The following reaction occurs in aqueous acid solution: NO3– +I– →IO3– +NO2
In the balanced equation, using the smallest whole-number coefficients, what is the coefficient of NO3–?
If you run into trouble I will be glad to help you through. Explain exactly what you don't understand.
So, after I did the half-reactions, I still only got a coefficient for NO_3- to be 1, yet in the answer key provided (this is a practice midterm question), it says the coefficient is 6....I'm confused as to how you are supposed to figure that out.
Well, I once heard that NO3– is a bit of a diva. It always wants to be the star of the show, so it demands a coefficient of 1. That way, it can confidently say, "I'm number one!"
To determine the coefficient of NO3– in the balanced equation, we need to balance the equation by ensuring that the number of atoms of each element is the same on both sides of the equation.
Let's start by counting the number of atoms for each element involved in the reaction:
On the reactant side:
NO3–: 1 nitrogen (N) atom, 3 oxygen (O) atoms
I–: 1 iodine (I) atom
On the product side:
IO3–: 1 iodine (I) atom, 3 oxygen (O) atoms
NO2: 1 nitrogen (N) atom, 2 oxygen (O) atoms
Now we need to balance the equation by adjusting the coefficients to make the number of atoms equal on both sides.
Let's start with the nitrogen (N) atoms:
On the reactant side, we have 1 N atom from NO3–, and on the product side, we have 1 N atom from NO2. The nitrogen atoms are already balanced.
Next, let's balance the iodine (I) atoms:
On the reactant side, we have 1 I atom from I–, and on the product side, we have 1 I atom from IO3–. The iodine atoms are also already balanced.
Finally, let's balance the oxygen (O) atoms:
On the reactant side, we have 3 O atoms from NO3–, and on the product side, we have 3 O atoms from IO3– and 2 O atoms from NO2. To balance the O atoms, we need to multiply the NO3– on the reactant side by 3:
NO3– + 3I– → IO3– + NO2
Therefore, the coefficient of NO3– in the balanced equation, using the smallest whole-number coefficients, is 3.
NO3^- + I^- > IO3^- +NO2
N (on the reactant) oxidation# is 5 (on the product) oxidation# is 4
there will be 1 mole of e- on the oxidation 1/2 reaction
I (on the reactant) oxidation# is 1 (on the product) oxidation#is 5
there will be 6 moles of e- on the reduction 1/2 reaction
In order to balance that given reaction, we have to balance the number of electrons on the redox 1/2 reactions.
6(3H^+ + NO3^- > NO2 + e- + 3H2O)
3H20 + 6e- +I^- > IO3^- 3H^+
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15H^+ + 6NO3^- I^- > 6NO2 + IO3^- + 15H2O
Now, in the balanced equation, the coefficient of NO3- is 6.
You need to learn to do these by yourself. Here is a good site that will talk about all of the things you need to know to balance redox equations.
http://www.chemteam.info/Redox/Redox.html
Here is something that will help you get started. N changes from and oxidation state of +5 on the left to +4 on the right; I changes from -1 on the left to +5 on the right.