Let us consider the following reaction

$$\ce{I2 + 2 Na2S2O3 -> 2 NaI + Na2S4O6}$$

Now, in order to calculate the equivalent mass of $\ce{Na2S2O3}$, first I need to calculate it's $n$-factor which turns out to be $0.5$ because the oxidation state of $\ce{S}$ in $\ce{Na2S2O3}$ is $+2$ whereas in $\ce{Na2S4O6}$ it is $+2.5$.And the $n$-factor of $\ce{I2}$ is $1$. So, the equivalent mass of $\ce{Na2S2O3}$ is

$$\frac{\text{molecular mass}}{0.5}$$ And Iam confused with the following half-reactions: $$\ce{I2 +2e->2I-}$$ $$\ce{2S2O3^{2-}->S4O6^{2-} +2e}$$ I appreciate any help in order to solve a problem using milliequivalents?

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    $\begingroup$ The first and third reactions are not balanced in atoms. $\endgroup$ – user55119 May 14 '19 at 0:53
  • $\begingroup$ @user55119Thanks for correction $\endgroup$ – Adnan AL-Amleh May 14 '19 at 1:09

N-factor Is the change of atom oxidation state multiplied by number of atoms changing this state.

Therefore, n-factor of $\ce{I2}$ is $2$,
n-factor of $\ce{Na2S2O3}$ is $1$.

So, equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass, for molecular iodine it is half of its molar mass.

  • $\begingroup$ thanks for the help $\endgroup$ – Adnan AL-Amleh May 14 '19 at 17:23
  • $\begingroup$ excuse me: Is the equivalent mass of $\ce{KIO3}$ equal 5.$$\ce{KIO3 + 5 KI + 3 H2SO4 -> 3 I2 + 3 K2SO4 + 3 H2O}$$ $\endgroup$ – Adnan AL-Amleh May 14 '19 at 17:50
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    $\begingroup$ Well, this is somewhat tricky. For the mentioned reaction, releasing iodine, the equivalent mass is M/5. But if the iodine is just an intermediate product to be titrated by Na2S2O3, the equivalent mass is M/6. As in the former, the oxidation number change is by 5, in the latter by 6. $\endgroup$ – Poutnik May 14 '19 at 18:01

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