Question

The 'n' factor for Ferrocyanide ion in the
below given reaction is?
[Fe(CN)61-4 → Fe+3.+ CO2 + NC
(A) 56
(B) 32
(C) 2
(D) 61​

Answer 1

Answer:

For [Fe(CN)6]4−, n=61.

Explanation:

Note: The question is incomplete. Please refer to the attached picture for the complete question.

Given:  [Fe(CN)6]4− + MnO4− →  Fe3+ + CO2 + NO3− + Mn2+

Find: n factor.

Solution:

[Fe(CN)6]4− + MnO4− →  Fe3+ + CO2 + NO3− + Mn2+

1. Divide the equation into two halves:

[Fe(CN)6]4− → Fe3+ + CO2 + NO3−

MnO4− → Mn2+

2. Balance all atoms other than O and H

[Fe(CN)6]4− → Fe3+ + 6CO2 + 6NO3−

MnO4− → Mn2+

3. Balance O by adding H2O

[Fe(CN)6]4− + 30H2O → Fe3+ + 6CO2 + 6NO3−

MnO4− → Mn2+ +4H2O

4. Balance H by adding H+

[Fe(CN)6]4− + 30H2O → Fe3+ + 6CO2 + 6NO3− +60H+

MnO4− +8H+ → Mn2+ +4H2O

5. Neutralise charge by adding electrons

[Fe(CN)6]4− + 30H2O  → Fe3+ + 6CO2 + 6NO3− +60H+ +61e−

MnO4− +8H+ +5e− → Mn2+ +4H2O

From the first equation, we find that [Fe(CN)6]4− is undergoing a change of 61 electrons.

Therefore for [Fe(CN)6]4−, n=61.

Option D is the answer.

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