# Balancing redox reactions

This has gotten me confused since long. Not all equations have clean half- reactions, some cations are in more than one compound while some anions are in some other compounds. Some reactions don't even have any charges and just look like normal equations (if you don't separate ions), should we just balance them like normal reactions then? That takes too much time and we don't even know which side to put $$\ce{H2O}$$ on.

Also when can you cancel a spectator ions? Beacuse we can't always be sure that after balancing rest of reaction the spectator ions will fit in nicely again without any disturbance to equation.

Can someone give a strategy to balance any and all redox reactions, using this one as an example- $$\ce{KIO3 + KI + HCl -> KCl + H2O + I2}$$

(I have seen other answers on stackexchange but none tells a method for all reactions I have doubts in. Some just aren't balanced by the methods I have read).

• Do you know how to calculate the $n$ or the $z$ factor of the reaction? And the change in the oxidation states of the reactant species? Mar 28 at 5:53
• Yes I know that. But what after that? Mar 28 at 5:59
• Then there is this shortcut method, of "cross multiplication" of the change in oxidation state. For example, the change in oxidation state of $\ce{I}$ of $\ce{KIO_3}$ is +5, so write 5 in front of $\ce{KI}$ and since there is a change of 1 in oxidation state of $\ce{I}$ in $\ce{KI}$, then write 1 in front of $\ce{KIO_3}$, (which means write nothing). Now, if you see, writing 6 infront of $\ce{KCl}$ and 3 in front of $\ce{I_2}$ does the job. This method is simply based upon writing the individual half reactions and then multiplying by a factor to cancel the electrons. Mar 28 at 6:05
• 1 pack contains n sweets. 1 child requests m sweets. You need m/n * o packs for o children. Many things are trivially easy, if you can forget it is about chemistry. // Spectator ions and molecules do just fine outside the equation. Involve only what participates. K+ does not take part in redox reactions in water solutions. There is no KI nor KIO3 nor HCl nor KCl in solutions. There are just hydrated ions. Mar 28 at 6:48

The usual way of solving such a problem is to write two half-equations, and make the sum of them, after multiplying each one by a factor so as to get the same number of electrons. Each equation is about one of the atoms that changes its oxidation number.

Let's start from the first such atom, iodine in the ion iodate $$\ce{IO3-}$$. We will build its own half-equation, step by step. It is a long process, made of four steps : balancing the atom that changes its oxidation number, balancing then the oxygen, then the hydrogen, then the charges.

First half-equation.

Here iodine goes from iodate ion $$\ce{IO3-}$$ whose oxidation number is $$+5$$ to zero in $$\ce{I2}$$. Write iodate on the left and iodine on the right, like $$\ce{IO3- + ... -> I2 + ...}$$ Now manage first to get the same number of atoms changing of oxidation number before going further. This gives : $$\ce{2IO3- + .. -> I2 + ...}$$ When it is done, take care of the oxygen atoms, and add enough $$\ce{H2O}$$ molecules to compensate the excess of oxygen atoms present on one side of the equation. Here there are $$2·3$$ oxygen atoms on the left. So add $$\ce{6H2O}$$ on the right. this makes $$\ce{2 IO3- + ... -> I2 + 6 H2O + ..}$$ Now take care of the $$\ce{H}$$ atoms, and add enough $$\ce{H+}$$ ions on the side where $$\ce{H}$$ atoms are missing. Here $$\ce{12H}$$ are missing at left. This gives : $$\ce{2IO3- + 12 H+ + ... -> I2 + 6H2O + ...}$$ Now all atoms are balanced. The last thing to do is balance the charges. There are none on the righthand side. But on the left there $$12+$$ and $$2-$$ This makes $$+10$$. 10 electrons are needed on the left to get the same total charges equal on both sides of the equation. This gives the final half-equation $$\ce{2IO3- + 12H+ + 10 e- -> I2 + 6 H2O}$$

Second half-equation.

Here it is quicker, because iodine ion $$\ce{I-}$$ yields $$\ce{I2}$$ without bringing any hydrogen or oxygen atom. The four steps gives the end half-equation $$\ce{2 I- -> I2 + 2 e-}$$

Here the two equations must be added in such a way as to make electrons disappear. So the second half-equation must first be multiplied by $$5$$ to get the same number of electrons at left and at right. Adding the first half-equation plus $$5$$ times the second gives $$\ce{2IO3- + 12 H+ + 10 e- + 10 I- -> I2 + 6 H2O + 5 I2 + 10 e-}$$ which can be simplified and gives the final redox equation : $$\ce{2IO3- + 12 H+ + 10 I- -> 6I2 + 6 H2O }$$
$$12$$ spectator $$\ce{K+}$$ cations must still be added on both sides to compensate the $$\ce{IO3-}$$ and $$\ce{I-}$$ ions at left. Then $$\ce{12 Cl-}$$ are to be added to compensate for the $$\ce{H+}$$ ions at left. This gives $$\ce{2KIO3 + 10 KI + 12 HCl -> 6I2 + 6 H2O + 12 KCl }$$ which can still be divided by $$2$$ yielding the final equation $$\ce{KIO3 + 5 KI + 6 HCl -> 3I2 + 3 H2O + 6 KCl }$$