Number of electrons to balance redox reaction

[terryds]

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

 

[Borek]

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^-

 

[terryds]

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

 

[Borek]

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.

 

[terryds]

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??

 

[Borek]

Yes, that's correct.

I₂ + 12OH^- → 2IO₃^- + 6H₂O + 10e^-