By: Chem Man

Chem Man — Sat, 16 Jan 2010 10:05:29 +0000

a) Standard reduction potentials are:
Mg2+ + 2e- —-> Mg Eo = -2.37 V
Pb2+ + 2e- —–> Pb Eo = -0.13 V
The half reaction with the most positive voltage is more likely to occur.
To determine the voltage for Mg + Pb2+, reverse the first half-reaction and add to the second.
Mg —–> Mg2+ + 2e- Eo = +2.37
Pb2+ + 2e- —–> Pb Eo = -0.13 V
Mg + Pb2+ —-> Mg2+ + Pb Eo = 2.24 V
Since the voltage is positve, the reaction will occur. Mg will displace Pb2+ from aqueous solution.
Mg is oxidized because it loses e-; Pb2+ is reduced because it gains e-.

b)
Sn2+ + 2e- —–> Sn Eo = -0.14 V
2H+ + 2e- ——> H2 Eo = 0.00 V
To determine if Sn will react with H+, reverse the first half reaction and add to the second.
Sn —–> Sn2+ + 2e- Eo = +0.14 V
2H+ + 2e- —–> H2 Eo = 0.00 V
Sn + 2H+ —–> Sn2+ + H2 Eo = + 0.14 V
Since the voltage is positive, the reaction between Sn and H+ will occur.
H+ is reduced because it gains electrons; Sn is oxidized because it loses electrons.

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By: Chem Man