Ratio of moles of Fe(||) oxidised by equal volumes of equimolar KMnO4 and K2Cr2O7 solutions in acidic medium will be:
1 mole of KMnO4 oxidises 5 moles of Fe2+ as ( 8 H+ + 5 Fe2+ + MnO4- --> 5 Fe3+ + Mn2+ + 4H2O)
and 1 mole of K2Cr2O7 oxidises 6 moles of Fe2+ ( 6 Fe+2 + Cr2O7^-2 + 14H+ --> 6 Fe+3 + 2Cr+3 + 7H2O)
Fe2+ :KMnO4 = 5:1 ; Fe2+ : K2Cr2O7 = 6:1
the explanation is:
The redox reaction between Fe2+ ions and KMnO4 in acidic medium is"
MnO4^- + Fe^2+ --> Fe^3+ + Mn^2+
Balancing the above reaction The unbalanced half equations:
(oxidation): Fe2+(aq) → Fe3+(aq)
(reduction): MnO4-(aq) → Mn2+(aq)
The two half-reactions (balanced) are shown
oxidation: Fe2+ --> Fe3+ + e-
reduction: 5e- + 8H+ + MnO4- --> Mn2+ + 4H2O
To balance the complete reaction the electrons lost must equal the electrons gained. Multiplying the components of the oxidizing reaction by 5 and adding to the reducing reaction we get a balanced net-reaction. (The 5 electrons on both side cancel.)
8 H+ + 5 Fe2+ + MnO4- --> 5 Fe3+ + Mn2+ + 4H2O
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The redox reaction between Fe2+ ions and K2Cr2O7 in acidic medium is:
Cr2O7^2- + Fe^2+ --> Cr^3+ + Fe^3+
Balancing the above reaction:
The unbalanced half equation are:
oxidation half-reaction: Fe+2 --> Fe+3
reduction half-reaction: Cr2O7^-2 --> Cr+3
The balanced half reactions are:
Oxidation: Fe+2 --> Fe+3 + e-
Reduction : Cr2O7^-2 + 14 H+ + 6e- --> 2Cr+3 + 7 H2O
the number of electrons lost must be equal to number of electrons gained. Therefore, multiply oxidation half by 6. and adding the two equations, the electrons gets cancelled.
6 Fe+2 + Cr2O7^-2 + 14H+ --> 6 Fe+3 + 2Cr+3 + 7H2O
and 1 mole of K2Cr2O7 oxidises 6 moles of Fe2+ ( 6 Fe+2 + Cr2O7^-2 + 14H+ --> 6 Fe+3 + 2Cr+3 + 7H2O)
Fe2+ :KMnO4 = 5:1 ; Fe2+ : K2Cr2O7 = 6:1
the explanation is:
The redox reaction between Fe2+ ions and KMnO4 in acidic medium is"
MnO4^- + Fe^2+ --> Fe^3+ + Mn^2+
Balancing the above reaction
The unbalanced half equations:
(oxidation): Fe2+(aq) → Fe3+(aq)
(reduction): MnO4-(aq) → Mn2+(aq)
The two half-reactions (balanced) are shown
oxidation: Fe2+ --> Fe3+ + e-
reduction: 5e- + 8H+ + MnO4- --> Mn2+ + 4H2O
To balance the complete reaction the electrons lost must equal the electrons gained. Multiplying the components of the oxidizing reaction by 5 and adding to the reducing reaction we get a balanced net-reaction. (The 5 electrons on both side cancel.)
8 H+ + 5 Fe2+ + MnO4- --> 5 Fe3+ + Mn2+ + 4H2O
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The redox reaction between Fe2+ ions and K2Cr2O7 in acidic medium is:
Cr2O7^2- + Fe^2+ --> Cr^3+ + Fe^3+
Balancing the above reaction:
The unbalanced half equation are:
oxidation half-reaction: Fe+2 --> Fe+3
reduction half-reaction: Cr2O7^-2 --> Cr+3
The balanced half reactions are:
Oxidation: Fe+2 --> Fe+3 + e-
Reduction : Cr2O7^-2 + 14 H+ + 6e- --> 2Cr+3 + 7 H2O
the number of electrons lost must be equal to number of electrons gained. Therefore, multiply oxidation half by 6. and adding the two equations, the electrons gets cancelled.
6 Fe+2 + Cr2O7^-2 + 14H+ --> 6 Fe+3 + 2Cr+3 + 7H2O
