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. $\left(I\right),\left(R\right)$ and $\left(III\right),\left(P\right)$
• B. $\left(I\right),\left(U\right)$ and $\left(III\right),\left(T\right)$
• C. $\left(I\right),\left(S\right)$ and $\left(III\right),\left(P\right)$
• D. $\left(I\right),\left(R\right)$ and $\left(III\right),\left(T\right)$

Answer 1

The correct option is D $\left(I\right),\left(R\right)$ and $\left(III\right),\left(T\right)$

(I) In $K_4\left[Fe\right(CN{)}_6]$ oxidation state of $Fe$ is 2+.
All the redox reactions involved in $K_4\left[Fe\right(CN{)}_6]\to F{e}^{3+}+C{O}_2+N{O}_3^{-}$ are:
$F{e}^{2+}\to F{e}^{3+}+e^{-}$.........(1)
$6C{N}^{-}\to 6C{O}_2+12e^{-}$........(2)
Since here oxidation state of carbon is getting changed from +2 to +4
$6C{N}^{-}\to 6N{O}_3^{-}+48e^{-}$..........(3)
Since here oxidation state of nitrogen is getting changed from -3 to +5
So the n-factor for this reaction will be $=48+12+1=61$

(III) In the reaction $B{r}_2+2NaOH\to NaBr{O}_3+NaBr+H_{2}O$
$B{r}_2$ is undergoing both oxidation and reduction.
$B{r}_2\to 2Br{O}_3^{-}+10e^{-}\cdots \left(1\right)$
$[B{r}_2+2e^{-}\to 2B{r}^{-}]\times 5\cdots \left(2\right)$
$\left(1\right)+\left(2\right)\Rightarrow 6B{r}_2\to 2Br{O}_3^{-}+10B{r}^{-}$
Equivalent weight of $B{r}_2=\frac{6\left[B{r}_2\right]}{10}=\frac{3}{5}$
$\therefore$ n-factor $=\frac{3}{5}$