What is the value of n-factor of [fe(cn)6]4 in the given reaction?
Answers
Answered by
12
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.
Attachments:
Similar questions