Electrolysis and electroplating electrolytes and oxidation states

[sgstudent]

When electrplating copper with a dilute NaCl aqueous solution and the anode is iron while cathode is platinum what will the results be? Will copper form cations or will the OH- be oxidised? Also, if we use concentrated NaCl will there be any difference? I'm thinking that the copper shouldn't be able to oxidised as it should not be able to exist as ions alone? So only the Cl-/OH- should oxidise. But if the copper is replaced with zinc then it could happen as at the cathode, H+ ions will be reduced. So the Zn2+ ions will always have an anion to have an equilibrium. But for the copper case, there is nothing to fill up the equilibrium as in there us no anion "counterpart" to it. But I'm not very sure about this.

When a reaction AB-->BA at the reactants side A oxidation: +1 B is -1 at product A:-1 B:+1 is AB both oxidised and reduced?

When a reaction AB+CD-->AD+CB and reactants oxidation A:+1 B:-1 C:+1 D:-1 while at the products side A:+2 D:-2 C:+2 B:-2
What is oxidised and what is reduced?

Thanks for help guys!

 

[jbsteele]

In the first scenario, copper will form cations, and the OH- will be oxidized. If a concentrated NaCl solution is used, the reaction rate of the electrolysis will be faster, but the results will still be the same. For the second scenario, A and B are both being oxidized and reduced. This is known as a redox reaction. The oxidation states of A and B change from +1 and -1 respectively, to -1 and +1 respectively. For the third scenario, A and D are both being oxidized, and B and C are both being reduced. A and C are oxidized from +1 to +2, while B and D are reduced from -1 to -2.