equilibrium - How to calculate the equivalent mass of Na2S2O3?

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?