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According to my book, 1 mole ClO- is equivalent to 1 mole Cl2. How is that even possible? 1 mole ClO- contains 1 mole Cl atoms. On the contrary, 1 mole Cl2 gas contains 2 mole Cl atoms. So, how are they equivalent?

Let me put it in this way. If I have 257.5g (5 mole) of ClO- ion, will it be correct to say that I have 355g (5 mole) Cl2 gas?

I have provided some context to my question. It's not required to be read in order to answer this question.

Context (not required to be read):

3.04g bleaching powder is dissolved in water to make a 400mL solution. To titrate 25mL of this solution, 40mL 0.075M sodium thiosulphate solution is needed.

Q) Calculate the amount of available active Cl2 from the mentioned bleaching powder.

My book started by finding the molarity of liberated I2 from KI (0.06M). Then my book wrote the following equation:

$$\ce{ClO− + 2I− + 2 H+ → I2 + H2O + Cl−}$$

Then my book said, "Here, two electrons from iodide ion (I-) are accepted by a single hypochlorite ion (ClO-), and oxidation number of Cl (+1) in ClO- is reduced to -1 in Cl- ion. Hence, the oxidation number of Cl is reduced by 2 units. [$\ce{Cl2 + 2e- -> 2Cl-}$] (I couldn't write the oxidation numbers in latex as my book did for this reaction). So, one mole ClO- ion is equivalent to 1 mole Cl2. The molarity of liberated I2 = Molarity of active Cl2 = 0.06M." Then my book went on to find that supplied bleaching powder contains 1.704g of active Cl2, which they claimed was the answer to the given question.

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    $\begingroup$ In terms of redox reactions in aqueous solutions, they are equivalent. In fact, chlorine in basic solutions turns quantitatively into hypochlorite while retaining its reducing power: $$\ce{Cl2 + 2OH- -> OCl- + Cl- + H2O}$$ When you acidify the solution, some of the hypochlorite and chloride turn into chlorine again. $\endgroup$
    – Karsten ♦
    Commented May 12, 2022 at 11:32
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    $\begingroup$ @KarstenTheis, Did you mean "chlorine in basic solutions turns quantitatively into hypochlorite while retaining its oxidizing power"? Only the hypochlorite ion releases chlorine upon acidification. The other chloride is just a spectator. $\endgroup$
    – ACR
    Commented May 12, 2022 at 21:29
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    $\begingroup$ @AChem Yes, oops, that is what I meant to say. In a reaction, either the two chlorines in dichlorine change oxydation state from zero to -1, or the chlorine in hypochlorite changes by two steps, from +1 to -1. In either case, two electrons are transferred $\endgroup$
    – Karsten ♦
    Commented May 12, 2022 at 23:39

3 Answers 3

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You will find the "equivalency" relations in many classical volumetric or gravimetric analysis. Active chlorine and active oxygen are such representative terms; these terms are academically old but they are still used in industrial analysis around the world.

When we say that a certain bleach contains a certain amount of active chlorine (old word for hypochlorite), all it means that this bleach will also liberate the same amount of iodine from potassium iodide if the same amount of chlorine gas were used.

I think your confusion stems from counting the atoms in chlorine and hypochlorite. This equivalency is not based on atom counting, it is based on the causing the chemical reactions as Poutnik showed in the equations.

For example do not be surprised if someone says,

1 mol $\ce{KMnO4}$ $\equiv$ 5 mol $\ce{Fe^{2+}}$. Now there is no atomic equivalency.

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    $\begingroup$ Your answer makes a great deal of sense to me. This is what I thought initially too. However, my book muddied the concept for me: when asked to find the amount of active Cl2 in a given amount of Ca(OCl)Cl, they found that 1.704g of Cl2 is present. They should've made it clear that 1.704g of Cl2 gas is not actually present in the given amount of bleaching powder, but rather the given amount of bleaching powder only behaves (liberates the same amount of Iodine from potassium iodide) as if 1.704g of Cl2 gas was present instead of the given amount of bleaching powder. Am I making sense? $\endgroup$
    – tryingtobeastoic
    Commented May 12, 2022 at 16:33
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    $\begingroup$ @tryingtobeastoic, Yes you are right. $\endgroup$
    – ACR
    Commented May 12, 2022 at 18:26
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If we consider reaction $$\ce{ClO- + Cl- + 2 H+ -> H2O + Cl2},$$ there is 1:1 ratio, 1 $\ce{ClO-}$ is equivalent to 1 $\ce{Cl2}$.

The equivalence does not mean the atom counts must match, as they can come from other sources (from $\ce{Cl-}$ in this case).

Similarly

\begin{align} \ce{ClO- + 2 I- + 2 H+ &-> Cl- + I2 + H2O}\\ \ce{Cl2 + 2 I- &-> 2 Cl- + I2 } \end{align}

The equivalent molar amounts or masses are those, where one form can be in respective quantity actually or formally converted to the other, with eventual participation of other components. Or, equivalent molar amounts or masses may in an applicable context substitute each other.

Note that the below clarifies the very frequent confusion of mole vs mol.

quantity unit name unit symbol
mass kilogram kg
molar amount mole mol
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Working with Bleaches like NaOCl and Ca(OCl2) and some organic reagents etc, I can tell you that Cl2 has very low free chlorine. Cl2 in water makes ClO- as the active specie. So, its fair to say 1 mol of Cl2 has same free chlorine as 1 mol of ClO- though its highly inaccurate. Because bleaching happens in the solution and not air, I'd say 1 mol of ClO- has more free chlorine compared to 1 mol of Cl2

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