What is the balanced half reaction for MnO4 and VO in redox reactions?

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SUMMARY

The balanced half-reaction for the reduction of permanganate (MnO4-) in the presence of vanadium(II) (VO2+) is MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. The oxidizing agent is MnO4-, while the reducing agent is VO2+, which is oxidized to V(OH)4+. This discussion clarifies the oxidation states and the balancing of redox reactions involving complex ions.

PREREQUISITES
  • Understanding of redox reactions and oxidation states
  • Familiarity with half-reaction method for balancing equations
  • Knowledge of common oxidation states of manganese and vanadium
  • Basic chemistry concepts related to acids and bases (H+ ions)
NEXT STEPS
  • Study the half-reaction method for balancing redox reactions
  • Learn about the properties and reactions of manganese compounds
  • Explore the oxidation states of transition metals, focusing on vanadium
  • Investigate the role of acids in redox reactions, particularly H+ ions
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Chemistry students, educators, and anyone studying redox reactions in inorganic chemistry, particularly those focusing on transition metals like manganese and vanadium.

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Homework Statement


Identify the oxidizing agent and the reducing agent and write a balanced equation for each half reaction.
MnO4(1-) + VO(2+) ----> Mn(2+) + V(OH)4 (1+)


Homework Equations


None.


The Attempt at a Solution


I know how to do these when there is just one atom. However, when there are molecules such as MnO4 I get confused.

So would the oxidizing agent be VO or just V?

Would the Half reaction for that be:

MnO4(1-) + 3 e- ----> Mn(2+) ?
 
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the half reaction for MnO4- is: MnO4- (aq) + 8H+ (aq) + 5e- <-> Mn2+ (aq) + 4H2O (l)

i would say V4+ is the reducing agent as its oxidation state goes from 4+ to 5+ thus being oxidised and the oxygen has the same -2 oxidation state through the eqn, although I am not 100% sure
 
What is ON for manganese before and after the reaction?
 

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