Relative Half-Cell Potentials Assuming standard conditions, answer the following
ID: 495703 • Letter: R
Question
Relative Half-Cell Potentials Assuming standard conditions, answer the following questions. (Use the table of Standard Reduction Potentials for common Half-reactions from your text. If hydrogen is one of the reagents, assume acidic solution.) cf Table 11.1 on p 480 of Zumdahl 7th ed. Is Fe2+(aq) capable of reducing Cr3+(aq) to Cr2+(aq)? Is H2(g) capable of reducing Ag+(aq)? Is Sn metal capable of reducing Fe3+(aq) to Fe2+(aq)? Is VO2+(aq) capable of oxidizing Fe2+(aq)? Is Cr3+(aq) capable of oxidizing Fe2+(aq)? Is Sn2+(aq) capable of oxidizing Fe2+(aq)?
Explanation / Answer
1) Consider the two half reactions:
Fe3+ (aq) + e- ------> Fe2+ (aq); E0 = +0.771 V
Cr3+ (aq) + e- -------> Cr2+ (aq); E0 = -0.41 V
The more positive the reduction potential of an element, the more strongly oxidizing is the element. Consequently, the element with the more positive reduction potential oxidizes the other element and itself gets reduced. Therefore, Fe3+ can oxidize Cr2+ (aq) to Cr3+ (aq) and itself get reduced to Fe2+ (aq). Therefore, the reverse reaction cannot occur under standard conditions, i.e., Fe2+ (aq) cannot reduce Cr3+ (aq) to Cr2+ (aq).
2) Consider the standard reduction reactions:
Ag+ (aq) + e- ------> Ag (s) ; E0 = +0.7994 V
2 H+ (aq) + 2 e- ------> H2 (g); E0 = 0.00 V
Ag+ (aq) has a higher reduction potential than H2 and hence is the better oxidizing agent. Consequently, Ag+ (aq) can oxidize H2 to H+. In other words, H2 is oxidized and Ag+ is reduced. Therefore, H2 can reduce Ag+ to Ag.
3) Consider the half reactions:
Fe3+ + e- ------> Fe2+ (aq); E0 = +0.771 V
Sn2+ + 2 e- ------> Sn (s); E0 = -0.14 V
Fe3+/Fe2+ system has the more positive reduction potential and therefore, Fe3+ is the oxidizing agent and gets reduced to Fe2+. Sn, having a more negative reduction potential will have a tendency to get oxidized. Therefore, Sn can reduce Fe3+ to Fe2+.
4) The half-reactions are
VO2+ (aq) + 2 H+ (aq) + 2 e- ------> VO+ (aq) + H2O (l); E0 = +0.99 V
Fe3+ (aq) + e- -----> Fe2+ (aq); E0 = +0.771 V
VO2+ (aq) has a higher standard reduction potential then Fe3+ and consequently is the oxidizing agent. Therefore, VO2+ can oxidize Fe2+ to Fe3+, itself being reduced to VO+.