Predicting the Electrolysis Products of Aqueous
Ionic Solutions
Problem: What products form during electrolysis of aqueous solutions of
the following salts? (a) KBr (b) AgNO3 (c) MgSO4
Plan: We identify the reacting ions and compare their electrode potentials
with those of water, taking the 0.4 to 0.6 overvoltage into consideration.
Whichever half-reaction has the higher (less negative) electrode potential
occurs at that electrode.
Solution:
(a) K+(aq) + e- K(s) Eo = -2.93 V
2 H2O(l) + 2 e- H2(g) + 2 OH -(aq) Eo = -0.42 V
Despite the overvoltage, which makes E for reduction of water between
-0.8 and -1.0 V, H2O is still easier to reduce than K+, so H2 (g) forms
at the cathode.
2 Br -(aq) Br2 (l) + 2 e- -Eo = -1.07 V
2 H2O(l) O2 (g) + 4 H+(aq) + 4 e- -Eo = -0.82 V
Because of the overvoltage, it is easier to oxidize Br - than water, so Br2
forms at the anode.
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