cmoney asked:
I’m stuck on this problem, which involves predicting oxidation and reduction reactions by using their standard electrode potentials…
Predict whether, to any significant extent,
a) Mg (s) will displace Pb^2+ from aqueous solution;
b) Sn(s) will react with and dissolve in 1M HCl
I dont understand how one can tell which is being oxidized and reduced from a)… and also, what does it take for Sn to react and/or dissolve in HCl?
thanks in advance,
January 16th, 2010 at 2:05 am
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.