Q.Calculate the equivalent weight of potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$, molar mass $294\ \text{g mol}^{-1}$) when it acts as an oxidant in acidic medium, given $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6e^{-} \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$.

[!TLDR] Equivalent weight = molar mass ÷ n-factor, with $n=6$ for dichromate as an oxidant. [!ANSWER] Equivalent weight of $\text …

The n-factor of a species in a redox reaction is the number of electrons it gains (as an oxidant) or loses (as a reductant) per formula unit. From $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6e^{-} \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$, one formula unit of dichromate gains $6$ electrons (2 Cr atoms each gaining 3 electrons, $+6\to+3$), so its n-factor as an oxidant in acidic medium is $n=6$. Equivalent weight $= \df …

Identify the n-factor as the total number of electrons transferred per formula unit from the half-equation …

Using $n=3$ (the per-chromium-atom electron change) instead of $n=6$ (the per-formula-un …