I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Report (any type)/Brief report
22/09/2021
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:
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^-##.
Here:
##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:
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+.
The oxidation number of hydrogen in a compound is 1+, except in metal hydrides, such as NaH, where it is -1.
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.
The oxidation number of an atom in uncombined, or elemental form is zero.
For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal zero.
For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion.
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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