# Balancing Redox Reactions (oxidation number method)

I'm confused about balancing redox reactions using the oxidation number method. I know that in order to get my coefficients for the two reactants I need to find the simplest whole number. However, when I come across certain reactions I'm not getting the same answer as the answer key.

For example, $$\ce{Cu(NH3)4^2+ + S2O4^2- + SO3^2- + Cu + NH3}$$

I know that $$\ce{S2}$$ is oxidized: $$+3 \rightarrow +4$$; Change= -1 but since there are two sulfur I know that I need to multiply -1 by 2, so I get -2 electrons for the reactant $$\ce{S2O4^2-}$$. $$\ce{Cu}$$ is reduced: +2 => 0; Change= +2

My total change would be -2 and 2. If I multiply the two numbers I get 4 and the simplest common factor for 4 is 2 and 2. So my coefficient for $$\ce{Cu(NH3)4^2+}$$ should be a 2 and my coefficient for $$\ce{S2O4^2-}$$ should also be a 2. Right? However, on my answer key both $$\ce{Cu(NH3)4^2+}$$ and $$\ce{S2O4^2-}$$ have a coefficient of 1. I'm not sure how they got that answer.

• Simplest whole number is not a thing at all. You need to find the least common multiple of 2 and 2; what can it be? Not 4, mind you. – Ivan Neretin Nov 20 '18 at 15:12
• Correction the least common multiple. I think I made a mistake, the total change should be -4 since I had to multiply the change for copper (which was 2) and the change for sulfur (which was -2). So shouldn’t the coefficients have been 2 for Cu(NH3)4^2+ and 2 for S2O4^2-? – Cece Nov 20 '18 at 15:31
• Welcome to Chemistry.SE! Please note that formulas can be better expressed with \$\ce{ }\$ for chemical formulas/equations, \$\$ for math term/equations, and \$\pu{ }\$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies. – A.K. Nov 20 '18 at 15:36
• You don't have to multiply 2 and 2. You have to find their least common multiple. – Ivan Neretin Nov 20 '18 at 15:38
• Oh gosh. I’ve been doing that for some of the other reactions and have gotten the correct answer. No wonder I was confused. Thank you Ivan. – Cece Nov 20 '18 at 15:51

It seems that you have the answer, but I'll work it out anyways.

Given the reaction:

$$\ce{Cu(NH3)4^2+ + S2O4^2- -> SO3^2- + Cu + NH3}\tag{1}$$

You can separate it into two half reactions so that you don't have to solve the whole equation at once.

$$\ce{Cu(NH3)4^2+ -> Cu + NH3}\tag{2}$$

$$\ce{S2O4^2- -> SO3^2-}\tag{3}$$

Balance equation (2)

First just balance the stoichiometry... 4 molecules of $$\ce{NH3}$$ are needed

$$\ce{Cu(NH3)4^2+ -> Cu + 4NH3}\tag{4}$$

Now notice that copper is going from $$\ce{Cu2+ -> Cu}$$ so we have to add two electrons on the left to get the charges to balance.

$$\ce{2e- + Cu(NH3)4^2+ -> Cu + 4NH3}\tag{5}$$

That gives the copper half-cell reaction.

Balance equation (3)

First we need to double the $$\ce{SO3^2-}$$ on the right

$$\ce{S2O4^2- -> 2SO3^2-}\tag{6}$$

Notice that copper starts of as the $$\ce{Cu(NH3)4^2+}$$ complex, so the solution is basic. We have more oxygen atoms on the right, so in water you can use the mythical reaction $$\ce{2OH- -> H2O + O}$$ to balance. We need to add $$\ce{4OH-}$$ to the left, and $$\ce{2H2O}$$ on the right to get the stoichiometry right.

$$\ce{S2O4^2- + 4OH- -> 2SO3^2- + 2H2O}\tag{7}$$

Now the atoms balance, but the left side has a -6 charge, but the right has a -4 charge. So we need to add two electrons on the right to balance charges for the equation.

$$\ce{S2O4^2- + 4OH- -> 2SO3^2- + 2H2O + 2e-}\tag{8}$$

That gives the $$\ce{S2O4^2-}$$ half-cell reaction.

Finish

Equation (5) has two electrons on the left, and equation(8) has two electrons on the right. So we can just add those equations and cancel out the common electrons from both sides to get:

$$\ce{Cu(NH4)^2+ + S2O4^2- + 4OH- -> Cu + 4NH3 + 2SO3^2- + 2H2O}\tag{9}$$