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Balancing Redox ½ Reaction (Basic)

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Presentation on theme: "Balancing Redox ½ Reaction (Basic)"— Presentation transcript:

1 Balancing Redox ½ Reaction (Basic)

2 Balancing Basic Balancing in basic solution is just like acidic with one additional step When balancing in acidic you will have H+ at the end In basic you don’t want H+, you want OH-

3 To change to basic To change to basic, simply add an equivalent number of OH- to each side and simplify

4 Example Balancing the following by using the ½ reaction method in a basic solution AsO4-3 + Zn  H3As + Zn +2 Acidic: 11H+ + AsO Zn  H3As + 4 H2O + 4Zn+2 Basic: +11 OH OH- 11 H2O + AsO Zn  H3As + 4 H2O + 4Zn+2+ 11OH- 7 H2O + AsO Zn  H3As + 4Zn+2+ 11OH-

5 Another Example Ag + NO3-  Ag + + N+2 Acidic:
3Ag +6 H+ + NO3-  3Ag + + N+2+ 3H2O Basic: + 6OH OH- 3Ag +6H2O + NO3-  3Ag + + N+2+ 3H2O + 6OH- 3Ag +3H2O + NO3-  3Ag + + N+2 + 6OH-

6 One More Mn(NO3)2 + NaBiO3 + HNO3  NaMnO4 + Bi(NO3)3 + H2O + NaNO3
Acidic: 2Mn H+ + 5 BiO3-  2MnO4- + 5Bi H2O Basic: + 14 OH OH- 2Mn H2O + 5 BiO3-  2MnO4- + 5Bi H2O+ 14 OH- 2Mn+2 + 7H2O + 5 BiO3-  2MnO4- + 5Bi OH-


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