| 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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