From the Electrolysis of concentrated copper(ii) sulphate using carbon electrodes, what product will I get ?

You have a mixture of `"Cu"^"2+", "SO"_4^"2-"`, and `"H"_2"O"`.

Something must be oxidized, and something must be reduced, so you consult a table of standard reduction potentials like the one below to find suitable candidates,

The only candidate for reduction in the table is

`"Cu"^"2+""(aq)" + 2"e"^"-" → "Cu(s)"; "+0.34 V"`

Now,we look for candidates for oxidation.

`"Cu"^"2+"` and `"SO"_4^"2-"` are already in their highest oxidation states, so the only candidate for oxidation is water.

We see two possibilities in the table:

`"H"_2"O"_2"(aq)" + 2"H"^"+""(aq)" + 2"e"^"-" → 2"H"_2"O"(l); "+1.77 V"`
`"O"_2"(g)" + 4"H"^"+""(aq)" + 4"e"^"-" → 2"H"_2"O(l)";color(white)(mm) "+1.23 V"`

The reaction that requires the least voltage is the formation of oxygen.

Thus the electrolytic cell half-reactions are:

`color(white)(mmmmmmmmmmmmmmmmmmmmmmmmmmmmm)ul(E^@"/V")`
`2×["Cu"^"2+""(aq)" + 2"e"^"-" → "Cu(s)"];color(white)(mmmmmmmmmmmm) "+0.34"`
`1×[2"H"_2"O(l)" → "O"_2"(g)" + 4"H"^"+""(aq)" + 4"e"^"-"] ;color(white)(mmmmmmmm) "-1.23"`
`color(white)(mm)"2Cu"^"2+""(aq)" + 2"H"_2"O""(l)" → "2Cu(s)" + "O"_2"(g)" + 4"H"^"+""(aq)"; "-0.89"`

Thus, if you electrolyze a solution of `"CuSO"_4` using graphite electrodes, you will see metallic carbon being deposited at the cathode and bubbles of oxygen released at the anode