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


 

3 years ago

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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
           
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