The reducing agent in this redox equation?

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SUMMARY

The reducing agent in the redox equation Cr2O7 2- + 2OH- --> 2CrO4 2- + H2O is Cr2O7 2-. Upon analyzing the oxidation states, chromium in Cr2O7 2- is reduced from +6 to +6 in CrO4 2-, indicating that it is the species undergoing reduction. The change in oxidation number confirms Cr2O7 2- acts as the reducing agent in this reaction.

PREREQUISITES
  • Understanding of oxidation states and how to assign them
  • Familiarity with redox reactions and their components
  • Knowledge of the dichromate ion (Cr2O7 2-) and chromate ion (CrO4 2-)
  • Basic skills in balancing chemical equations
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  • Study the process of assigning oxidation states in various compounds
  • Learn about the role of reducing and oxidizing agents in redox reactions
  • Explore the properties and reactions of dichromate and chromate ions
  • Practice balancing redox equations using the half-reaction method
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Chemistry students, educators, and anyone studying redox reactions and oxidation states in chemical equations.

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Homework Statement


Which is the reducing agent in this equation, if any?

Cr2O7 2- + 2OH- --> 2CrO4 2- +H2O


Homework Equations





The Attempt at a Solution



After giving the elements their oxidation numbers I found the reducing agent to be Cr2O7 2-

Is this correct?
 
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Which element changes oxidation number? From what to what?
 

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