In your experiment, the copper in the anode (positive electrode) will ionize and dissolve,
Cu(s) = Cu2+(aq) + 2e-
and copper(II) ions from the solution will be deposited on the cathode (negative electrode):
Cu2+ (aq) + 2e- = Cu(s)
With some care, the copper deposited on the electrode will be purer than the copper the anode was made of. This is called electrorefining and it's used to purify metals in industry.
For every mole of copper deposited at the cathode, you'll need to supply 2 moles of electrons (and one mole of Cu2+). So you'd expect that two factors in determining the mass of copper deposited would be the following:
Total charge transferred to the cathode. This will be equal to the current (charge per second) times the number of seconds the current is applied. Two ways to increase the amount of copper deposited, then are to
2H+(aq) + 2e- = H2(g)
The little bubbles of hydrogen gas forming at the cathode surface will produce a brittle, poorly adhering copper deposit. Stirring the solution during the electrolysis will minimize prevent hydrogen reduction at the cathode, but you'll find that there's a limit to the amount of current you can use (and the potential you can apply to the cathode).
If you add a little nitrate, it will be reduced to ammonia at the cathode. Reduction of nitrate will compete with and effectively prevent reduction of hydrogen on the copper surface, and the deposited copper will be smooth, shiny, and very pure.
Solution composition. The concentration of the Cu2+ ion will influence the amount of copper you see deposited. For the reduction of Cu2+ ion at the cathode, Cu2+ (aq) + 2e- = Cu(s). The Nernst equation implies that the cathode's potential is proportional to the log of the molarity of Cu2+:
E = Eo + (RT/nF) ln [Cu2+]
where Eo is the electrode potential under standard conditions (about +0.34 V for this reaction), n is the number of moles of electrons transferred (2 moles of electrons per mole of Cu deposited), F is the charge per mole of electrons transferred (F = 96487 Coulombs/mol), R is the gas law constant (8.31451 J/mol K; isn't it interesting how this constant crops up in places that have nothing to do with ideal gases?) and [Cu2+] is the molarity of the copper(II) ions in the solution.
When [Cu2+ ] is less than 1 M, the log term will be negative, and the electrode potential will be more negative. That means that the lower the concentration of Cu2+, the more negative potential (applied voltage) will be required. For a given applied voltage then, the more dilute the CuSO4 solution is, the less copper will be deposited on the cathode.
This isn't a bad thing, if your objective is purifying the copper. Using a highly concentrated copper(II) ion solution will deposit more copper on the cathode, but the deposit may be spongy or coarse and will not adhere well to the cathode. Commercial electroplating operations keep metal cation concentrations (and current density) low.
Author: Fred Senese firstname.lastname@example.org
Copyright © 1997-2010 by Fred Senese
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Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/redox/faq/print-cu-deposition.shtml