Which atom has a change in oxidation number of –3 in the following redox reaction? K2Cr2O7 + H2O + S - - > KOH + Cr2O3 + SO2
K
Cr***
O
S
2 answers
right
(Redn half-rxn; Cr2O7^2- is Oxidizing Agent)
16HOH + 2K^+ + 4Cr2O7^2- + 12e^- => 4Cr2O3 + 20OH^-
(Cr^+6 => Cr^+3 + 3e^-)
(Oxidation half-rxn; S^o is the Reducing Agent)
12OH^- + 3S^o => 2K^ + 3SO2 + 6HOH + 12e^-
(S^o => S^+4 + 4e^-)
Net Oxdn-Redn Rxn (K^+ = Spectator Ion):
4Cr2O7^-2 + 3S^o + 10HOH => 4Cr2O3 + 3SO2 + 20OH^-
for more DrReb048(at) g m a i l (dot) c o m
16HOH + 2K^+ + 4Cr2O7^2- + 12e^- => 4Cr2O3 + 20OH^-
(Cr^+6 => Cr^+3 + 3e^-)
(Oxidation half-rxn; S^o is the Reducing Agent)
12OH^- + 3S^o => 2K^ + 3SO2 + 6HOH + 12e^-
(S^o => S^+4 + 4e^-)
Net Oxdn-Redn Rxn (K^+ = Spectator Ion):
4Cr2O7^-2 + 3S^o + 10HOH => 4Cr2O3 + 3SO2 + 20OH^-
for more DrReb048(at) g m a i l (dot) c o m
2 answers