To balance the equation for the oxidation of the hydroperoxide ion (HO2^-(aq)) to oxygen gas (O2(g)) by permanganate ion (MnO4^-(aq)) in a basic solution, follow these steps:
Step 1: Write the unbalanced chemical equation:
HO2^-(aq) + MnO4^-(aq) → MnO2(s) + O2(g)
Step 2: Balance the atoms, excluding oxygen and hydrogen:
The manganese (Mn) is balanced initially, as there is only one Mn atom on each side of the equation.
MnO4^-(aq) → MnO2(s) + O2(g)
Step 3: Balance the oxygen atoms by adding water (H2O):
Since we are working in a basic solution, we can add OH^-(aq) ions to balance the hydrogen from the water.
MnO4^-(aq) + H2O(l) → MnO2(s) + O2(g)
Step 4: Balance the hydrogen atoms by adding hydroxide (OH^-(aq)):
For every hydrogen atom, add one OH^-(aq) ion on the other side of the equation.
MnO4^-(aq) + H2O(l) → MnO2(s) + O2(g) + OH^-(aq)
Step 5: Balance the charge by adding electrons (e^-):
Add electrons to the side with the greatest positive charge, which in this case is the left side.
MnO4^-(aq) + 4OH^-(aq) + 3e^- → MnO2(s) + 2H2O(l) + OH^-(aq)
Step 6: Balance the two-sided equation:
Multiply each half-reaction by a factor to make the number of electrons equal on both sides. In this case, we need 6 electrons.
2MnO4^-(aq) + 8OH^-(aq) + 6e^- → 2MnO2(s) + 4H2O(l) + 3OH^-(aq)
Step 7: Simplify the equation:
Combine the excess OH^-(aq) ions on the right side of the equation.
2MnO4^-(aq) + 8OH^-(aq) + 6e^- → 2MnO2(s) + 4H2O(l) + 2OH^-(aq)
Step 8: Remove the spectator ions:
Remove the excess OH^-(aq) ions on both sides of the equation.
2MnO4^-(aq) + 6e^- → 2MnO2(s) + 4H2O(l)
Step 9: Confirm the equation:
Finally, count the atoms on each side and make sure that they are balanced. In this case, we have:
On the left side: 2 Mn, 4 O, 6 e^-.
On the right side: 2 Mn, 4 O, 4 H, 6 e^-.
The equation is now balanced. The balanced equation for the oxidation of hydroperoxide ion to oxygen gas by permanganate ion in a basic solution is:
2MnO4^-(aq) + 6e^- → 2MnO2(s) + 4H2O(l)