Question

What is the value of n-factor of [fe(cn)6]4 in the given reaction?

Answer 1

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 C is the answer.