloride solution produces results at each electrode. At the cathode, water is reduced to $\ce{OH-}$ and makes the solution more alkaline. Copper hydroxide is a blue, bulky precipitate, feebly acidic, and soluble in concentrated alkali hydroxide solutions. So if you got copper ions into solution at the anode, you could form some cupric hydroxide which is reported to form blue polymeric species in solution [1].
Crystals from concentration of this solution will likely be $\ce{NaCl}$ contaminated with some $\ce{CuCl2}$. Examination with a microscope could show whether $\ce{CuCl2}$ has formed, as the blue-green rhombic dihydrate (because, after all, it is in water), or, if you actually did enough electrolysis to do more than color the water, whether you precipitated the yellow-green hexagonal basic salt $\ce{CuCl2.Cu(OH)2}$ or the hydroxide or some mixture [2].
Although cuprous chloride is highly insoluble in water, it does form complexes in concentrated chloride solutions: the main species in $1\rm{M}$ $\ce{KCl}$ is $\ce{CuCl2^-}$ [1]. So you really have a complex situation with anode/cathode, $\ce{OH-}$/$\ce{Cl2}$, $\ce{Cu^0}$/$\ce{Cu+}$/$\ce{Cu^{+2}}$, $\ce{CuCl2}\cdot2\ce{H2O}$/$\ce{Cu(OH)2}$, and polymeric complexes with $\ce{OH-}$ and $\ce{Cl-}$, and various combinations of products depending on how long you do the electrolysis and, of course, whether you emit $\ce{O2}$ or $\ce{Cl2}$.
Of course, that's what makes chemistry so much fun. There are many variables, some of which you can control.
[1]: Advanced Inorganic Chemistry, Cotton and Wilkinson, 1966
[2]: CRC Handbook, 62$^{\rm nd}$ ed.
Electrolysis of sodium chloride solution produces results at each electrode. At the cathode, water is reduced to $\ce{OH-}$ and makes the solution more alkaline. Copper hydroxide is a blue, bulky precipitate, feebly acidic, and soluble in concentrated alkali hydroxide solutions. So if you got copper ions into solution at the anode, you could form some cupric hydroxide which is reported to form blue polymeric species in solution [1].
Crystals from concentration of this solution will likely be $\ce{NaCl}$ contaminated with some $\ce{CuCl2}$. Examination with a microscope could show whether $\ce{CuCl2}$ has formed, as the blue-green rhombic dihydrate (because, after all, it is in water), or, if you actually did enough electrolysis to do more than color the water, whether you precipitated the yellow-green hexagonal basic salt $\ce{CuCl2.Cu(OH)2}$ or the hydroxide or some mixture [2].
Although cuprous chloride is highly insoluble in water, it does form complexes in concentrated chloride solutions: the main species in $1\rm{M}$ $\ce{KCl}$ is$\ce{CuCl2^-}$ [1]. So you really have a complex situation with anode/cathode, $\ce{OH-}$/$\ce{Cl2}$, $\ce{Cu^0}$/$\ce{Cu+}$/$\ce{Cu^{+2}}$, $\ce{CuCl2}\cdot2\ce{H2O}$/$\ce{Cu(OH)2}$, and polymeric complexes with $\ce{OH-}$ and $\ce{Cl-}$, and various combinations of products depending on how long you do the electrolysis and, of course, whether you emit $\ce{O2}$ or $\ce{Cl2}$.
Of course, that's what makes chemistry so much fun. There are many variables, some of which you can control.
[1]: Advanced Inorganic Chemistry, Cotton and Wilkinson, 1966
[2]: CRC Handbook, 62$^{\rm nd}$ ed.
More
VOTE