Balance Redox Equation by Ion-electron (Acidic medium)

Redox-reaction

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1. Balance the Given redox-reaction  by ion-electron Method

Cr₂O₇^{– –} +Fe⁺⁺+ H⁺ —-> Cr³⁺+Fe³⁺+H₂O

+6

Cr₂O₇^{– –} + Fe⁺⁺ + H⁺ —-> Cr³⁺ +Fe³⁺ +H₂O

+6

Cr₂O₇^{– –    ……>} Cr³⁺               (Reduction)

Fe²⁺ —-> Fe³⁺                     (Oxidation)

Balance reduction half rxn

Cr₂O₇^{– –}    —–> 2Cr³⁺
To balance oxygen atom

Cr₂O₇^{– –} —–> 2Cr³⁺+7H₂O

To balance H-atom

Cr₂O₇^{– –} + 14H⁺ —–> 2Cr³⁺ + 7H₂O

To balance charge

Cr₂O₇^{– –} + 14H⁺ 6e^– —–> 2Cr³⁺ + 7H₂O ……      Eq. (1)

To balance oxidation half rxn-

Fe⁺⁺ —–>Fe³⁺

To balance charge

Fe⁺⁺ —–>Fe⁺⁺⁺ + e^–                                          ……      Eq. (2)

Multiply Eq. (2) by 6 & add Eq (i) & (ii)

6Fe²⁺ —–> 6Fe³⁺ + 6e

Cr₂O₇^{– –} + 6Fe⁺⁺ + 14H⁺ —–> 2Cr³⁺+6Fe³⁺ +7H₂O

1. MnO₄^– + Fe⁺⁺ + H^+→ Mn⁺⁺ + Fe³⁺+H₂O

+7

MnO₄^– + Fe⁺⁺ + H⁺ →Mn⁺⁺ + Fe³⁺+H₂O

+7             +2

MnO₄^– → Mn⁺⁺                          (Reduction)

To balance reduction half rxn-

MnO₄^– → Mn⁺⁺

To balance O-atom

MnO₄^– → Mn⁺⁺+ 4H₂O

To balance H-atom

MnO₄^–+8H^+→ Mn⁺⁺+4H₂O

To balance charge

MnO₄^– + 8H⁺ +5e^– → Mn⁺⁺+4H₂O                                          (i)

To balance oxidation half rxn

Fe⁺⁺ →Fe⁺⁺⁺                    (Oxidation)

To balance charge

Fe⁺⁺ → Fe⁺⁺⁺ +e                                                              (2)

Multiply Eq.(2) by 5 and add both equations

5Fe⁺⁺ → 5Fe³⁺ + 5e

MnO₄^– +5Fe²⁺+8H⁺ → Mn⁺⁺+5Fe³⁺ + 4H_2O

D  MnO₄^– + SO₃^{– –} → Mn²⁺ + SO₄^2–

+7         +4           +2       +6

MnO₄^– + SO₃^{– –} → Mn⁺⁺ + SO₄^{– –}

+7            +2

MnO₄^{– –} → Mn⁺⁺                          (Reduction)

+4         +6

SO₃^{– –} → SO₄^{– –}                                        (Oxidation)

To balance reduction half rxn

MnO₄^– →Mn⁺⁺

To balance O-atom

MnO₄^– → Mn⁺⁺ + 4H₂O

To Balance H- atom

MnO₄^– + 8H⁺ → Mn⁺⁺ + 4H₂O

To balance Charge

MnO₄^– + 8H⁺ + 5e^– → Mn⁺⁺ + 4H₂O                                 (i)

To balance oxidation Half Rxn

SO₃^{– –} → SO₄^{– –}

To balance H-atom

SO₃^{– –} +H₂O → SO₄^{– –} + 2H⁺

To balance Charge

SO₃^{– –} + H₂O → SO₄^{– –} + 2H⁺ + 2e                         (ii)

Multiply eq(1) by 2 and eq. (2) by 5 & add both equations

2MnO₄^– + 16H⁺ +10e^– → 2Mn⁺⁺+ 8H₂O

5SO₃^{– –} + 5H₂O → 5SO₄^{– –} +10 H⁺ + 10e^–

2MnO₄^– +5SO₃^{– –} + 6H⁺ → 2Mn⁺⁺ + 5SO₄^{– –} + 3H₂O

To balance the given  redox-reaction

1. NO₃^– +I^– → NO + I

NO₃^– →   NO                        (Reduction)

-1     0

I ^– → I₂                                   (Oxidation)

To balance Reduction half rxn-

NO₃^–→ NO

To balance O-atom

NO₃^– → NO+2H₂O

To balance H-atom

NO₃^– +4H⁺ → NO + 2H₂O

To balance charge

NO₃^– + 4H⁺ + 3e^– → NO+2H₂O                                         (i)

To balance oxidation half rxn

I^– → I₂

To balance I-atom

2I^– → I₂

To balance Charge

2I^– → I₂ + 2e                                                                         (ii)

Multiply eq (1) by 2 & eq. (2) by 3 & add both equations

2NO₃^– + 8H⁺+6e → 2NO+4H₂O

6I^– → 3I₂ + 6e^–

2NO₃^– + 6I^– + 8H⁺ → 2NO + 3I₂ + 4H₂O

To balance the given redox-reaction

Cr₂O₇^{– –} + SO₂ → Cr³⁺ + HSO₄^–

+6

Cr₂O₇^{– –} → Cr³⁺         (reduciton)

+4       +6

SO₂ → HSO₄^–           (Oxidation)

To balance reduction half rxn

            +6

            Cr₂O₇^{– –} → Cr³⁺

To balance Cr atom

Cr₂O₇^{– –} → 2Cr³⁺

To balance O-atom

Cr₂O₇^{– –} →2Cr³⁺ + 7H₂O

To balance H-atom

Cr₂O₇^{– –} +14H⁺ → 2Cr³⁺ + 7H₂O

To balance charge

Cr₂O₇^{– –} + 14 H⁺ → 2Cr³⁺ + 7H₂O

To balance charge

Cr₂O₇^{– –} + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O                            (i)

To balance Oxidation half rxn

+4            +6

SO₂   → HSO₄^–

To balance O-atom

SO₂ + 2H₂O → HSO₄^–

To balance H-atom

SO₂+ 2H₂O → HSO₄^– + 3H⁺

To balance charge

SO₂ + 2H₂O → HSO₄^– + 3H⁺ +2e^–                                     (ii)

Multiply eq. (2) by 3 & add both equations

3SO₂ + 6H₂O → 3HSO₄^– + 9H⁺ + 6e^–

Cr₂O₇^{– –} + 14H⁺ +6e^– → 2Cr³⁺ + 7H₂O

Cr₂O₇^{– –} + 3SO₂ + 5H⁺ → 2Cr³⁺ + 3HSO₄^– + H₂O

f   To balance the given redox-reaction       

MnO₄^– + I^– →Mn⁺⁺ + I₂ + H₂O

+7               +2

MnO₄^– → Mn⁺⁺       (Reduction)

0

I^– →I₂                                     (Oxidation)

To balance reduction half rxn

MnO₄^– → Mn⁺⁺

To balance O-atom

MnO₄^– → Mn⁺⁺ + 4H₂O

To balance H-atom

MnO₄^– + 8H⁺ → Mn⁺⁺ + 4H₂O

To balance charge

MnO₄^– + 8H⁺ +5e^– → Mn⁺⁺+ 4H₂O                                                          (i)

To balance oxidation half rxn

            I^– →I₂

To balance I-atom

2I^– → I₂

To balance charge

2I^– → I₂ + 2e^–                                                                                                (ii)

Multiply eq(1) by 2 & eq. (2) by 5 & add both equations

2MnO₄^– +16H⁺+10e → 2Mn⁺⁺ + 8H₂O

10 I^– →5I₂+10e^–

2MnO₄^– +10I^– +16H⁺ → 2Mn⁺⁺ + 5I₂ + 8H₂O

1. g)To balance the given redox-reaction
2. IO₃^– + Fe⁺⁺ + Cl^– → ICl + Fe³⁺

+5              +1

IO₃^– +Cl^– → ICl (Reduction)

Fe⁺⁺ → Fe³⁺ (Oxidation)

To balance Reduction half rxn

+5                   +1

            IO₃^– + Cl^– → ICl

To balance O-atom

IO₃^– +Cl^– → ICl + 3H₂O

To balance H-atom

IO₃^– + Cl^– + 6H⁺ → ICl + 3H₂O

IO₃^– +Cl^– +6H⁺ + 4e^– → ICl + 3H₂O                                                          (i)

To balance Oxidation half r×n

            Fe⁺⁺ →Fe³⁺

To balance charge

Fe⁺⁺ → Fe³⁺ +e^–                                                                                (ii)

Multiply eq. (2) by 4 & add both equations

4Fe⁺⁺ → 4Fe³⁺ + 4e^–

IO₃^– +Cl^– + 6H⁺ + 4e^– → ICl + 3H₂O

IO₃^– + 4Fe⁺⁺ + Cl^– + 6H⁺ → ICl+ 4Fe³⁺ + 3H₂O

Balanced redox-reaction