Return to balancing half-reactions in base solution
1) NiO2 ---> Ni(OH)2
Step One: Balance the half-reaction AS IF it were in acid solution: 2e¯ + 2H+ + NiO2 ---> Ni(OH)2
Step Two: Convert all H+ to H2O: 2e¯ + 2H2O + NiO2 ---> Ni(OH)2 + 2OH¯
Step Three: Remove any duplicate molecules or ions: none to be removed in this example. Other problems below will have duplicates.
2) BrO4¯ ---> Br¯
Step One: Balance the half-reaction AS IF it were in acid solution: 8e¯ + 8H+ + BrO4¯ ---> Br¯ + 4H2O
Step Two: Convert all H+ to H2O: 8e¯ + 8H2O + BrO4¯ ---> Br¯ + 4H2O + 8OH¯
Step Three: Remove any duplicate molecules or ions: 8e¯ + 4H2O + BrO4¯ ---> Br¯ + 8OH¯
3) SbO3¯ ---> SbO2¯
2e¯ + H2O + SbO3¯ ---> SbO2¯ + 2OH¯
4) Cu2O ---> Cu
2e¯ + H2O + Cu2O ---> 2Cu + 2OH¯
5) S2O32¯ ---> SO32¯
Step One: 3H2O + S2O32¯ ---> 2SO32¯ + 6H+ + 4e¯ (note the 2 in front of the SO32¯)
Step Two: 6OH¯ + 3H2O + S2O32¯ ---> 2SO32¯ + 6H2O + 4e¯
Step Three: 6OH¯ + S2O32¯ ---> 2SO32¯ + 3H2O + 4e¯
6) Tl+ ---> Tl2O3
6OH¯ + 2Tl+ ---> Tl2O3 + 3H2O + 4e¯
7) Al ---> AlO2¯
4OH¯ + Al ---> AlO2¯ + 2H2O + 3e¯
8) Sn ---> HSnO2¯
3OH¯ + Sn ---> HSnO2¯ + H2O + 2e¯
9) CrO42¯ ---> Cr(OH)3
3e¯ + 4H2O + CrO42¯ ---> Cr(OH)3 + 5OH¯
10) HfO(OH)2 ---> Hf
The results of the "fake acid" method are:4e¯ + 4H+ + HfO(OH)2 ---> Hf + 3H2O
Convert to base with 4 hydroxides on each side; eliminate three water molecules for the final answer:
4e¯ + H2O + HfO(OH)2 ---> Hf + 4OH¯