How do you balance equations using the half reaction method?

Question

I can do the easy, obvious ones but the more complicated ones stump me. Here is an example of one I can't get the answer to:

1) CrI3 Cl2 OH- H2O --> (CrO4)2- IO4- Cl- OH- H2O

If you could show me how to do this one in explaining, that would be great. Thanks!

Answer 1

CrI3 (has Cr+3 )   --> (CrO4)2-  (has Cr+6) & 3 e- lost

CrI3 (has 3 I's @ -1 ea) -->3 IO4- (has 3 I @ +7 each) 24 e- lost

that's a total of 27 electrons lost

Cl2 ( has Cl @ zero each)  takes 2 e- --> 2Cl-  ( has Cl @ -1 each)

to balance the electrons taken & lost, we double the top & 27 times  the bottom:

2 CrI3  --> 2 (CrO4)2-   & 6 IO4-  & 54 e- lost

27 Cl2  takes 554 e- --> 54Cl-

now combine the two:

2CrI3 & 27 Cl2  --> 2 (CrO4)2- & 54 Cl- & 6 (IO4)-

there is an excess of 64 "-" on the right, to balance the charges  we add 64 OH- 's to the left side:

2 CrI3 & 27Cl2 & 64 OH- --> 2 (CrO4)2- & 54 Cl- & 6 (IO4)-

to use up the 64 H's in the 64 OH- on the left, we will make 32 H2O's on the right:

2 CrI3 & 27Cl2 & 64 OH- --> 2 (CrO4)2- & 54 Cl- & 6 (IO4)- & 32 H2O

everything balances

54 e- taken = 54 e- lost

64 "-" charges = 64 "-" charges

2 Cr & 6 I's = 2 Cr & 6 I's

54 Cl's = 54 Cl's

64 H's & 64 O's =  64 H's & 64 O's

Answer 2

CrI3 + Cl2 + OH- + H2O --> CrO4^2- + IO4- + Cl- + OH- + H2O

start by assigning oxidation states to each element.

Cr is oxidized from +3 to +6

iodide is oxidized from -1 to +7

chlorine is reduced from 0 to -1

With basic solution we use a different "trick" to balance O and H.  Instead of balancing oxygen with water, as when it is acidic, in a basic solution you balance oxygen with twice as many OH- as you need, then balance the hydrogen with water.

In this case we have an oxidation half-reaction that contains two elements which are being oxidized, Cr and I.  Write the two half-reactions and balance them, then multiply each half reaction by coefficients which will make the number of electrons gained and lost the same.  Add the two half-reactions and you're done.   But not to trivialize it, this is a difficult problem, and about as hard as they get.

2(CrI3  +  32OH-  --->  CrO4^2-  +  3IO4- + 16H2O + 27e-

27(Cl2 + 2e- --> 2Cl-

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2CrI3 + 64OH- + 27Cl2 --> 2CrO4^2- + 6IO4- + 54Cl- + 32H2O