The hydroperoxide ion, HO2–(aq), reacts with permanganate ion, MnO4–(aq) to produce MnO2(s) and oxygen gas. Balance the equation for the oxidation of hydroperoxide ion to O2(g) by permanganate ion in a basic solution.
Why are the 3e + 2H2O added ?
Well, chemistry and humor are a great mix! Let's balance this equation in a fun and witty way.
First, let's start by writing the unbalanced equation:
HO2– + MnO4– → MnO2 + O2
Now, let's get this equation balanced, and make some chemical magic happen:
2 HO2– + MnO4– → MnO2 + 2 O2
Voila! We've achieved chemical equilibrium in a positively hilarious way. Just watch out for those flying oxygen molecules – they can be quite breathtaking!
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)
To balance the equation for the oxidation of the hydroperoxide ion (HO2-) to oxygen gas (O2) by the permanganate ion (MnO4-) in a basic solution, you need to follow these steps:
Step 1: Separate the equation into half-reactions.
Consider the reaction as an oxidation-reduction (redox) reaction consisting of two half-reactions: the oxidation half-reaction and the reduction half-reaction.
Oxidation half-reaction: HO2- → O2
Reduction half-reaction: MnO4- → MnO2
Step 2: Balance the atoms other than hydrogen and oxygen in each half-reaction.
Start by balancing the atoms other than hydrogen and oxygen in each half-reaction.
In the oxidation half-reaction, the hydroperoxide ion (HO2-) is already balanced.
In the reduction half-reaction, we need to balance the manganese atoms by adding water molecules.
Reduction half-reaction:
MnO4- → MnO2
Step 3: Balance the oxygen atoms by adding water molecules.
Balance the oxygen atoms in each half-reaction.
Oxidation half-reaction:
HO2- → O2 + H2O
Reduction half-reaction:
MnO4- + H2O → MnO2
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+).
Balance the hydrogen atoms in each half-reaction.
Oxidation half-reaction:
HO2- → O2 + H2O + 2H+
Reduction half-reaction:
MnO4- + H2O → MnO2 + 4H+
Step 5: Balance the charges by adding electrons.
To balance the charges, add electrons (e-) to each half-reaction.
Oxidation half-reaction:
HO2- + 2H+ → O2 + H2O + 2e-
Reduction half-reaction:
MnO4- + H2O + 4e- → MnO2 + 4H+
Step 6: Multiply each half-reaction by an appropriate factor to make the number of electrons gained equal to the number of electrons lost.
To make the number of electrons gained equal to the number of electrons lost, multiply the oxidation half-reaction by 4 and the reduction half-reaction by 2.
Oxidation half-reaction:
4HO2- + 8H+ → 4O2 + 4H2O + 8e-
Reduction half-reaction:
2MnO4- + 2H2O + 8e- → 2MnO2 + 8H+
Step 7: Combine the half-reactions and cancel out common terms.
Add the two multiplied half-reactions together and cancel out common terms.
Final balanced equation in basic solution:
4HO2- + 8H+ + 2MnO4- → 4O2 + 4H2O + 2MnO2
HO2^- = + OH^- => O2 + 2e + H2O
MnO4^- + 3e + 2H2O==> MnO2 + 4OH^-
Here are the two balanced half equations. Multiply eqn 1 by 3 and multiply eqn 2 by 2 then add the two equations. Finally, cancel anything that appears on both sides of the final equation. Post your work if you get stuck.