What is the minimum voltage needed to produce a reaction in the electrolysis of copper(II) sulfate using inert electrodes?
any help would be much appreciated, thanks!
any help would be much appreciated, thanks!
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The Cathode reaction in this process is:
Cu^2+ + 2e => Cu
The Anode reaction in this process is:
2H2O => O2 + 4H^1+ +4e
The standard reduction potentials for these are:
-.34 (Sign change because it is backward from the table)
-1.23 (Assuming this is at pH of 0, and since water is oxidized, and it is negative because it is backward from the table)
You need to multiply the Cu equation by 2 to cancel out the electrons, which means the potential
needs to be multiplied by 2 as well. So the standard reduction for the cathode ends up being (-.34*2)
From here, it's just
Ecell=Cathode-Anode
Hope you can get the answer from that!
Cu^2+ + 2e => Cu
The Anode reaction in this process is:
2H2O => O2 + 4H^1+ +4e
The standard reduction potentials for these are:
-.34 (Sign change because it is backward from the table)
-1.23 (Assuming this is at pH of 0, and since water is oxidized, and it is negative because it is backward from the table)
You need to multiply the Cu equation by 2 to cancel out the electrons, which means the potential
needs to be multiplied by 2 as well. So the standard reduction for the cathode ends up being (-.34*2)
From here, it's just
Ecell=Cathode-Anode
Hope you can get the answer from that!