How to calculate the equivalent mass of Na2S2O3? [closed]








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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?




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.


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