Determine Electrolysis Products via E Value
Let's predict the electrolysis products of dilute KOH using platinum electrodes.
The species to consider are:
Cathode: K+ and water
Anode: OH- and water
1. Reduction at Cathode
Since K+ and water can be reduced at the cathode, we have to use their electrode potentials or E value to compare their ease of reduction.
Species reduced are found on the left hand side of the half equations in the Data Booklet, so we focus on finding K+ and H2O on the left hand side of half equation.
Water has a more positive E value, more likely reduced hence will be reduced at the cathode.
Product at cathode is H2 gas.
2. Oxidation at Anode
OH- and water can be oxidised at the anode and we can use E value to compare their ease of oxidation.
Species oxidised are on the right hand side of half equation, so we focus on finding OH- and H2O on the right hand side of half equation.
OH- has a more negative E value, more likely oxidised hence will be oxidised at the anode.
Product at anode is O2 gas.
Topic: Electrochemistry, Physical Chemistry, A Level Chemistry, Singapore
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