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