BALANCING A CHEMICAL EQUATION BY ION-ELECTRON METHOD |in basic medium|

 

 

Redox equation is in which oxidation and reduction occur simultaneously. Balancing of such equation in basic medium have two extra steps as compare to acidic medium.steps are given below: 

1. Assigning the oxidation numbers to each element in whole given chemical equation. 

2. Figure out those elements in which oxidation and reduction is occurring. 

3. Split the equation in two halves i.e oxidation half reaction and reduction half reaction. 

4. Balance the number of atoms first then balance the charges by balancing electrons.hydrogen and oxygen are balanced by H+ and H2O respectively. 

5. Add two half reactions and deduct same items like electrons which are present on both side of equation. 

6. Now add one OH- for each H+ on both side of sum equation. 

7. Now deduct waters present both side of equation. check the equation is balanced or not by counting atoms and charges.

 Practice chemical equation:

Zn + NO₃ ^-1    → Zn⁺² + NO₂

Step1: Assign oxidation numbers to each element in given chemical equation. 

Zn⁰ + N⁺⁵O₃^{-2 -1}    → Zn⁺² + N⁺⁴O₂^-2

Step2: Figure out those elements in which oxidation and reduction is occurring.

 Zn⁰ + N⁺⁵O₃ ^-1    → Zn⁺² + N⁺⁴O₂

_{Zn and N are changing its oxidation numbers.}

_{Zn = 0 to +2}

_{N = +5 to +4}

 

Recall increase in oxidation number is oxidation Zn is increasing its oxidation number from 0 to +2 so Zn is oxidizing and N is decreasing its oxidation number from +5 to +4 so N is reducing in above practice equation.

Step3 & 4: Split the equation in two halves i.e oxidation half reaction and reduction half reaction.Balance the number of atoms first then balance the charges by balancing electrons.Hydrogen and oxygen are balanced by H+ and H2O respectively.

Oxidation half reaction:( Removal of electrons and increase of Oxidation number) 

 

in Zn oxidation is occurring (increasing oxidation number 0 to +2) form oxidation half reaction.

Zn⁰  → Zn⁺²

^{Now balance number of  Zn atoms at both side of chemical equation which are already equal i.e 1, now balance the number of electrons. Oxidation. Number is increasing by 1 i.e 0 to +2. Zn is losing 2 electrons and getting oxidized.}

 

 Zn^{0  →}   Zn^{+2  +} 2e^- 

Reduction half reaction:( gain of electrons and decrease in oxidation number)

N⁺⁵O₃ ^-1→N⁺⁴O₂

There is one N at both side and 3 O at reactants and 2 O at products side. Oxygens are not balanced hence we are allowed to balance O by adding H2O at product side.

 N⁺⁵O₃ ^-1→N⁺⁴O₂ + H₂O

Now by adding H2O now we have introduced 2 H at product side, to balance the H add 2H+ at reactant side. 

2H⁺ +N⁺⁵O₃ ^-1→N⁺⁴O₂ + H₂O

 ^{Now balance number of electrons to balance charges at both side. N oxidation number is changing from +5 to +4 hence it is gaining 1 electron ( reduction occurring).}

1e^- + 2H⁺ +N⁺⁵O₃^-1→N⁺⁴O₂ + H₂O

 

Look at oxidation half reaction Zn is losing 2 electrons, while in reduction half reaction N is gaining 1 electron that is not possible, loss and gain of electron is always equal in redox equations. So to balance the loss and gain of electron multiply reduction half reaction to 2.

1e^- +2H⁺ +N⁺⁵O₃ ^-1→N⁺⁴O₂ + H₂O) x 2

2e^- +4H⁺ +2N⁺⁵O₃ ^-1→2N⁺⁴O₂ +2 H₂O

Step 5: Now we have equal electron number at both side of equation. Now add both halves reaction.

Oxidation half  reaction:  Zn^{0  →}   Zn^{+2  +} 2e^-   
Reduction half reaction:2e^- +4H⁺ +2N⁺⁵O₃ ^-1→2N⁺⁴O₂ +2 H₂O

Sum equation: Zn +2NO₃^-1 + 4H⁺ → Zn⁺² + 2NO₂ + 2H₂O

Now look at the sum equation everything is balanced now but as we are balancing it in basic medium we have to do two extra steps.we have added H+ which have made the medium acidic now we perform two extra steps. 

step 6: Add OH- for each H+ at both side of equation.

4OH^-+Zn +2NO₃^-1 + 4H⁺ → Zn⁺² + 2NO₂ + 2H₂O+ 4OH^-

Now combine H+ and OH- to form H2O. 4H+ and 4OH- at reactant side combine together to form 4H2O.

Zn +2NO₃^-1 + 4H₂ O→ Zn⁺² + 2NO₂ + 2H₂O+ 4OH^-

Step 7: Now the last step is to subtract H2O from both side of equation.

Zn +2NO₃^-1 + 2 4H₂ O→ Zn⁺² + 2NO₂ + 2H₂O+ 4OH^-            

-2H₂O                               - 2H₂O

Final balanced equation:Zn +2NO₃^-1 + 2H₂ O→ Zn⁺² + 2NO₂ + 4OH^- 

After subtracting H2O from both side of equation two H2O left in final equation.Now we have OH- in equation which is showing medium is basic.Lastly check again the number of atoms and charges.

 ^{Now the equation is balanced in all ways i-e atoms and charges.}