Question

Match the half reaction in List 1 with equivalent mass (molar mass = $M$) in List 2.

Answer 1

(A) ${C{r}_2{O}_7}^{2-}\to C{r}^{3+}$
The change in oxidation number per chromium atom is $+6$ to $+3$.
The net change in the oxidation number for two chromium atoms is $6$.
Thus, $equivalentmass=\frac{molarmass}{6}$
(B) ${C_2{O}_4}^{2-}\to C{O}_2$
The change in oxidation number per carbon atom is $1$.
The net change in the oxidation number for two carbon atoms is $2$.
Thus, $equivalentmass=\frac{molarmass}{2}$
(C) ${Mn{O}_4}^{-}\to Mn{O}_2$
The change in oxidation number per manganese atom is $+7$ to $+4$.
Thus, $equivalentmass=\frac{molarmass}{3}$
(D) ${H{C}_2{O}_4}^{-}\to {C_2{O}_4}^{2-}$
In this reaction, only one hydrogen atom is lost.
Thus, $equivalentmass=molarmass$