In the reaction (acidic medium): , the -factor of is: — Redox Reactions and Volumetric Analysis Chemistry Question
Question
In the reaction (acidic medium): $K_4[Fe(CN)_6] \rightarrow Fe^{3+} + CO_2 + NO_3^-$, the $n$-factor of $K_4[Fe(CN)_6]$ is:
Answer: 61
💡 Solution & Explanation
**1.** $Fe^{2+} \rightarrow Fe^{3+}$ involves loss of $1 e^-$. <br>**2.** $6C^{2+} \rightarrow 6C^{4+}$ involves loss of $6 \times 2 = 12 e^-$. <br>**3.** $6N^{3-} \rightarrow 6N^{5+}$ involves loss of $6 \times 8 = 48 e^-$. <br>**Total electrons lost** per molecule = $1 + 12 + 48 = 61$.
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