Section Experiment 30 Advance Study Assignment: Determination of Iron by Reactio

Question

Section Experiment 30 Advance Study Assignment: Determination of Iron by Reaction with Perman Redox Titration 1. write the balanced net ionic equation for the reaction between Mno, ion and Fe" ion in acidic solution. 2. How many moles of Fe ion can be oxidized by IA x 10 moles Mno, ion in the reaction in Question 1? 3. A solid sample containing some Fe ion had a total mass of 0.9791 g. It required 18.2 mL 002034 M KMno, to titrate the Fe in the dissolved sample to a pink end point. a How many moles of Mao. ion were required? moles How many moles of Fe" were present in the sample? moles (continued on following page)

Explanation / Answer

Q.1: When MnO4-(aq) reacts with Fe2+(aq), the former is reduced to Mn2+(aq) and the later is oxidized to Fe3+(aq). We follow the below steps while balancing a chamical reaction in acidic medium.

Step-1: Write the oxidation-half and reduction-half reaction separately and balance all atoms except O and H.

Reduction-half: MnO4-(aq) ----> Mn2+(aq)

Oxidation-half: Fe2+(aq) -----> Fe3+(aq)

Step-2: Balance O atoms by adding suitable number of H2O molecules.

Reduction-half: MnO4-(aq) ----> Mn2+(aq) + 4H2O

Oxidation-half: Fe2+(aq) -----> Fe3+(aq)

Step-3: Balance H atoms by adding suitable number of H+(aq) ions in acidic medium.

Reduction-half: MnO4-(aq) + 8H+(aq) ----> Mn2+(aq) + 4H2O

Oxidation-half: Fe2+(aq) -----> Fe3+(aq)

Now all the atoms are balanced.

Step-4: Balance electrons by adding suitable number electrons

Reduction-half: MnO4-(aq) + 8H+(aq) + 5e- ----> Mn2+(aq) + 4H2O

Oxidation-half: Fe2+(aq) -----> Fe3+(aq) + 1e-

Step-5: Cross multiply both the half reactions by a suitable number so that the electrons cancel on both sides. Then add both the half-reactions to get the final balanced equation.

Reduction-half: MnO4-(aq) + 8H+(aq) + 5e-  ----> Mn2+(aq) + 4H2O

Oxidation-half: [Fe2+(aq) -----> Fe3+(aq) + 1e- ] x 5

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MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) -------> Mn2+(aq) + 4H2O + 5Fe3+(aq) (answer)

Q.2: From the above balanced reaction it is clear that

1 mol MnO4-(aq) oxidizes 5 mol Fe2+(aq).   

Hence 1.4x10-2 mol MnO4-(aq) will oxidize the moles of Fe2+(aq)

= 1.4x10-2 mol MnO4-(aq) x [5 mol Fe2+(aq) / 1 mol MnO4-(aq)]

= 7.0x10-2 mol Fe2+  (answer)

Q.3:(a) Moles of MnO4-(aq) required = 0.02034 mol/L x 0.0182 L = 3.7x10-4 mol (answer)

(b):Moles of Fe2+ present = 3.7x10-4 mol x 5 = 1.85x10-3 mol  (answer)

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