Now, balance the e- in the reduction half-reaction: 3е¯ + MnО¯ + 4H+ MnO2 + 2H2O Re-writing the equations over each other: [CN + H₂O [3e- + MnO4¯ + 4H+ CNO- + 2H+ + 2e-] x3 MnO2 + 2H2O] x2 Balance the electrons in each half-rxn so that they can be added: 3CN- + 3H2O- 3CNO + 6H++ 6e- 6-+2MnO4 + 8H+ 2MnO2 + 4H₂O 3CN + 2MnO4 + 2H+ 3CNO + 2MnO2 + H2O CN + MnO CNO + MnO2 The two half-reactions are: CN- CNO- MnO4¯ MnO2 (oxidation) (reduction) Balance each half-reaction as if it occurred in acidic solution: The oxidation half-reaction: CN- + H₂O CNO + 2H+ CN- + H₂O - CNO + 2H+ + 2e- Balance the reduction of permanganate by adding O atoms via H₂O molecules, then balance using H+: MnO4 + 4H+ MnO2 + 2H2O
Now, balance the e- in the reduction half-reaction: 3е¯ + MnО¯ + 4H+ MnO2 + 2H2O Re-writing the equations over each other: [CN + H₂O [3e- + MnO4¯ + 4H+ CNO- + 2H+ + 2e-] x3 MnO2 + 2H2O] x2 Balance the electrons in each half-rxn so that they can be added: 3CN- + 3H2O- 3CNO + 6H++ 6e- 6-+2MnO4 + 8H+ 2MnO2 + 4H₂O 3CN + 2MnO4 + 2H+ 3CNO + 2MnO2 + H2O CN + MnO CNO + MnO2 The two half-reactions are: CN- CNO- MnO4¯ MnO2 (oxidation) (reduction) Balance each half-reaction as if it occurred in acidic solution: The oxidation half-reaction: CN- + H₂O CNO + 2H+ CN- + H₂O - CNO + 2H+ + 2e- Balance the reduction of permanganate by adding O atoms via H₂O molecules, then balance using H+: MnO4 + 4H+ MnO2 + 2H2O
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Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.88QP: Calculate the cell potential of a cell operating with the following reaction at 25C, in which...
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Question
![Now, balance the e- in the reduction half-reaction:
3е¯ + MnО¯ + 4H+
MnO2 + 2H2O
Re-writing the equations over each other:
[CN + H₂O
[3e- + MnO4¯ + 4H+
CNO- + 2H+ + 2e-] x3
MnO2 + 2H2O] x2
Balance the electrons in each half-rxn so that they can be added:
3CN- + 3H2O-
3CNO + 6H++ 6e-
6-+2MnO4 + 8H+
2MnO2 + 4H₂O
3CN + 2MnO4 + 2H+
3CNO + 2MnO2 + H2O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637a8243-98d2-45ca-8dd1-4e9d7d0f97ca%2Fb708a1da-a218-45ba-8c9f-d87717f16a4c%2Fvaemkea.jpeg&w=3840&q=75)
Transcribed Image Text:Now, balance the e- in the reduction half-reaction:
3е¯ + MnО¯ + 4H+
MnO2 + 2H2O
Re-writing the equations over each other:
[CN + H₂O
[3e- + MnO4¯ + 4H+
CNO- + 2H+ + 2e-] x3
MnO2 + 2H2O] x2
Balance the electrons in each half-rxn so that they can be added:
3CN- + 3H2O-
3CNO + 6H++ 6e-
6-+2MnO4 + 8H+
2MnO2 + 4H₂O
3CN + 2MnO4 + 2H+
3CNO + 2MnO2 + H2O
![CN + MnO CNO + MnO2
The two half-reactions are:
CN-
CNO-
MnO4¯
MnO2
(oxidation)
(reduction)
Balance each half-reaction as if it occurred in acidic solution:
The oxidation half-reaction:
CN- + H₂O
CNO + 2H+
CN- + H₂O - CNO + 2H+ + 2e-
Balance the reduction of permanganate by adding O atoms
via H₂O molecules, then balance using H+:
MnO4 + 4H+
MnO2 + 2H2O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637a8243-98d2-45ca-8dd1-4e9d7d0f97ca%2Fb708a1da-a218-45ba-8c9f-d87717f16a4c%2Fjv3ds9f6.jpeg&w=3840&q=75)
Transcribed Image Text:CN + MnO CNO + MnO2
The two half-reactions are:
CN-
CNO-
MnO4¯
MnO2
(oxidation)
(reduction)
Balance each half-reaction as if it occurred in acidic solution:
The oxidation half-reaction:
CN- + H₂O
CNO + 2H+
CN- + H₂O - CNO + 2H+ + 2e-
Balance the reduction of permanganate by adding O atoms
via H₂O molecules, then balance using H+:
MnO4 + 4H+
MnO2 + 2H2O
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