CHEMISTRY HOMEWORK DUE THURSDAY EVENING, KINDLY SHOW ALL WORKS!!! (5). Examine t
ID: 968472 • Letter: C
Question
CHEMISTRY HOMEWORK DUE THURSDAY EVENING, KINDLY SHOW ALL WORKS!!!
(5). Examine the following half-reactions and circle the strongest oxidizing agent among the substances.
[PtCl4]2– (aq) + 2e– Pt (s) + 4Cl– (aq) E° = 0.755 V
RuO4 (s) + 8H+ (aq) + 8e– Ru (s) + 4H2O (l) E° = 1.038 V
FeO42– (aq) + 8H+ (aq) + 3e– Fe3+ (aq) + 4H2O (l) E° = 2.07 V
H4XeO6 (aq) + 2H+ (aq) + 2e– XeO3 (aq) + 3H2O (l) E° = 2.42 V
(6). Calculate the mass of palladium produced by the reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes. The molecular weight of palladium is 106.4 g/mol.
Pd2+ + 2 e- Pdo
Explanation / Answer
(5). Are you concerning about reduction potential (E°) value. The greater the reduction potential (towards negative value), the greater tendecy to be reduced(means it is a strong oxidizing agent)
Hence [PtCl4]2– (aq) + 2e– Pt (s) + 4Cl– (aq) E° = 0.755 V has greater reduction potential among given 4 half-reactions. Therefore this is strong oxidizing agent.
(6) According to Faraday's 1st law W = Act/ZF
W -> amount of substance in grams, A -> atomic weight of element in grams, c-> current in amperes, t-> time in seconds, Z-> atomic number of metal element, F-> faraday value 96500 coulombs
W = [106.4 g/mol X 3.2 amp X (30X60)]/46X 96500
W = 0.1381 grams