# Number of electrons to balance redox reaction

1. Nov 29, 2015

### terryds

1. The problem statement, all variables and given/known data

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

3. 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

2. Nov 29, 2015

### Staff: Mentor

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. Nov 29, 2015

### terryds

So, the number of electron required to balance the equation is 5 electron, right? Since the oxidation state increases by 5

4. Nov 29, 2015

### Staff: Mentor

5. Nov 29, 2015

### terryds

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. Nov 29, 2015

### Staff: Mentor

Yes, that's correct.

I2 + 12OH- → 2IO3- + 6H2O + 10e-