What is the value of n-factor of [Fe(CN)6]4- in the given reaction [Fe(CN)6]-4 .....in presence of MnO4 - / H+ --> Fe3+ + CO2 + NO3 - a) 18 b) 31 c) 61 d) 28 Share with your friends Share 4 Vartika Jain answered this Dear Student, [Fe(CN)6]4- + MnO4- → Fe3+ + CO2 + NO3- + Mn2+Step 1: Divide the equation into two halves[Fe(CN)6]4- → Fe3+ + CO2 + NO3-MnO4- → Mn2+Step 2: Balance all atoms other than O and H[Fe(CN)6]4- → Fe3+ + 6CO2 + 6NO3-MnO4- → Mn2+Step 3: Balance O by adding H2O[Fe(CN)6]4-+30H2O → Fe3+ + 6CO2 + 6NO3-MnO4- → Mn2+ +4H2OStep 4: Balance H by adding H+[Fe(CN)6]4-+30H2O → Fe3+ + 6CO2 + 6NO3- +60H+MnO4- +8H+ → Mn2+ +4H2OStep 5: Neutralise charge by adding electrons[Fe(CN)6]4-+30H2O → Fe3+ + 6CO2 + 6NO3- +60H++61e-MnO4- +8H+ +5e-→ Mn2+ +4H2ONow, since in the first reaction, [Fe(CN)6]4- is undergoing a change of 61 electrons, Therefore for [Fe(CN)6]4-,n=61Hence, correct answer is (c) 1 View Full Answer
