How to balance the reaction between ethanol and chlorine using oxidation number method?

Question

Balance the following chemical equation using oxidation number method:

$$\ce{C2H5OH + Cl2 -> CHCl3 + HCO2- + Cl- + H2O}~\text{(In basic medium)}$$

Chlorine in $\ce{Cl2}$ is in oxidation state $0$ and it is changing from $0$ to $-1$ in both $\ce{CHCl3}$ and $\ce{Cl-}.$

$\ce{C2H5OH}$ has two carbons and both of them have different oxidation states: one has $-3,$ the other has $-1,$ and they are changing to $+2$ in both the products $\ce{CHCl3}$ and $\ce{HCO2-}.$

So, it is definitely a redox reaction and thus I want to use oxidation number method to balance this equation.

I am finding it hard to balance because of multiple products of $\ce{C}$ and $\ce{Cl}$ in the same equation.

Answer 1

This problem is rather difficult to solve, because some hypotheses have to be made, as will be explained in the following text.

Let's first admit that, in the molecule $\ce{C2H5OH}$, the two carbon atoms are, on average, at oxidation state $-2$. It helps solving the problem. By chance, these carbon atoms are going to the same oxidation state $+2$ at the end, both in chloroform $\ce{CHCl3}$ and in the formate ion $\ce{HCOO^-}$. So each carbon atom has a leap of $+4$ in their oxidation number. In total it makes a leap of $+8$. And this may be carried out by $\ce{Cl2}$ if $4~\ce{Cl2}$ do the job, introducing $3$ $\ce{Cl}$ atoms in the $\ce{CHCl3}$ so that the $5$ remaining chlorine atoms are transformed into chloride ions. We also have to admit here that the first carbon atom will always produce $\ce{CHCl3}$ and that the second carbon atom will always producing the $\ce{HCOO-}$ ion. As a consequence the relative amount of $\ce{CHCl3}$ and $\ce{HCOO-}$ are equal. At this step the equation is $$\ce{C2H5OH + 4 Cl2 + ... -> CHCl3 + HCOO- + 5 Cl- + ...}$$ The problem of the oxidation is solved, but some oxygen, hydrogen, and charges are still missing. Six OH- ions have to be added at left to satisfy the stoichiometry. It gives the final equation $$\ce{C2H5OH + 4 Cl2 + 6 OH- -> CHCl3 + HCOO- + 5 Cl- + 5 H2O}$$ I would be interested to know if this problem can be solved without the present hypotheses.

Answer 2

Using whole molecules as oxidizing or reducing agents is my favorite method for dealing with redox reactions where atomic oxidation numbers are not straightforward. Here we use ethanol as a reducing reagent. With that let us design our half reactions:

Oxidation of ethanol to formate ion

We suppose ethanol is oxidized to give formate ion according t

$\ce{C2H5OH\to 2 HCO2^-}$

where we have balanced the carbon atoms. Introducing the water, hydroxide ions (in basic solution), and electrons we would have the fully balanced equation

$\ce{C2H5OH + (a)OH^-\to 2 HCO2^- + (b)H2O + (c)e^-}$

We expect the hydroxide ions to be on the left to balance all the negative charges on the right. To balance the hydrogen atoms we need $a+6=2b+2$, for oxygen we need $a+1=b+4$. Thus $b=7,a=10$ and when we put these results in the charge balance will give $(-10)=(-2)-c$ so $c=8$. Then

$\ce{C2H5OH +10 OH^-\to 2 HCO2^- + 7 H2O + 8 e^-}$

Reduction of chlorine

This is straightforward sincevinly chlorine atoms are reduced:

$\ce{Cl2 + 2e^-\to 2 Cl^-}$

Combine this with the oxidation half-reaction in the usual way and you have a balanced equation ... without the chloroform.

What happened?

We have a combination of two reactions, one oxidizing the alcohol to formate ion and the other oxidizing it to chloroform. We balanced just the oxidation to formate ion. Let's now look at the component where the alcohol is oxidized to chloroform.

Oxidation of ethanol to chloroform

Again we use ethanol as axwhole-molecule reducing agent. We must combine it with some chloride ions to balance carbon and chlorine, leading to an equation with the form

$\ce{C2H5OH + 6 Cl^- + (a)OH^-\to 2 CHCl3 + (b)H2O + (c)e^-}$

Balancing hydrogen requires $a+6=2b+2$ (again), and the oxygens give $a+1=b$. So $b=3,a=2$ and the charge balance then gives $(-6)-2=-c$ or $c=8$:

$\ce{C2H5OH + 6 Cl^- + 2 OH^-\to 2 CHCl3 + 3 H2O + 8 e^-}$

Combine this with the chlorine reduction given earlier and you have the balanced equation for the oxidation to chloroform. Note that in both component reactions each ethanol molecule gives off the same number of electrons and therefore should react with the same amount of chlorine. Thus the ratio of chlorine to ethanol is independent of how you combine these component reactions. (The actual value should emerge when you combine each of the oxidation half-reaction with the reduction.)

How to combine the two balanced equations

So you have two balanced equations, one with formate and one with chloroform. How do you combine them? Since any combination satisfies all the laws of physics, the answer is you can't with just pencil and paper. You have to do an experiment to determine the relative amounts of formate and chloroform you get, and match your combination of the two equations with those relative amounts. Don't forget to include your reaction conditions, as changing conditions could easily change the formate/chloroform ratio.

Answer 3

The key is to remembering that $n \cdot m = m \cdot n$.

It seems obvious, but you would not believe how much struggles people suffer when trying to apply this trivial equation to chemical reactions.

In this special case, The total change of oxidation numbers is zero, otherwise, it would mean we have lost some electrons somewhere.

$$\sum{n_i \cdot \Delta ON_i } = 0$$.

The stoichiometric coefficients are in the reciprocal ratio to ( absolute value of ) the oxidation number change ratio.

So the sum of the oxidation number change of carbons in 1 ethanol molecule is $( (+2 - -3) + ( +2 - -1) ) = +5 + +3 = +8$

The sum of the oxidation number change of chlorine in 1 chlorine molecule is $2 \cdot (-1 - 0) = -2$

We have then:

$$n_{\ce{C2H5OH}} \cdot 8 + n_{\ce{Cl2}} \cdot -2 = 0$$

$$n_{\ce{Cl2}} = 4 \cdot n_{\ce{C2H5OH}} $$

Note that you are missing negative ions of the left side for charge balance, also overall mass balance is missing.

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Enumeration of stoichiometric coefficients have also a general, analytical mathematical solution, but personally, I never used it, as it would take much longer time than just enumerating it on the fly by the chemist's heart:

• You are solving a set of linear algebraic equations for multiple variables.
• Each variable is a stoichiometric coefficient for a reagent or a product.
• Each equation represents some kind of balance.
• 1 equation for the atom count balance for each of the elements
• 1 equation for the charge balance, if applies.
• In most cases there will remain 1 degree of freedom, so 1 stoichiometric coefficient must be chosen and some others will depend on the choice.
• E.g. there can be $\ce{n H2 + n F2 -> 2n HF}$, so we choose $\ce{n=1}$ for $\ce{H2}$ and rest will follow.