Asked by kat
Today I began an electroplating lab between a copper strip and a coin in electrolyte solution with a battery. I am struggling to figure out the copper half-reactions that occured, and whether they are reduction or oxidation. I only need two, and one is Cu2+ + 2e- -> Cu please help me with the final one and help with understand which is reduction and which is oxidation
Answers
Answered by
DrBob222
The Cu strip is going into solution. That one is
Cu ==> Cu^2+ + 2e
Then the Cu^2+ ions in solution are plated onto the penny as pure copper. That one is
Cu^2+ + 2e ==> Cu
If you remember the definitions you can't go wrong with which is which.
Oxidation is the loss of electrons. That occurs, by definition, at the anode.
The gain of electrons is reduction.
Cu ==> Cu^2+ + 2e
Then the Cu^2+ ions in solution are plated onto the penny as pure copper. That one is
Cu^2+ + 2e ==> Cu
If you remember the definitions you can't go wrong with which is which.
Oxidation is the loss of electrons. That occurs, by definition, at the anode.
The gain of electrons is reduction.
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