Electrolysis of Aqueous Salt Solutions |
Key Concepts
- Dissloving a salt in water enables it to conduct electricity.
- The species present in the electrolyte are:
water (as the solvent)
cations (from the salt)
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 oxidised 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 non-spontaneous reaction: Eo for the electrolytic cell is negative.
- Applied emf must be greater than the emf for the cell, ie greater than -Eo.
- Mass of substance produced electrolytically is proportional to the quantity of electricity flowing.
Electrolytic Cell
M+ →
← X-
← H2O →
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For the electrolysis of an aqueous salt solution, MX(aq) using inert electrodes:
The electrolyte contains:
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The possible electrolytic cell reactions are:
| Anode (+) |
cathode (-) |
|---|
| oxidation of water | H2O(l) → ½O2(g) + 2H+ + 2e E0 = -1.23V |
Reduction of water | H2O(l) + e → ½H2(g) + OH- E0 = -0.83V |
| oxidation of anions | X- → X + e E0x |
Reduction of cations | M+ + e → M E0m |
Determining which oxidation and reduction reactions occur based on E0 values:
At the anode:
- If E0x > -1.23V, X- will be oxidized to X
Br- → ½Br2(l) + e E0 = -1.08V
-1.08V > -1.23V so the Br- anion will be oxidized at the anode NOT water.
- If E0x < -1.23V, water will be oxidized to oxygen gas
F- → ½F2(g) + e E0 = -2.89V
-2.89V < -1.23V so water will be oxidized at the anode NOT the F- anion.
At the cathode:
- If E0m > -0.83V, M+ will be reduced to M
| Zn2++2e→Zn |
Fe2++2e→Fe |
Ni2++2e→Ni |
Sn2++2e→Sn |
Pb2++2e→Pb |
Cu2++2e→Cu |
| E0=-0.76V |
E0=-0.44V |
E0=-0.24V |
E0=-0.14V |
E0=-0.13V |
E0=+0.34V |
| For all these metal cations, E0 > -0.83V so the cation will be reduced at the cathode NOT water.
|
- If E0m < -0.83V, water will be reduced to hydrogen gas
| K++e→K |
Ba2++2e→Ba |
Ca2++2e→Ca |
Na++e→Na |
Mg2++2e→Mg |
Al3++3e→Al |
| E0=-2.94V |
E0=-2.91V |
E0=-2.87V |
E0=-2.71V |
E0=-2.36V |
E0=-1.68V |
| For all these active metal cations, E0 < -0.83V so water will be reduced at the cathode NOT the cation.
|
Example : Electrolysis of KI(aq)
K+ →
← I-
← H2O → |
| Possible Cathode Reactions |
|---|
| (a) |
K+ + e |
→ |
K |
E0 = -2.94V
|
| (b) |
H2O + e |
→ |
½H2(g) + OH- |
E0 = -0.83V |
E0 for the reduction of water is greater than E0 for the reduction of K+
Water is a stronger oxidant than K+
Water will be reduced at the cathode NOT K+.
|
| Possible Anode Reactions |
|---|
| (a) |
I- |
→ |
½I2(s) + e |
E0 = -0.54V
|
| (b) |
I- |
→ |
½I2(aq) + e |
E0 = -0.62V
|
| (c) |
H2O |
→ |
½O2(g) + 2H+ + 2e |
E0 = -1.23V |
E0 for the oxidation of I- is greater than E0 for the oxidation of water.
I- is a stronger reductant than water
I- will be oxidized at the anode NOT water.
|
|
| Possible REDOX Reactions |
|---|
| anode |
I- |
→ |
½I2(s) + e |
E0 = -0.54V
|
| cathode |
H2O + e |
→ |
½H2(g) + OH- |
E0 = -0.83V |
|
|
| Overall |
I- + H2O |
→ |
½I2(s) + ½H2(g) + OH- |
E0 = -1.37V |
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A minimum of 1.37V would need to be supplied by an external power supply in order for this electrolysis reaction to proceed.
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AUS-e-TUTE Class and School Group members (Teachers and their classes) can access the interactive learning activity for this topic.
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