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Balancing Redox Reaction

Dr. Nishant Khandelwal
26/05/2021 0 0

Redox reaction is an Oxidation-Reduction Reaction. Balancing a redox reaction in an essential medium requires an additional step when we finish of which involves the addition of an equal number of hydroxyl ions to both sides of the equation, followed by combining the H+ and OH- to make H2O and cancelling out the water molecule that is present on both sides.

Balancing involves the following steps:

Step 1: Write the unbalanced ionic equation. Fe(s) + CrO42- (aq) → Fe2O3 (s) + Cr2O3 (s)

Step 2: Write separate half-reactions for the oxidation and the reduction processes.

Oxidation : Fe (s) → Fe2O3  (aq)          Fe is in +3 oxidation state here Fe2O3  (aq)

Fe0 (s) → Fe+3 (aq)

The oxidation state of iron changes from 0 to +3; therefore, Iron is Oxidised.

Reduction  CrO42- (aq) → Cr2O3 (s); Cr+6 (aq)  → Cr+3 (s)

The oxidation state of Chromium changes from +6 to +3; therefore, Chromium is reduced.

Step 3: Balance the atoms in the half-reactions other than the hydrogen and oxygen.

2 Fe (s)  → Fe2O3 (aq); 2 CrO42- (aq) → Cr2O3 (s) 

Step 4: Balance oxygen atoms by adding water molecules to the respective side of the equation.

2 Fe (s) + 3H2O(l) → Fe2O3 (aq); 2 CrO42- (aq) → Cr2O3 (s) + 5H2O(l)

Step 5: Balance Hydrogen atoms by adding H+ ions to the respective side of the equation.

2 Fe (s) + 3H2O(l) → Fe2O3 (aq) + 6 H+(aq); 10 H+(aq) + 2 CrO42- (aq) → Cr2O3 (s) + 5H2O(l)

Step 6: Balance the charges by adding electrons to each half-reaction and then equalise the electrons by multiplying with a coefficient.

2 Fe (s) + 3H2O (l)→ Fe2O3 (aq) + 6 H+(aq) + 6e- ( Fe0 --> Fe3+ + 3e- ) x 2

6e- + 10 H+ (aq) + 2 CrO42- (aq) → Cr2O3 (s) + 5H2O(l) (  Cr+6 (aq) + 3e- → Cr+3 (s) ) x 2

Step 7: a) Add the two half-reactions together. b) The electrons must cancel.

2 Fe (s) + 3H2O(l) → Fe2O3 (aq) + 6 H+ (aq)+ 6e-                  Oxidation 

 6e-   + 10 H+ (aq)+ 2 CrO42- (aq) → Cr2O3 (s) + 5H2O(l) Reduction


2 Fe (s) + 3H2O(l) + 10 H+ (aq)+ 2 CrO42- (aq) → Fe2O3 (aq) + 6 H+ (aq)+ Cr2O3 (s) + 5H2O(l)

c) Balance any remaining substances by inspection, cancel out H2O or H+ that appear on both sides if necessary.

2 Fe (s) +  4 H+ (aq)+ 2 CrO42- (aq) → Fe2O3 (aq) +Cr2O3 (s) + 2H2O(l)

Step 8: Adding OH- hydroxide ions on both sides equivalent to H+ ions.

 4 OH-(aq)  + 2 Fe (s) +  4 H+ (aq)+ 2 CrO42- (aq) → Fe2O3 (aq) +Cr2O3 (s) + 2H2O(l) + 4 OH-(aq)

Step 9: Combining OH- ions and H+ ions to form water.

2 Fe (s) +  4 H2 O(l) + 2 CrO42- (aq) → Fe2O3 (aq) +Cr2O3 (s) + 2H2O(l) + 4 OH-(aq)

Step 10:  Cancelling out water molecule, if any.

2 Fe (s) +  4 H2 O(l) + 2 CrO42- (aq) → Fe2O3 (aq) +Cr2O3 (s) + 2H2O(l) + 4 OH-(aq) gives 2 Fe (s) +  2 H2 O(l) + 2 CrO42- (aq) → Fe2O3 (aq) + Cr2O3 (s) + 4 OH-(aq)

Now we have our Balanced Redox reaction in basic medium:

2 Fe (s) +  2 H2 O(l) + 2 CrO42- (aq) → Fe2O3 (aq) + Cr2O3 (s) + 4 OH-(aq)

The equation is finally balanced by atoms and by charge, The presence of hydroxide ions means that the reaction takes place in basic solution. 

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