Problem: Balance each skeleton reaction, use Appendix D to calculate E°cell, and state whether the reaction is spontaneous:(a) Cu+(aq) + PbO2(s) + SO42−(aq) ⟶ PbSO4(s) + Cu2+(aq) [acidic]

FREE Expert Solution

When balancing redox reactions under acidic conditions, we will follow the following steps.

Step 1: Separate the whole reaction into two half-reactions.

Step 2: Balance the non-hydrogen and non-oxygen elements first.

Step 3: Balance oxygen by adding H2O to the side that needs oxygen. (1 O: 1 H2O)

Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen. (1 H: 1 H+)

Step 5: Balance the charges: add electrons to the more positive side (or less negative side).

Step 6: Balance electrons on the two half-reactions.

Step 7: Get the overall reaction by adding the two reactions.


a. Cu+(aq) + PbO2(s) + SO42−(aq) ⟶ PbSO4(s) + Cu2+(aq)

(1-2): Separate and balance non-H and O atoms

Cu+(aq) →  Cu2+(aq)

 PbO2(s) + SO42−(aq) ⟶ PbSO4(s)


(3): Add H2O

Cu+(aq)   Cu2+(aq)

PbO2(s) + SO42−(aq)  PbSO4(s) + 2 H2O (l)


(4): Add H+

Cu+(aq)   Cu2+(aq)

4H+ (aq) + PbO2(s) + SO42−(aq)  PbSO4(s) + 2 H2O (l)    


(5): Balance charges

Cu+(aq)   Cu2+(aq) + 1 e-

2e- + 4H+ (aq) + PbO2(s) + SO42−(aq)  PbSO4(s) + 2 H2O (l)

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Problem Details

Balance each skeleton reaction, use Appendix D to calculate E°cell, and state whether the reaction is spontaneous:

(a) Cu+(aq) + PbO2(s) + SO42−(aq) ⟶ PbSO4(s) + Cu2+(aq) [acidic]

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