 # why in electrolysis of an aqueous solution of copper sulphate using copper electrodes, sulphate ions and hydroxide ion do not oxidises?

11 years ago

hi somy, in this table of copper sulphate reaction chart , you can check the reason ...for many sulphate reactions ...

 Electrolyte cathode product cathode equation anode product anode equation comments molten aluminium oxide Al2O3(l) molten aluminium Al3+(l) + 3e- ==> Al(l) oxygen gas 2O2-(l) - 4e- ==> O2(g) or  2O2-(l)  ==> O2(g) + 4e- Method 2 in principle: The industrial method for the extraction of aluminium its ore aqueous copper(II) sulfate CuSO4(aq) copper deposit any conducting electrode e.g. carbon rod, any metal including copper itself Cu2+(aq) + 2e- ==> Cu(s) oxygen gas inert electrode like carbon (graphite rod) or platinum (i) 4OH-(aq) - 4e- ==> 2H2O(l) + O2(g) or  4OH-(aq) ==> 2H2O(l) + O2(g) + 4e- (ii) 2H2O(l) - 4e- ==> 4H+(l) + O2(g) or 2H2O(l) ==> 4H+(l) + O2(g) + 4e- Adapt Method 3 - e.g. use carbon rods: The blue colour of the copper ion will fade as the copper ions are converted to the copper deposit on the cathode aqueous copper (II) sulphate CuSO4(aq) copper deposit any conducting electrode e.g. carbon rod, any metal including copper itself Cu2+(aq) + 2e- ==> Cu(s) copper(II) ions - the copper anode dissolves copper anode Cu(s) - 2e- ==> Cu2+(aq) or  Cu(s) ==> Cu(s) + 2e- See Method 3: This is the basis of the method of electroplating any conducting solid with a layer of copper. When using both copper cathode and anode, the blue colour of the copper ion does not decrease because copper deposited at the (-) cathode = the copper dissolving at the (+) anode. molten sodium chloride NaCl(l) molten sodium Na+(l) + e- ==> Na(l) chlorine gas 2Cl-(l) - 2e- ==> Cl2(g)  or  2Cl- ==> Cl2 + 2e- Method 2 in principle: This a method used to manufacture sodium and chlorine. See Down's Cell aqueous sodium chloride solution (brine) NaCl(aq) hydrogen 2H+(aq) + 2e- ==> H2(g) or 2H3O+(aq) + 2e- ==> H2(g) + 2H2O(l) or 2H2O(l) + 2e- ==> H2(g) + 2OH-(aq) chlorine gas 2Cl-(aq) - 2e- ==> Cl2(g)  or  2Cl- ==> Cl2 + 2e- Method 1a: This is the process by which hydrogen, chlorine and sodium hydroxide are manufactured hydrochloric acid HCl(aq) hydrogen gas 2H+(aq) + 2e- ==> H2(g) chlorine gas 2Cl-(aq) - 2e- ==> Cl2(g)  or  2Cl- ==> Cl2 + 2e- Method 1a: All acids give hydrogen at the cathode sulphuric acid sulfuric acid H2SO4(aq) hydrogen gas 2H+(aq) + 2e- ==> H2(g) oxygen gas (i) 4OH-(aq) - 4e- ==> 2H2O(l) + O2(g) or  4OH-(aq) ==> 2H2O(l) + O2(g) + 4e- (ii) (+) 2H2O(l) - 4e- ==> 4H+(l) + O2(g) or 2H2O(l) ==> 4H+(l) + O2(g) + 4e- Method 1b: All acids give hydrogen at the cathode. Whereas hydrochloric acid gives chlorine at the anode, the sulfate ion does nothing and instead oxygen is formed. This is the classic 'electrolysis of water' molten lead(II) bromide PBr2(l) molten lead Pb2+(l) + 2e- ==> Pb(l) bromine vapour 2Br-l) - 2e- ==> Br2(g)  or  2Br- ==> Br2 + 2e- Method 2: A good demonstration in the school laboratory - brown vapour and silvery lump provide good evidence of what's happened molten calcium chloride CaCl2(l) solid calcium Ca2+(aq) + 2e- ==> Ca(s) chlorine gas 2Cl-(aq) - 2e- ==> Cl2(g)  or  2Cl- ==> Cl2 + 2e- Method 2 in principle: The basis of the industrial method for the manufacture of calcium metal ************** ************ **************************************************** ********** ******************************************************* *************************************