Please balance the following equation?? MnO4- + I- = MnO2 + IO3- (basic solution)

The best way to do this is by using the half-reaction method. No matter what redox equation you need balanced, know that if you use the half-reaction method, though it may be a bit more work than other ways, it will always give you the right answer (that is, as long as you do it right).

First we want to split this equation into two relative equations. One will be the oxidation part of the equation, and the other will focus on the reduction.

To identify the oxidation equation you should first write the equation in ionic form to identify which element is being reduced and which is being oxidized. In this case, the Manganese is being reduced, and the Iodine is being oxidized.

To explain this, in the reactants, the Manganese has an oxidation number of +7. It has to have this charge to counteract the 8- from the oxygen (Again, if you need this explained as well, I will include it at the bottom of my answer). The iodine is being oxidized because it goes from having a charge of 1- to a charge of 5+ (It has a 5+ charge due to the 6- charge of the oxygen).

Now that we have done this, the oxidation equation would be:

##I^(-) -> IO_3^(-)##

And the reduction equation would be:

##MnO_4^(-) -> MnO_2##

In this equation, both the reduction and oxidation equations have oxygen in them. Because of this we must add molecules of water to the side of the equation that needs oxygen to balance.

However, once we add water, since water also consists of hydrogen, we must add ##H^(+)## ion to whichever side is lacking them. I think its easier if I just show you:

##3H_2O + I^(-) -> IO_3^(-)##

Here we added three molecules of ##H_2O## because that will give us the three atoms of oxygen that the product side of this equation has. But… we are still missing something. Now the reactant side has six extra hydrogens that the product side does not.

When dealing with this step in the balancing process, it is best to remember:

  • What you do to one side, you must do to the other.

Now that we’ve covered the oxygen, we must cover the hydrogen part of this:

##3H_2O + I^(-) -> IO_3^(-) + 6H^(+)##

One more thing needs to be done here. Ask yourself, is the on both sides of the equation equal? The answer is no, they are not.

To fix this issue, you must add a negative charge to the equation to balance the charges. You do this by adding electrons. For this equation, the left side already has a net charge of 1-. We cannot change this, or make it positive. Remember:

You can only add a negative charge.

This is what you would do:

##3H_2O + I^(-) -> IO_3^(-) + 6H^(+) + 6e^(-)##

Good now both sides have a charge of 1- and the oxidation equation is balanced.

On to reduction: Here you do the same thing. I am just going to give you the balanced equation here, and if you need reference, it is the exact same process that we did to the oxidation part of the equation:

##3e^(-) + 4H^(+) + MnO_4^(-) -> MnO_2 + 2H_2O##

Now that we have both equations balanced, we do something similar to the process of elimination in math equations.

##(3H_2O + I^(-) -> IO_3^(-) + 6H^(+) + 6e^(-))##

##(3e^(-) + 4H^(+) + MnO_4^(-) -> MnO_2 + 2H_2O) * 2##

We multiply the second equation by two so that:

*The electrons on both equations are equal. *

This means that we multiplied by two because the first equation has six electrons while the second only has three. ##(3e^(-) * 2 = 6e^(-))##

Now rewrite what we have:

##3H_2O + I^(-) -> IO_3^(-) + 6H^(+) + 6e^(-)##

##6e^(-) + 8H^(+) + 2MnO_4^(-) -> 2MnO_2 + 4H_2O##

We can now cancel out similarities so we are left with:

##2H^(+) + I^(-) + 2MnO_4^(-) -> 2MnO_2 + IO_3^(-) + H_2O##

Finally, if this were an acidic solution, we would be done and this would be the correct answer. However, we are in a basic solution.

So since we are, we must add ##OH^(-)## ions, or hydroxide ions, to replace the ##H^(+)## ions. We add 2 ##OH^-## ions to both sides of the equation because there are ##2H^+## ions.

When you add them to the reactant side of the equation, you get two molecules of ##H_2O##, and on the other side, you get an additional ##2OH^-##.


##2H_2O + I^(-) + 2MnO_4^(-) -> 2MnO_2 + IO_3^(-) + H_2O + 2OH^-##

The similar waters cancel out and…

##H_2O + I^(-) + 2MnO_4^(-) → 2MnO_2 + IO_3^(-) + 2OH^(-)##

We are now done with the equation, and the above equation would be your final answer. I made sure that everything was balanced out, but if you see a mistake please tell me!:) Hope this helps!

Addition: I told you I would include a segment about the if you didn’t understand. The general rules for oxidation numbers are:

  1. The oxidation number of a monatomic ion is equal in magnitude and sign to its ionic charge. For example, the oxidation number of the bromide ion (Br1-) is 1-. That of the Fe3+ ion is 3+.

  2. The oxidation number of hydrogen in a compound is 1+, except in metal hydrides, such as NaH, where it is -1.

  3. The oxidation number of oxygen in a compound is 2-, except in peroxides, such as H202, where it is -1, and in with the more electronegative fluorine, where it is positive.

  4. The oxidation number of an atom in uncombined, or elemental form is zero.

  5. For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal zero.

  6. For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion.

Calculate Your Essay Price
(550 words)

Approximate price: $22

Calculate the price of your order

550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
The price is based on these factors:
Academic level
Number of pages
Basic features
  • Free title page and bibliography
  • Unlimited revisions
  • Plagiarism-free guarantee
  • Money-back guarantee
  • 24/7 support
On-demand options
  • Writer’s samples
  • Part-by-part delivery
  • Overnight delivery
  • Copies of used sources
  • Expert Proofreading
Paper format
  • 275 words per page
  • 12 pt Arial/Times New Roman
  • Double line spacing
  • Any citation style (APA, MLA, Chicago/Turabian, Harvard)

Our guarantees

Delivering a high-quality product at a reasonable price is not enough anymore.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.

Money-back guarantee

You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.

Read more

Zero-plagiarism guarantee

Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.

Read more

Free-revision policy

Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.

Read more

Privacy policy

Your email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems.

Read more

Fair-cooperation guarantee

By sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language.

Read more

Order your essay today and save 10% with the coupon code: best10