Balancing redox reactions occurring in acidic solutions

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The discussion focuses on balancing the redox reaction PbO2(s) + I-(aq) → Pb2+(aq) + I2(s) in acidic solution. The half-reactions are identified as the reduction of I- to I2 and the oxidation of PbO2 to Pb2+. To balance the reaction, participants emphasize the importance of balancing oxygen and hydrogen atoms first, followed by adding electrons to equalize charge. The challenge lies in correctly inserting the number of electrons on both sides to achieve balance. Properly balancing atoms and charges is crucial for a valid redox equation.
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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.
 
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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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