Calculate the equivalent weight of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ in an acidic medium where it acts as an oxidizing agent.

Solution

1. The reduction half-reaction is ${\text{Cr}}_2{\text{O}}_7^{2-}+14{\text{H}}^{+}+6{\text{e}}^{-}\to 2{\text{Cr}}^{3+}+7{\text{H}}_2\text{O}$.
2. The change in oxidation state of chromium is from $+6$ to $+3$, so for two chromium atoms, the total change is $2\times (6-3)=6$.
3. The equivalent weight is defined as $\text{Molar Mass}/\text{n-factor}$.
4. Hence the answer is (C).