Balancing equation for creation of Potassium benzoate

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SUMMARY

The balanced chemical equation for the preparation of potassium benzoate (C6H5CO2K) from benzyl alcohol (C6H5CH2OH) and potassium permanganate (KMnO4) in aqueous KOH involves the conversion of manganese to manganese dioxide (MnO2). The presence of KOH indicates the need to incorporate hydroxide ions (OH-) and water molecules when balancing the equation. The process can be effectively managed by applying the half-reaction method to separate oxidation and reduction reactions, ensuring accurate balancing of hydrogen and oxygen atoms.

PREREQUISITES
  • Understanding of redox reactions
  • Familiarity with half-reaction method
  • Knowledge of balancing chemical equations
  • Basic chemistry concepts regarding oxidation states
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  • Study the half-reaction method for balancing redox reactions
  • Learn about the role of hydroxide ions in aqueous reactions
  • Explore the properties and reactions of potassium permanganate
  • Review oxidation states and their significance in chemical reactions
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Homework Statement



Potassium benzoate C6H5CO2K can be prepared from benzyl alcohol, C6H5CH2OH, and potassium permanganate KMnO4 in aqueous KOH. The manganese is converted to MnO2

The Attempt at a Solution



So, my attempt to balance has left me scratching my head

C6H5CH2OH + KMnO4 [tex]\rightarrow[/tex] C6H5CO2K +MnO2

Question: Since it states aqueous KOH, where does that apply in balancing out this?

What would the equation be balanced?

Thanks in advance for your help.
 
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KOH means you need to use OH- and water when balancing hydrogen and oxygen.

Do you know how to balance by half reactions? Can you split the system into reduction and oxidation half reactions?

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