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where M is the molar mass of the substance (usually given in SI units of grams per mole) and v is the valency of the ions. For Faraday's first law, M, F, v are constants; thus, the larger the value of Q, the larger m will be.
Copper(II) sulfate is an inorganic compound with the chemical formula Cu SO 4.It forms hydrates CuSO 4 ·nH 2 O, where n can range from 1 to 7. The pentahydrate (n = 5), a bright blue crystal, is the most commonly encountered hydrate of copper(II) sulfate, [10] while its anhydrous form is white. [11]
m is the mass of the substance produced at the electrode (in grams), Q is the total electric charge that passed through the solution (in coulombs), n is the valence number of the substance as an ion in solution (electrons per ion), M is the molar mass of the substance (in grams per mole), F is Faraday's constant (96485 coulombs per mole).
Important examples of electrolysis are the decomposition of water into hydrogen and oxygen, and bauxite into aluminum and other chemicals. Electroplating (e.g., of copper, silver, nickel, or chromium) is done using an electrolytic cell. Electrolysis is a technique that uses a direct electric current (DC).
Important examples of electrolysis are the decomposition of water into hydrogen and oxygen, and of bauxite into aluminium and other chemicals. Electroplating (e.g. of Copper, Silver, Nickel or Chromium) is done using an electrolytic cell. Electrolysis is a technique that uses a direct electric current (DC). [citation needed]
The applied voltage which is just sufficient to overcome the back EMF due to polarization and also to bring about the electrolysis of an electrolyte without any hindrance is known as decomposition potential. The decomposition potential Ed is composed of various potentials and is given by: Ea (min)= Ed= Eb+ Es+ Ev. where: Ea = applied potential
The exact relationship depends on the nature of the reactions at the two electrodes. For the electrolysis of aqueous copper(II) sulfate (CuSO 4) as an example, with Cu 2+ (aq) and SO 2− 4 (aq) ions, the cathode reaction is the reduction Cu 2+ (aq) + 2 e − → Cu(s) and the anode reaction is the corresponding oxidation of Cu to Cu 2+.
"It can also be prepared by electrolysis of magnesium sulfate [Epsom salts] solution at moderate voltage with a copper anode: this reaction produces hydrogen, copper sulfate solution, and copper hydroxide precipitate." what happens to the magnesium? MgSO4(aq) + Cu(S) --dc--> h2(g) + Mg(OH)2(s) + CuSO4(aq) would make more sense.