Table 27.1. Preparation of Several Oxidation-Reduction Reactions Test Addition o
ID: 502374 • Letter: T
Question
Table 27.1. Preparation of Several Oxidation-Reduction Reactions Test Addition of Solution B Tube No. Solution A Chlorine bleach and 2 drops of 6M HCI Drops of 0.1 M Fe(NH (the active ingredient in chlorine bleach is CIO Chlorine bleach and 2 drops of 6 M HC1 Add 1 drop of starch solution followed by drops of 0.1 M Ku 0.01 M KMnoa and 2 drops of 6 M Hason Drops of 0.1 M Fe(NHL):Sos (Figure 273) 0.01 M KMno, and 2 drops of6MHzSO, Drops of M KrCo. Deionized water and 2 drops of 6MH SO. Add 1 drop of starch solution followed by Deionized water and 2 drops of 6 M H so. drops of 0.1 M Ku Drops of o.1 M Fe(NHaorso, 0.1 M H on and 2 drops of 6MH so. Add 1 drop of starch solution followed by 0.1 M Htoh and 2 drops of 6 M HSO. drops of 0.1 MKlExplanation / Answer
The half reactions are written in terms of the ions.
1) Reduction half: OCl- +2 H+ + 2 e- -----> Cl- + H2O
Oxidation half: Fe2+ -------> Fe3+ + e-
Overall: 2 Fe2+ + OCl- + 2 H+ ------> 2 Fe3+ + Cl- + H2O
2) 1) Reduction half: OCl- +2 H+ + 2 e- -----> Cl- + H2O
Oxidation half: 2 I- -------> I2 + 2 e-
Overall: 2 I- + OCl- + 2 H+ ------> I2 + Cl- + H2O
3) Reduction half: MnO4- + 8 H+ + 5 e- ------> Mn2+ + 4 H2O
Oxidation half: Fe2+ -------> Fe3+ + e-
Overall: 5 Fe2+ + MnO4- + 8 H+ ------> 5 Fe3+ + Mn2+ + 4 H2O
4) Reduction half: MnO4- + 8 H+ + 5 e- ------> Mn2+ + 4 H2O
Oxidation half: C2O42- + 8 H+ ------> 2 CO2 + 4 H2O + 2 e-
Overall: 2 MnO4- + 5 C2O42- + 48 H+ -----> 2 Mn2+ + 10 CO2 + 24 H2O
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.