For the following reaction:
$K_{4} [F e {(C N)}_{6}] ⟶ {F e}^{3 +} + C O_{2} + N O_{3}^{-}$

If the $n$ - factor is $(60 + x)$ then the value of $x$ is _________.

Open in App

Solution

Oxidation state of $C$ in $K_{4} [F e (C N)_{6}] = + 2$

Oxidation state of $C$ in $C O_{2} = + 4$

$n_{f}$ due to $C = 6 (4 - 2) = 12$

Oxidation state of $N$ in $K_{4} [F e (C N)_{6}] = - 3$

Oxidation state of $N$ in $N O_{3}^{-} = + 5$

$n_{f}$ due to $N = 6 [5 - (- 3)] = 48$

Oxidation state of $F e$ in $K_{4} [F e (C N)_{6}] = + 2$

Oxidation state of $F e$ in $F e^{3 +} = + 3$

$n_{f}$ due to $F e = (3 - 2) = 1$

Therfore, $n_{f}$ of overalll reaction $= (12 + 48 + 1) = 61 = (60 + 1)$

Hence, the value of $x$ is $1$ .

Suggest Corrections

Similar questions

K_{4} [F e {(C N)}_{6}] ⟶ {F e}^{3 +} + C O_{2} + N O_{3}^{⊖}

K_{4} [F e {(C N)}_{6}] O x i d a t i o n - ------------- \to {F e}^{3 +} + C O_{2} + N O_{3}^{⊖}

[F e (C N)_{6}]^{4 -} \to F e^{3 +} + C O_{2} + N O_{3}^{-}

3.74

= \frac{M o l . w t .}{61}

K_{4} [F e (C N)_{6} O x i d a t i o n - ----- \to F e^{3 +} + C O_{2} + N O_{3}^{-}

\begin{matrix} List - I & List - II (I) For the reaction & (P) \frac{3}{5} K_{4} [F e (C N)_{6}] \to F e^{3 +} + C O_{2} + N O_{3}^{-} the n-factor is: (II) For the reaction & (Q) 28 F e S_{2} \to F e^{3} + + S O_{2} the n-factor is: (III) For the reaction & (R) 61 B r_{2} + 2 N a O H \to N a B r O_{3} + N a B r + H_{2} O the n-factor is: (IV) For the reaction & (S) 1 A s_{2} S_{3} \to A s^{5 +} + S O_{4}^{2 -} the n-factor is: (T) \frac{5}{3} (U) 11 \end{matrix}

Join BYJU'S Learning Program