Answers to Ten More Examples
Half-reactions: basic solution

Return to balancing half-reactions in base solution

Return to Redox Menu


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¯


Return to balancing half-reactions in base solution

Return to Redox Menu