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Electrolysis of Aqueous Salt Solutions Chemistry Tutorial
salt in water to produce an aqueous solution enables it to conduct electricity.
2O (l) → M
+ (aq) + X - (aq) The species present in the electrolyte are:
(i) water (as the
(ii) cations (from the salt)
(iii) anions (from the salt)
The use of inert electrodes, electrodes made of a material that will not take part in the reactions, means the only species present that can take part in the
electrolytic cell reactions are the anions and cations of which the salt is composed and water.
At the anode,
either water is oxidized or the anions of the salt are oxidized.
As a first approximation it is likely that if:
(i) water is a stronger reductant than the anions, water will be oxidized at the anode.
(ii) the anion is a stronger reductant than water, anions will be oxidized at the anode.
At the cathode,
either water is reduced or the cations of the salt are reduced.
As a first approximation it is likely that if
(i) water is a stronger oxidant than the cations, water will be reduced at the cathode.
(ii) the cation is a stronger oxidant than water, cations will be reduced at the cathode.
Electrolysis is a
E for the electrolytic cell is negative
o Applied voltage (emf) must be greater than the emf for the cell, ie greater than -E
o (electrolytic cell).
Mass of substance produced electrolytically is proportional to the quantity of electricity flowing.
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No ads = no money for us = no free stuff for you! Electrolytic Cell for Electrolysis of Aqueous Salt Solution
For the electrolysis of an aqueous salt solution,
M X (aq) using inert electrodes:
The electrolyte contains:
The possible electrolytic cell reactions are:
oxidation of water
H → ½O 2O (l) 2(g) + 2H + + 2e - E 0 = -1.23 V Reduction of water
H + e 2O (l) - → ½H 2(g) + OH - E 0 = -0.83 V
oxidation of anions
X → X + e - - E 0 x Reduction of cations
M + e + - → M E 0 m
Determining which oxidation and reduction reactions occur based on E
0 x > -1.23 V then X - will be oxidized to X
- → ½Br 2(l) + e - E 0 = -1.08 V
-1.08 V > -1.23 V so the Br - anion will be oxidized at the anode NOT water.
0 x < -1.23V then water will be oxidized to oxygen gas
- → ½F 2(g) + e - E 0 = -2.89 V
-2.89 V < -1.23 V so water will be oxidized at the anode NOT the F - anion.
0 m > -0.83 V then M + will be reduced to M
2++2e -→Zn Fe
2++2e -→Fe Ni
2++2e -→Ni Sn
2++2e -→Sn Pb
2++2e -→Pb Cu
0=+0.34V For all these metal cations, E 0 > -0.83V so the cation will be reduced at the cathode NOT water.
0 m < -0.83V, water will be reduced to hydrogen gas
++e -→K Ba
2++2e -→Ba Ca
2++2e -→Ca Na
++e -→Na Mg
2++2e -→Mg Al
0=-1.68V For all these active metal cations, E 0 < -0.83 V so water will be reduced at the cathode NOT the cation.
Example : Electrolysis of KI
Consider the electrolytic cell shown below for the electrolysis of a standard aqueous solution of potassium iodide, KI
Possible Cathode Reactions
+ + e - →
E 0 = -2.94 V
2O + e - →
2(g) + OH - E
0 = -0.83 V
0 for the reduction of water is greater than E 0 for the reduction of K +
Water is a stronger oxidant than K +
Water will be reduced at the cathode NOT K +.
Possible Anode Reactions
2(s) + e - E 0 = -0.54 V
2(aq) + e - E 0 = -0.62V
2(g) + 2H + + 2e - E
0 = -1.23 V
0 for the oxidation of I - is greater than E 0 for the oxidation of water.
I - is a stronger reductant than water
I - will be oxidized at the anode NOT water.
Possible REDOX Reactions
2(s) + e - E
0 = -0.54 V
2O + e - →
2(g) + OH - E
0 = -0.83 V
- + H 2O →
2(s) + ½H 2(g) + OH - E
0 = -1.37 V
A minimum of 1.37 V would need to be supplied by an external power supply in order for this electrolysis reaction to proceed.
This discussion assumes that all species are present in their standard states so that the electrode potentials are standard electrode potentials.
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