Question

Redox equations are balanced either by the ion electron method or by the oxidation number method. Both methods leads to the correct form of the balanced equation.

List-I contains questions related to some redox equations and List-II contains their answers
$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I) For the reaction}& \text{(P)}\frac{3}{5}\\ K_4\left[Fe\right(CN{)}_6]\to F{e}^{3+}+C{O}_2+N{O}_3^{-}& \\ \text{the n-factor is:}& \\ \text{(II) For the reaction}& \text{(Q) 28}\\ Fe{S}_2\to F{e}^3++S{O}_2& \\ \text{the n-factor is:}& \\ \text{(III) For the reaction}& \text{(R)}61\\ B{r}_2+2NaOH\to NaBr{O}_3+NaBr+H_{2}O& \\ \text{the n-factor is:}& \\ \text{(IV) For the reaction}& \text{(S)}1\\ A{s}_2{S}_3\to A{s}^{5+}+S{O}_4^{2-}& \\ \text{the n-factor is:}& \\ & \text{(T)}\frac{5}{3}\\ & \text{(U)}11\end{array}$

Which of the following options has the correct combination considering Lists-I and List-II?

• A. $\left(II\right),\left(U\right)$ and $\left(IV\right),\left(Q\right)$
• B. $\left(II\right),\left(R\right)$ and $\left(IV\right),\left(P\right)$
• C. $\left(II\right),\left(T\right)$ and $\left(IV\right),\left(Q\right)$
• D. $\left(II\right),\left(S\right)$ and $\left(IV\right),\left(U\right)$

Answer 1

The correct option is A $\left(II\right),\left(U\right)$ and $\left(IV\right),\left(Q\right)$

(II) In the reaction $Fe{S}_2\to F{e}^{3+}+S{O}_2$
both $Fe$ and $S$ is getting oxidised.
$F{e}^{2+}\to F{e}^{3+}+e^{-}$
$S_2^{2-}\to 2S{O}_2+10e^{-}$
So n-factor for this reaction is 11

(IV) In the reaction both $A{s}^{2+}$ and $S_3^{2-}$ is getting oxidised.
$A{s}_2^{3+}\to 2A{s}^{5+}+4e^{-}$
$S_2^{3-}\to 3S{O}_4^{2-}+24e^{-}$
So n-factor for this reaction is 28