Half Reaction Problem: Get Help Here

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Discussion Overview

The discussion revolves around balancing half-reactions in redox chemistry, specifically focusing on the reduction of permanganate (MnO4) to manganese dioxide (MnO2). Participants explore the necessary steps for balancing elements and charges in the reaction.

Discussion Character

  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant suggests that equal amounts of oxygen (O) and hydrogen (H) must be present on both sides of the equation and recommends adding H2O to the left side.
  • Another participant notes that the oxidation state of MnO4 is 7 and that of MnO2 is 4, proposing that three electrons are needed for the reduction.
  • A question is raised about the necessity of including H2O on the left side of the reaction.
  • A later reply confirms the need for three electrons to reduce Mn and advises balancing the electrons for both half-reactions before addressing hydrogen and oxygen.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approach to balancing the half-reaction, with no clear consensus on the exact steps or necessity of certain components in the equation.

Contextual Notes

Some assumptions about the balancing process and the role of H2O remain unresolved, and participants do not fully agree on the methodology for achieving balance.

fasterthanwoo
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Thanks for the help guys.
 
Last edited:
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You need equal amounts of the elements O and H on both sides of the equation. Try adding H2O on the left side and balance the elements, then balance the charges.
 
Thanks for the help guys.
 
Last edited:
fasterthanwoo said:
I am still having some trouble with this.

The MnO4 is being reduced.

The oxidation state of MnO4 is 7. The oxidation state of MnO2 is 4. Therefore I need three e- to reduce?

Is it needed to have H2O on the left side of the reaction?

Thank you.

Yes. Three electrons to reduce Mn. Now balance the electrons for both half-reactions before worrying about the H's and the O's and the H2O's.
 
Thanks for the help guys.
 
Last edited:

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