Question

$C{r}_2{O}_7^{2-}\stackrel{H^{+}}{⟶}C{r}^{3+}$
Find the equivalent weight of $C{r}_2{O}_7^{2-}$ in the above redox half reaction.
(Given the molar mass of $C{r}_2{O}_7^{2-}$ is $Mg/mol$)

• A. $\frac{M}{3}$
• B. $\frac{M}{4}$
• C. $\frac{M}{5}$
• D. $\frac{M}{6}$

Answer 1

The correct option is D $\frac{M}{6}$

In acidic medium, the balanced reaction will be:
$\stackrel{+6}{C}{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2\stackrel{+3}{C}{r}^{3+}+7H_{2}O$
Formula used for the n-factor calculation,
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atoms}$
$n_{f}ofC{r}_2{O}_7^{2-}=|6-3|\times 2=6$
$\text{Equivalent weight of}{K}_{2}C{r}_2{O}_7=\frac{\text{Molecular weight of}{K}_{2}C{r}_2{O}_7}{6}=\frac{M}{6}$