# How to balance more complex redox reactions?

I'm in an advanced level chemistry course, and so far we've been introduced to balancing redox reactions. The basic ones like $$\ce{MnO4- + Cl- -> Mn^2+ + Cl2}\tag1$$ are pretty easy for me, but I'm having trouble with equations where it's not apparent what I exactly need to put in my half reactions. Some examples of these kind of equations are: \begin{align} \ce{Pb + PbO2 + SO4^2- &-> PbSO4}\tag2\\ \ce{Ag(s) + CN- (aq) + O2(g) &-> [Ag(CN)2]- (aq)}\tag3\\ \ce{Cu + HNO3(aq) &-> Cu(NO3)2(aq) + NO2(g) + H2O(l)}\tag4 \end{align}

I'm not looking for solutions to any of these problems. I'm just looking for ways to correctly construct my half reactions before balancing, adding water and hydrogen ions, and balancing the charges.
Like in the first reaction on the reactants side, the first $$\ce{Pb}$$ has a charge of $$0$$ while the next one has a charge of $$+4$$. The $$\ce{Pb}$$ on the products side however has a charge of $$+2$$. It seems like one is being oxidized and the other reduced. And I'm also not sure what to do with the sulfate.

• Your first reaction is an example of a comproportionation reaction. These are a little tricky to recognize and balance--you're absolutely correct in saying that one is oxidized and the other reduced. Sep 8 '15 at 18:35

Once you know the method, balancing any redox equation that you will get in AP chem will be quite easy. All you have to do is follow this simple method:

1. Balance all the elements in the equation except for $$\ce{O}$$ and $$\ce{H}$$. For polyatomic ions that are spectator ions, just think as them as one big element.
2. Check if $$\ce{O}$$ are balanced. If they aren't, add water molecules to the side, witch doesn't have enough $$\ce{O}$$.
3. Check if $$\ce{H}$$ are balanced. If they aren't, add $$\ce{H+}$$ ions to the side, witch doesn't have enough $$\ce{H}$$.
4. Now that all the elements are balanced, make sure that the charges are balanced. If not balanced, add electrons to the side which is more positive.
5. Check your equation now to make sure that everything is balanced correctly.

I will do the first equation as an example for you. You should be able to balance rest of the equations by following this procedure.

1. Balanced elements except for O and H: Here the sulfate ion is a spectator ion, so you can imagine it as one big element: $$\ce{Pb + PbO2 + 2SO4^2- -> 2PbSO4 }$$
2. Balance oxygen: $$\ce{Pb + PbO2 + 2SO4^2- -> 2PbSO4 + 2H2O }$$
3. Balance hydrogen: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -> 2PbSO4 + 2H2O }$$
4. Balance charge: charge is already balanced. Don't need to add electrons.
5. Check equation to make sure everything is balanced correctly: all the elements and charges are balanced correctly.

So your balanced redox equation for the first example is: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -> 2PbSO4 + 2H2O }$$

Note: If the question asks to provide the redox equation in basic conditions there are two more steps.

1. Add the same number of $$\ce{OH-}$$ ions as $$\ce{H+}$$ to the products and reactants. The $$\ce{OH-}$$ ions and $$\ce{H+}$$ will react to form water.
2. Cancel out the water molecules from both sides (a bit like algebra).

That should give the redox equation in basic conditions.

• Great answer. Balancing reactions is just a way to express that mass, charge, and moles of .atoms are conserved Sep 9 '15 at 2:41
• What do we do with the leftover electrons? May 28 '20 at 15:03
• @Aditya From the equations in the question, you should not get leftover elections since these represent redox reactions. However, if you were asked to write the half-equation for an oxidation or reduction equation, then it is perfectly fine to have leftover elections, in fact you must have leftover elections, otherwise you are doing something wrong, E.g. half-equation for oxidation of iron is: $\ce{Fe} \rightarrow \ce{Fe}^{2+} + 2e^-$. May 28 '20 at 21:22
• @Aditya In that case, you will need to use the above process to write the 2 half-equations (which should have electrons on the opposite side). Once you have done that, you will need to multiply each half-equation by an appropriate number so that when you had the 2 half-equations, the electrons cancel each other out. May 30 '20 at 11:57
• A quick example to illustrate: Consider the reduction of lead ions by aluminium. The oxidation and reduction half-equations are respectively, $\ce{Pb}^{2+} + 2e^- \rightarrow \ce{Pb}$ and $\ce{Al} \rightarrow \ce{Al}^{3+} + 3e^-$. As you can see, the electrons are on the opposite sides as expected. However we cannot add the 2 half-equations right away since the electrons will not cancel out. Instead, we need to multiply the first half-equation by $3$ and the second half-equation by $2$ so that the number of electrons for both half-equations are $6$. Now you can add them together. May 30 '20 at 11:59