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Electrolysis of Aqueous Salt Solutions Tutorial

Key Concepts

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Electrolytic Cell for Electrolysis of Aqueous Salt Solution


← X-

← H2O →

For the electrolysis of an aqueous salt solution, MX(aq) using inert electrodes:

The electrolyte contains:

  • H2O
  • M+
  • X-

The possible electrolytic cell reactions are:

Anode (+) cathode (-)
oxidation of water H2O(l) → ½O2(g) + 2H+ + 2e-     E0 = -1.23 V Reduction of water H2O(l) + e- → ½H2(g) + OH-     E0 = -0.83 V
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:

At the cathode:

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Example : Electrolysis of KI(aq)

Consider the electrolytic cell shown below for the electrolysis of a standard aqueous solution of potassium iodide, KI(aq):


← I-

← H2O →

Possible Cathode Reactions
(a) K+ + e- K E0 = -2.94 V
(b) H2O + e- ½H2(g) + OH- E0 = -0.83 V
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 = -0.54 V
(b) I- ½I2(aq) + e- E0 = -0.62V
(c) H2O ½O2(g) + 2H+ + 2e- E0 = -1.23 V
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 = -0.54 V
cathode H2O + e- ½H2(g) + OH- E0 = -0.83 V

Overall I- + H2O ½I2(s) + ½H2(g) + OH- E0 = -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.

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This discussion assumes that all species are present in their standard states so that the electrode potentials are standard electrode potentials.