Question

balance half rxns: PbO2 (s) = Pb2+ E=+1.455 V; Cr2O72- = Cr3+ E=+1.33

Answer 1

Separate in half reactions

PbO2 ---> Pb+2

Cr2O7-2 ---> Cr+3

NOTE: The E is given to know, which one is reducing and oxidizing.

The greater value will reduce, the other will oxidize... Cr3+ will be oxidized

1. Balance elements in the equation other than O and H.

PbO2 ---> Pb⁺²

2Cr⁺³ ---> Cr2O7^-2

2. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation.

PbO2 ---> Pb⁺² + 2H2O

7H2O + 2Cr⁺³ ---> Cr2O7^-2

3. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation.

4H+ PbO2 ---> Pb⁺² + 2H2O

7H2O + 2Cr⁺³ ---> Cr2O7^-2 + 14H+

4.Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.... this is not the case btw)

4H+ PbO2 + 2e- ---> Pb⁺² + 2H2O

7H2O + 2Cr⁺³ ---> CrO7^-2 + 14H+ + 6e-

5. The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.

12H+ 3PbO2 + 6e- ---> 3Pb⁺² + 6H2O

7H2O + 2Cr⁺³ ---> CrO7^-2 + 14H+ + 6e-

6. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.

7H2O + 2Cr⁺³ +12H+ 3PbO2 + 6e- ---> 3Pb⁺² + 6H2O + CrO7^-2 + 14H+ + 6e-

H2O + 2Cr⁺³ + 3PbO2---> 3Pb⁺² + CrO7^-2 + 2H+