Balancing Redox Reactions

Question

Complete and balance the following redox reaction in acidic solution. Be sure to include the proper phases for all species within the reaction.

Cr2O7 2-(aq)+Pb2+(aq) --> Cr3+(aq)+PbO2(s)

Answer 1

It looks like you have posted several questions dealing with balancing redox reactions. This one is in acidic solution, so follow these steps for others that are also in acidic solution.

Cr₂O₇^2-(aq) ==> Cr³⁺(aq) ... unbalanced reduction reaction (Cr goes from +6 to +3 so is reduced)

Cr₂O₇^2-(aq) ==> 2Cr³⁺(aq) + 7H₂O(l) .. balanced for Cr and O

Cr₂O₇^2-(aq) + 14H⁺(aq) ==> 2Cr³⁺(aq) + 7H₂O(l) .. balanced for Cr, O and H using acid (H⁺)

Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e- ==> 2Cr³⁺(aq) + 7H₂O(l) .. balanced for mass and charge = balanced eq.

Pb²⁺(aq) ==> PbO₂(s) ... unbalanced oxidation reaction (Pb goes from +2 to +4 so is oxidized)

Pb²⁺(aq) + 2H₂O(l) ==> PbO₂(s) .. balanced for Pb and O

Pb²⁺(aq) + 2H₂O(l) ==> PbO₂(s) + 4H⁺(aq) .. balanced for Pb, O and H using acid (H⁺)

Pb²⁺(aq) + 2H₂O(l) ==> PbO₂(s) + 4H⁺(aq) + 2e- .. balanced for mass and charge = balanced eq.

Multiply the oxidation reaction by 3 to equalize electrons and then add the two reactions together:

3Pb²⁺(aq) + 6H₂O(l) ==> 3PbO₂(s) + 12H⁺(aq) + 6e-

Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e- ==> 2Cr³⁺(aq) + 7H₂O(l)

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3Pb²⁺(aq) + 6H₂O(l) + Cr₂O₇^2-(aq) +14H⁺(aq) + 6e- => 3PbO₂(s) +12H⁺(aq) + 6e- + 2Cr³⁺(aq) + 7H₂O(l)

Combine/cancel like terms to end up with final balanced equation:

3Pb²⁺(aq) + Cr₂O₇^2-(aq) + 2H⁺(aq) ==> 3PbO₂(s) + 2Cr³⁺(aq) + H₂O(l) BALANCED REDOX EQ