Number of electrons to balance redox reaction

In summary, to balance the equation, we must first balance the atoms and then add enough electrons to balance the charge. In this case, 10 electrons are needed on the right side of the reaction.
  • #1
terryds
392
13

Homework Statement



What number of electrons to balance the equation
##I_2(s)+OH^-(aq)\rightarrow IO_3^-(aq)+H_2O(l)##

The Attempt at a Solution


I see that the oxidation state of I increases from 0 to +5
But, I don't see any reduction at the reaction..
So, I don't understand how to balance the reaction
And I don't know how to get the number of required electrons to balance the reaction
 
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  • #2
There is no reduction because this is only a half reaction, where electrons are listed as a reagent. Just like Fe(II) to Fe(III) oxidation can be written as

Fe2+ → Fe3+ + e-
 
  • #3
Borek said:
There is no reduction because this is only a half reaction, where electrons are listed as a reagent. Just like Fe(II) to Fe(III) oxidation can be written as

Fe2+ → Fe3+ + e-

So, the number of electron required to balance the equation is 5 electron, right? Since the oxidation state increases by 5
 
  • #5
Borek said:
No, it is not that easy. First, balance atoms - they are not balanced at the moment. After that, add enough electrons to balance the charge. See examples here: http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method

Don't look at oxidation numbers at all.

Balance all atoms
##I_2(s)+12OH^-(aq)\rightarrow 2IO_3^-(aq)+6H_2O(l)##charges in left : -12
charges in right: -2

So, there should be 10 electrons added to the right side of reaction, right??
 
  • #6
Yes, that's correct.

I2 + 12OH- → 2IO3- + 6H2O + 10e-
 

1. How do you balance a redox reaction using the number of electrons?

To balance a redox reaction using the number of electrons, you need to first identify the oxidation and reduction half-reactions. Then, determine the change in oxidation numbers for each element in the reaction. The difference in oxidation numbers will give you the number of electrons needed for the reaction to be balanced. Add this number to the side of the reaction with the lower number of electrons and multiply the half-reactions by the appropriate coefficient to balance the number of electrons on both sides.

2. Can the number of electrons be fractional in a balanced redox reaction?

Yes, the number of electrons can be fractional in a balanced redox reaction. This can occur when the change in oxidation numbers for the elements is not a whole number. In this case, the number of electrons can be expressed as a fraction to balance the reaction.

3. What is the purpose of balancing the number of electrons in a redox reaction?

The purpose of balancing the number of electrons in a redox reaction is to ensure that the reaction follows the law of conservation of mass. This means that the number of atoms of each element on both the reactant and product sides of the reaction must be equal. Balancing the number of electrons also ensures that the overall charge of the reaction remains neutral.

4. Is it necessary to balance the number of electrons in both the oxidation and reduction half-reactions?

Yes, it is necessary to balance the number of electrons in both the oxidation and reduction half-reactions. This is because the overall redox reaction involves the transfer of electrons from the oxidation half-reaction to the reduction half-reaction. If the number of electrons is not balanced in both half-reactions, the reaction will not be balanced as a whole.

5. How do you know when the number of electrons in a redox reaction is balanced?

You know that the number of electrons is balanced in a redox reaction when the total number of electrons on both sides of the reaction is equal. Additionally, the change in oxidation numbers for each element should also be balanced. This means that the same number of electrons must be gained or lost by each element in the reaction. A balanced redox reaction should also have the same number of atoms of each element on both sides of the reaction.

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