Can you balance this redox equation in basic solution?

Question

Pb(OH)4^2- ClO^- --> PbO2 Cl^-

Answer 1

First, find which species are being oxidized and reduced.  Determine the charge on each atom, and see if it changes for any of the atoms when going from reactants to products.  If the oxidation state increases, it is being oxidized, and if it decreases, it is being reduced.

Here, you should see that Pb is being oxidized from 2+ to 4+ and that Cl is being reduced from Cl+ to Cl-.  A shortcut to get this (rather than finding all of the oxidation states) is that compounds with halogens and oxygen are usually oxidizers (as are permanganates, chromates, and peroxides, among others).

Next, we make half reactions - don't bother balancing the O and H atoms yet:

Pb(OH)4 2- ------> PbO2 + 2e-

ClO- + 2e- ------> Cl-

Since we are in basic solution, we can add either water or OH- ions to the products side to make the O and H balance.  I like to think of this as follows; bear in mind that this is not actually what happens, but it helps me balance these equations.  Two hydroxide ions can react to make an oxygen ion and water.  A water can react with an oxygen ion to form 2 hydroxides.  So when you must get rid of an oxygen, add water, and when you must form an oxide ion, add hydroxide.

Pb(OH)4 2- -------> PbO2 + 2e- + 2H2O

Here, the hydroxide ions are already attached to the lead compound; we don't need to add any more.  This supplies the oxide ions for the PbO2, making water as well.

H2O + ClO- + 2e- -------> Cl- + 2OH-

We have an extra oxide ion on that chlorine atom, so add water to form two hydroxides instead.

Now add your two half reactions such that the electrons will cancel out.  You may need to multiply one or both equations by a factor to get the total number of electrons involved in each half reaction to be the same.  Here, we see that there are 2 electrons in both of the reactions, so we can just add them:

Pb(OH)4 2- + ClO- + 2e- + H2O ------> PbO2 + 2H2O + Cl- + 2OH- + 2e-

We can see that there are the same number of electrons on either side, so these can cancel.  In addition, there is one H2O as a reactant, and 2 as a product: cancel the one on the reactant side with one of the product ones to leave only 1 H2O as a product.  The final, balanced equation should be:

ClO- + Pb(OH)4 2- ------> PbO2 + Cl- + H2O + 2OH-