Chemistry - Redox Reactions Question with Solution | TestHub

(A) $\to$ (s): Balancing the reaction gives $5P_2{H}_4\to 6P{H}_3+P_4{H}_2$. Since $6e^{-}$ are transferred per $5moles$ of $P_2{H}_4$, the $n$-factor is $\frac{6}{5}$, so $E=\frac{5M}{6}$.

(B) $\to$ (q): Balancing the reaction gives $3I_2\to 5I^{-}+I{O}_3^{-}$. Since $5e^{-}$ are transferred per $3moles$ of $I_2$, the $n$-factor is $\frac{5}{3}$, so $E=\frac{3M}{5}$.

(D) $\to$ (p): Balancing the reaction gives $3H_{3}P{O}_2\to P{H}_3+2H_{3}P{O}_3$. Since $4e^{-}$ are transferred per $3moles$ of $H_{3}P{O}_2$, the $n$-factor is $\frac{4}{3}$, so $E=\frac{3M}{4}$.

(C) $\to$ (r): By elimination, matching (A)-(s), (B)-(q), and (D)-(p) uniquely corresponds to Option (B).