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Balancing Redox Equations Using Oxidation Numbers

  • #1
Question
Balance the following redox using oxidation number. Please show all steps.

NaIO3 + NaI + HCl -> NaCl + I2 + H2O


Attempted Solution
So first I assigned oxidation number:
NaIO3 + NaI + HCl -> NaCl + I2 + H2O
1+ 5+ 2- | 1+ 1- | 1+ 1- | 1+ 1- | 0 | 1+ 2-

Then, I am not sure about this step, I know that when an element loses electrons it is oxidized and thus that compound is the reducing agent. Similarly, the element that gains electron reduced and thus that compound is the oxidizing agent.

Therefore, if oxidation number goes up, it is an oxidization. If oxidatin number goes down, it is reduction.

Now, here comes the problem, the oxidation number of Na, O, H, and Cl did not change during the reaction, only I changed. So is I2 both part of the reduction and oxidation?

I would like to clarify this before moving forward in balancing this question. Any help is strongly appreciated. Thanks in advance.
 

Answers and Replies

  • #2
Borek
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Yes, iodine gets both oxidized and reduced.
 
  • #3
148
2
Question
Balance the following redox using oxidation number. Please show all steps.

NaIO3 + NaI + HCl -> NaCl + I2 + H2O


Attempted Solution
So first I assigned oxidation number:
NaIO3 + NaI + HCl -> NaCl + I2 + H2O
1+ 5+ 2- | 1+ 1- | 1+ 1- | 1+ 1- | 0 | 1+ 2-

Then, I am not sure about this step, I know that when an element loses electrons it is oxidized and thus that compound is the reducing agent. Similarly, the element that gains electron reduced and thus that compound is the oxidizing agent.

Therefore, if oxidation number goes up, it is an oxidization. If oxidatin number goes down, it is reduction.

Now, here comes the problem, the oxidation number of Na, O, H, and Cl did not change during the reaction, only I changed. So is I2 both part of the reduction and oxidation?

I would like to clarify this before moving forward in balancing this question. Any help is strongly appreciated. Thanks in advance.
You are quite right -- I is the only element that changes its oxidation number in this reaction. The key to the solution is that on the reactant side of the equation there are two different I-containing substances with two different oxidation numbers for I.
 
  • #4
Yes, iodine gets both oxidized and reduced.
You are quite right -- I is the only element that changes its oxidation number in this reaction. The key to the solution is that on the reactant side of the equation there are two different I-containing substances with two different oxidation numbers for I.
Thanks for the quick reply Borek and JohnRC!

Just as I thought so, so I shall proceed...

NaIO3 + NaI + HCl -> NaCl + I2 + H2O
1+ 5+ 2- | 1+ 1- | 1+ 1- | 1+ 1- | 0 | 1+ 2-

As shown from the oxidation number, I in NaIO3 has decreased its oxidation number from 5+ to 0, that means a reduction occurred and NaIO3 is the oxidizing agent. On the other hand, the I in NaI has increased its oxidation number from 1- to 0, that means an oxidization occurred and NaI is the reducing agent.

To summarize:
NaIO3 ---RED: gain 5e-/I, gain 5e-/NaIO3---> I2
NaI ---OX: lost 1e-/I, lost 1e-/NaI---> I2

(I am going in syllabus here as I only started learning this topic and would like to make things clear for myself, so please bear with me)

Now according to the Electron Transfer Theory (ETT), the number of electrons lost must equals to the number of electrons gained, as such, a coefficient must be used so that each elements within the compounds that are involved in REDOX are balanced. In this case, the number would be 5.

Two questions at this step:
1) How and when do I apply this ETT coefficient, is it a must or only when it is necessarily? Do I have to apply ETT on both side of the "bridge" always? My teacher had an example that only apply the coefficient on one side of the "bridge". (By bridge I mean both on the reactant side and the product side)
2) What if the electron lost already equal to the electron gained? Is there still a ETT coefficient that I should be aware of? Example: Cu + H2SO4 -> CuSO4 + H2O + SO2
 
  • #5
AGNuke
Gold Member
455
9
1) If you want to balance a Redox equation, you must balance the electron transfer (charge transfer). This is done by equating half reactions and multiply them.

Then we must balance the H+ or OH- too, depending on the reaction medium. And then we can balance the rest of the equation stoichiometrically.

2) If the electron transfer is equal in both the half reactions, they just cancel out. See in your example,[tex]Cu \rightarrow Cu^{2+} + 2e^-[/tex][tex]SO_4^{2-} + 2e^- + 4H^+ \rightarrow SO_2 + 2H_2O[/tex][tex]Cu+SO_4^{2-}+4H^+\rightarrow Cu^{2+}+2H_2O+SO_2[/tex][tex]Cu+2H_2SO_4\rightarrow CuSO_4+2H_2O+SO_2[/tex]
 
  • #6
1) If you want to balance a Redox equation, you must balance the electron transfer (charge transfer). This is done by equating half reactions and multiply them.

Then we must balance the H+ or OH- too, depending on the reaction medium. And then we can balance the rest of the equation stoichiometrically.

2) If the electron transfer is equal in both the half reactions, they just cancel out. See in your example,[tex]Cu \rightarrow Cu^{2+} + 2e^-[/tex][tex]SO_4^{2-} + 2e^- + 4H^+ \rightarrow SO_2 + 2H_2O[/tex][tex]Cu+SO_4^{2-}+4H^+\rightarrow Cu^{2+}+2H_2O+SO_2[/tex][tex]Cu+2H_2SO_4\rightarrow CuSO_4+2H_2O+SO_2[/tex]


Thanks for replying AGNuke.
From my notes, I was told to use coefficients based on the electron transfer theory so that the number of electrons lost equals the number of electrons gained. So in the case of the above reaction, would one simply multiply 5 on both NaI and I2? But then this would cause imbalance on the number of I atoms (1+5 on the reactant side vs 5x2 on the product side) Also doesn't the compound involved in the REDOX reaction needs to be balanced? In other words, NaIO3 needs to balance with I2 and NaI needs to balance with I2?

As for 2) Can you elaborate on how they cancel?

I apologize for any inconvience this may have caused.
 
  • #7
Borek
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28,296
2,681
1) How and when do I apply this ETT coefficient, is it a must or only when it is necessarily? Do I have to apply ETT on both side of the "bridge" always? My teacher had an example that only apply the coefficient on one side of the "bridge". (By bridge I mean both on the reactant side and the product side)
2) What if the electron lost already equal to the electron gained? Is there still a ETT coefficient that I should be aware of? Example: Cu + H2SO4 -> CuSO4 + H2O + SO2
You want number of electrons lost to equal number of electrons gained. Simplest approach is to cross multiply - that is, multiply each half reaction by the number of electrons exchanged in the other reaction.

Say reduction consumes 2 electrons, oxidation produces 3 - you multiply reduced substance by 3 and oxidized by 2, and 2*3=3*2.

In the reaction between iodate and iodine you have 5 electrons involved in the reduction and 1 electron involved in oxidation - that means you multiply iodate by 1 and iodide by 5. 5*1=1*5 - technically multiplying by 1 is equivalent to no multiplication, so you can assume iodate was taken as it was.

But there is a trap in this approach. Imagine reduction takes 2 electrons, and oxidation produces 4. Cross multiplication gives 2*4=4*2 - but it will produce a balanced equation that is slightly incorrect, as coefficients in a balanced reaction should be the smallest possible. So in this case it is enough to multiply reduced part by 2. Note that it is not an error to use cross multiplication at this stage, you just have to remember to check in the final step if the reaction is balanced using the smallest coefficients. Which is a step you should do always.
 
  • #8
AGNuke
Gold Member
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Look at the first reaction, it is the Oxidation half reaction; 2 electrons liberated.
Look at the second reaction, it is the Reduction half reaction; 2 electrons consumed.

Add them, get the Redox reaction. 2 electrons on LHS, 2 electrons on RHS, they cancel out.

In your original question, it is true that you multiply the oxidation half reaction by 5, but you must also balance it first. I'll do it[tex]\left[NaI \rightarrow Na^++\frac{1}{2}I_2+e^-\right]\times 5[/tex][tex]HIO_3+5e^-+5H^+\rightarrow \frac{1}{2}I_2+3H_2O[/tex][tex]5NaI+HIO_3+5HCl\rightarrow 5NaCl + 3I_2 + 3H_2O[/tex]

You see, the source of H+ in the second equation was HCl, and I added 5Cl- in the first reaction, so that it can form NaCl with Na+.

We first balance the electron transfer, then other elements (Na and Cl) then last, Hydrogen and Oxygen.
 
  • #9
Borek
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AGN, what you did is technically a half reaction method, which doesn't require use of oxidation numbers (which is why it is better if you ask me). When using ON we usually use them to find initial ratio of reduced and oxidized substances (in this case: NaIO3 + 5NaI -> 3I2) and then balance everything else.
 
  • #10
AGNuke
Gold Member
455
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Just explaining how electrons cancel out, and he was having some problems in stoichiometry as well, considering he said 6 I on one side and 5 I2 on the other.

I myself use my own short-hand trick based on oxidation numbers to solve these problems. Can be messy but really fast though
 
  • #11
Thanks for all the help, I really appreciated it!
I am now able to solve this problem.

NaIO3 + NaI + HCl -> NaCl + I2 + H2O
1+ 5+ 2- | 1+ 1- | 1+ 1- | 1+ 1- | 0 | 1+ 2-

which I will rewrite as:
NaIO3 + NaI + HCl -> NaCl + I2 + I2 + H2O

Reduction : gain 5e-/NaIO3
Oxidation : lost 1e-/NaI

Electron Transfer Theory, Lowest Common Multiple = 5
NaIO3 + 5NaI + HCl -> NaCl + (1/2)I2 + (5/2)I2 + H2O

Balance metals and non-metals and simplify
NaIO3 + 5NaI + 6HCl -> 6NaCl + 3I2 + H2O

Balance oxygen and hydrogen
O: 3 vs 1
NaIO3 + 5NaI + 6HCl -> 6NaCl + 3I2 + 3H2O

H: 6 vs 6
No changes necessarily.

Finally, check if atomically and electronically balanced.
NaIO3 + 5NaI + 6HCl -> 6NaCl + 3I2 + 3H2O

Oxidation Number : 0 vs 0

Balanced.
 
  • #12
Borek
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