F5 chem

2008-10-07 4:37 am
Balance redox equations using ionic half equations:

Put down the clear steps!

1) MnO4(-) + MnO2 --> MnO4 (2-) (acidic)
2) Mn(3+) ----> MnO2 + Mn(2+) (acidic)
3)Cl2 ---> Cl(-) + ClO3(-) (basic)
4)Br2 ----> Br(-) + BrO3(-) (basic)

回答 (2)

2008-10-07 10:35 am
✔ 最佳答案
1)
+) MnO4- + e- → MnO42- 2- + 2H2O + 2e- (x2
+) MnO2 + 2H2O → MnO42- + 4H+ + 2e- (x1

2MnO4- + MnO2 + 2H2O → 3MnO42- + 4H+


Actually, the reaction must be done in a strong alkaline medium.
The equations should be:
+) MnO4- + e- → MnO42- 2- + 2H2O + 2e- (x2
+) MnO2 + 4OH- → MnO42- + 2H2O + 2e- (x1

2MnO4- + MnO2 + 4OH- → 3MnO42- + 2H2O

=====
2)
+) Mn3+ + 2H2O → MnO2 + 4H+ + e- (x1
+) Mn3+ + e- → Mn2+ O2 + 4H+ + e- (x1

2Mn3+ → MnO2 + Mn2+ + 4H+

=====
3)
+) Cl2 + 2e- → 2Cl- ClO3- + 6H2O + 10e- (x5
+) Cl2 + 12OH- → 2ClO3- + 6H2O + 10e- (x1

6Cl2 + 12OH- → 10Cl- + 2ClO3- + 6H2O

Divide both sides by 2:
3Cl2 + 6OH- → 5Cl- + ClO3- + 3H2O

=====
4)
+) Br2 + 2e- → 2Br- ClO3- + 6H2O + 10e- (x5
+) Br2 + 12OH- → 2BrO3- + 6H2O + 10e- (x1

6Br2 + 12OH- → 10Br- + 2BrO3- + 6H2O

Divide both sides by 2:
3Br2 + 6OH- → 5Br- + BrO3- + 3H2O
=
2008-10-07 6:15 am
1.) MnO4- + MnO2 --> MnO42-
MnO4- + MnO2 + 2H2O --> 2MnO42- + 4H+ + e-

2.) Mn3+ ----> MnO2 + Mn2+
2Mn3+ + 2H2O ----> MnO2 + Mn2++ 4H+

3.) Cl2 ---> Cl- + ClO3-
Cl2 + 3H2O ---> Cl- + ClO3- + 6H+ + 4e-
Cl2 + 3H2O + 6OH- ---> Cl- + ClO3- + 6H+ + 4e- + 6OH-
Cl2 + 3H2O + 6OH- ---> Cl- + ClO3- + 6H2O + 4e-
Cl2 + 6OH- ---> Cl- + ClO3- + 3H2O + 4e-

4.) Br2 ----> Br- + BrO3-
Br2 + 3H2O ----> Br- + BrO3- + 6H+ + 4e-
Br2 + 3H2O + 6OH- ----> Br- + BrO3- + 6H+ + 4e- + 6OH-Br2 + 3H2O + 6OH- ----> Br- + BrO3- + 6H2O + 4e-
Br2 + 6OH- ----> Br- + BrO3- + 3H2O + 4e-


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