What would be the product formed for electrolysis of copper sulphate between the electrodes (a) Platinum (b) Copper and NaCl (aq) between (a)Platinum and (b) Copper electrodes?
To determine the products formed during the electrolysis of copper sulfate between different electrodes, we need to consider the reduction potentials of the available species.
In the electrolysis of copper sulfate between platinum electrodes, we should first determine the species present in the solution. Copper sulfate (CuSO4) dissociates into copper ions (Cu2+) and sulfate ions (SO4^2-). During electrolysis, Cu2+ ions will be attracted to the cathode (negative electrode) to undergo reduction.
At the cathode: Cu2+ (aq) + 2e- → Cu (s)
The reduction of Cu2+ ions at the cathode leads to the formation of copper metal (Cu) on the electrode surface.
At the anode: 4OH- (aq) → 2H2O (l) + O2 (g) + 4e-
At the anode, water (H2O) is present as it is usually present in any aqueous solution, and it will undergo oxidation, generating oxygen gas (O2), water (H2O), and releasing electrons (e-).
Therefore, the product formed at the cathode (platinum electrode) will be copper metal, while oxygen gas will be formed at the anode.
In the electrolysis of a solution containing copper sulfate and sodium chloride (NaCl) between a copper cathode and a platinum anode, the species present in the solution are copper ions (Cu2+), sulfate ions (SO4^2-), sodium ions (Na+), and chloride ions (Cl-).
At the cathode (copper electrode): Cu2+ (aq) + 2e- → Cu (s)
Similar to the previous case, copper ions (Cu2+) will be reduced at the cathode, leading to the formation of copper metal (Cu) on the electrode surface.
At the anode (platinum electrode): 2Cl- (aq) → Cl2 (g) + 2e-
At the anode, chloride ions (Cl-) will be oxidized, producing chlorine gas (Cl2) and releasing electrons (e-).
Therefore, the product formed at the cathode (copper electrode) will be copper metal, while chlorine gas will be formed at the anode (platinum electrode).
In summary:
(a) Platinum electrodes:
Cathode: Cu2+ (aq) + 2e- → Cu (s) (Copper metal is formed)
Anode: 4OH- (aq) → 2H2O (l) + O2 (g) + 4e- (Oxygen gas is formed)
(b) Copper (cathode) and NaCl (aq) with platinum (anode) electrodes:
Cathode: Cu2+ (aq) + 2e- → Cu (s) (Copper metal is formed)
Anode: 2Cl- (aq) → Cl2 (g) + 2e- (Chlorine gas is formed)