Download Introduction to electrolysis MAIN POINTS • Electrolysis is

Introduction to electrolysis
MAIN POINTS
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Electrolysis is the decomposition of an electrolyte by an electric current. This is carried out in an
electrolytic cell.
An electrolyte is a molten compound or the solution of a compound, which conducts an electric
current and is decomposed by it.
Electrolytes contain ions which are charged atoms, or charged radicles (groups of atoms) which
+
carry the electric current. Example Na , OH .
Strong electrolytes consist totally of ions, i.e., are fully ionised, e.g., all salts, strong acids and alkalis.
These have high electrical conductivity.
Weak electrolytes consist mainly of molecules with relatively few ions, for example, weak acids and weak
bases such as ethanoic acid and aqueous ammonia. In solution a large proportion of the molecules
remain undissociated.
+
NH3 (aq) + H2O (l) = NH4 (aq) + OH-(aq)
The electrodes in the electrolytic cell are the anodes, which is the positive electrode and the cathode
which is the negative electrode. During electrolysis, the anode gains electrons which flow from the anode
to the cathode, which loses electrons.
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Anions are negative ions, they travel towards the anode during electrolysis, for example, OH , Cl where
they may give up electrons. (2Cl (aq) = Cl2(g) + 2e)
+
+
Cations are positive ions, they travel towards the cathode during electrolysis, for example, H , Na , Cu
2+
where they may gain electrons. (Cu + 2e = Cu).
2+
II To identify the ions present in electrolytes:
(i) Write the formula of the compound, for example, Al2(SO4)3
(ii) Write a balanced equation to separate it into atoms or radicles eg Al2(SO4)3 = 2Al + 3SO4
(iii) Put the appropriate charge on the atoms or radicles (numerically equal to the valency), that is,
3+
2Al2(SO4)3 = 2Al + 3SO4
(iv) Check to see that the total charge on the ions is zero (2x3+) + (3x2-) = 0
In the case of aqueous solutions, relatively very small numbers of H+ and OH- ions are present, due to
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the extremely slight ionisation of water H2O = OH (aq) + H (aq)
Question
1 (a) Write equations for the ionisation of:
(i) H2SO4
(ii) CuSO4
(iii) FeCl3
(iv) Al2O3
(v) AlPO4
(vi) (NH4)3PO4
(b) Give the formulae of the ions and or molecules present in:
(i) CuSO4(aq)
(ii) NaCl(aq)
(iii) NH3(aq)
(iv) CH3CO2H(aq)
III Factors affecting the formation of the products during electrolysis:
(i) The degree of electropositiveness and electronegativeness of the ions. At the cathode, if more than
one type of positive ion arrives at the cathode the one which gives up its charge most readily, i.e., the
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+
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least electropositive ion is discharged, e.g., H and Na , H is preferentially discharged 2H + 2e = H2.
Thus, the ion derived from the element lower in the electrochemical series is discharged. At the anode,
the least electronegative ion is discharged, that is, the ion which gives up its charge more readily. Thus, in
the electrolysis of dilute aqueous sodium chloride OH- ions are discharged in preference to Cl- ions, ie,
4OH-(aq) = 2H2O(l) + O2(g) + 4e.
There are exceptions to this rule:
(ii) Due to high concentration of an electrolyte, its anion may be discharged in preference to a less
electronegative ion, for example, with concentrated aqueous sodium chloride Cl- ions are discharged in
preference to OH- ions (from water) 2Cl = Cl2 + 2e
(iii) The type of electrode. In some cases, that is, for active electrodes, the anode may dissolve to
provide electrons, for example, a copper electrode in the electrolysis of aqueous copper sulphate. Cu(s) =
2+
Cu (aq) + 2e (left at the anode) since this occurs more readily than the discharge of the anions. Platinum
and carbon (graphite) are inert electrodes, they do not form ions during electrolysis.