Predicting the Electrolysis Products of Aqueous
Ionic Solutions
Solution: Cont.
(b) Ag+(aq) + e- Ag(s) Eo = 0.80 V
2 H2O(l) + 2 e- H2 (g) + 2 OH-(aq) Eo = -0.42 V
As the cation of an inactive metal, Ag+ is a better oxidizing agent than
H2O, so Ag forms at the cathode. NO3- cannot be oxidized, because N is
already in its highest (+5) oxidation state. Thus, O2 forms at the anode:
2 H2O(l) O2 (g) + 4 H+(aq) + 4 e-
(c) Mg2+(aq) + 2 e- Mg(s) Eo = -2.37 V
Like K+ in part (a), Mg2+ cannot be reduced in the presence of water, so
H2 forms at the cathode. The SO42- ion cannot be oxidized because S is
in its highest (+6) oxidation state. Thus, H2O is oxidized, and O2 forms
at the anode:
2 H2O(l) O2 (g) + 4 H+(aq) + 4 e-
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