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Balance Redox Equation by Ion-electron (Acidic medium)

Redox-reaction

source:chemistry.tutorvista.com

  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+ + 4H2O

D  MnO4 + SO3– –  Mn2+ + SO42–

+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– – +H2 SO4– – + 2H+

To balance Charge

SO3– – + H2 SO4– – + 2H+ + 2e                         (ii)

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

2MnO4 + 16H+ +10e  2Mn+++ 8H2O

5SO3– – + 5H2 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 + 2H2 HSO4

To balance H-atom

SO2+ 2H2 HSO4 + 3H+

To balance charge

SO2 + 2H2 HSO4 + 3H+ +2e                                     (ii)

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

3SO2 + 6H2 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

Saroj Bhatia: Dr. Saroj Bhatia is an Ph.D in chemistry who has been teaching chemistry for over a decade. Currently she is a respected principal of a renowned college in her hometown. She took this medium for online users. Her proudest achievement is helping people learn chemistry.
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