Balancing redox reactions occurring in acidic solutions

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SUMMARY

The discussion focuses on balancing the redox reaction PbO2(s) + I-(aq) → Pb2+(aq) + I2(s) in acidic solutions. The process involves identifying half-reactions for reduction and oxidation, specifically I- to I2 and PbO2 to Pb2+. The balancing of oxygen and hydrogen is achieved by adding 4H+ and 2H2O, followed by the introduction of electrons to equalize charge. The final balanced equation is PbO2 + 4H+ + 4I- → Pb2+ + 4I2 + 2H2O, emphasizing the necessity of balancing both atoms and charge.

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  • Understanding of redox reactions and their components
  • Knowledge of balancing chemical equations in acidic solutions
  • Familiarity with half-reaction method for redox balancing
  • Basic chemistry concepts including oxidation states and electron transfer
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  • Study the half-reaction method for balancing redox reactions
  • Learn about oxidation states and how to determine them in reactions
  • Explore examples of balancing redox reactions in both acidic and basic solutions
  • Review the principles of charge conservation in chemical equations
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Chemistry students, educators, and anyone looking to deepen their understanding of redox reactions and their balancing techniques in acidic environments.

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Homework Statement


Balance the following redox reactions occurring in an acidic solution.


Homework Equations


PbO2(s)+I-(aq)---->Pb2+(aq)+I2(s)


The Attempt at a Solution


1.) half reactions:
a.) reduction: I-(aq)----->I2(s)
b.) oxidation: PbO2(s)------>Pb2+(aq)
2.) balance Oxygens and Hydrogens
PbO2+4H+------>Pb2++2H2O
3.) add electrons to balance equations
a.)PbO2+4H++4e---->Pb+2+2H2O
b.) 4I--->4I2+4e-
4.) cancel out the electrons and combine the half reactions
PbO2+4H++4I--->Pb2++4I2+2H2O
 
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shikam08 said:
b.) 4I--->4I2+4e-

Nope.
 
Last edited by a moderator:
I don't understand how to insert the number of electrons on both sides to balance the equations and cancel out the electrons.
 
Balance atoms first, then use electrons to balance charge in half reaction. As long as charge on both sides is different, equation is not balanced.
 
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