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

Redox equation is in which oxidation and reduction occur simultaneously. 

Balancing of such equation occurs in 5 steps. 

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 check the equation is balanced or not by counting atoms and charges.

 Practice Equation

Ag + NO₃^-1  →   Ag⁺¹ + NO

Step1: According to step 1 assign oxidation numbers to each element in given chemical equation.  

     Ag⁰ + N⁺⁵ O₃^-1  →   Ag⁺¹ + N⁺² O^-2 

^{Rules for Assigning Oxidation Numbers:}

Silver Ag has  zero 0 oxidation number being an element.

Nitrogen in NO3-1  has +5 oxidation number( rule 8)

Nitrogen in NO has +2 oxidation number (rule7)

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

Ag⁰ + N⁺⁵O₃^-1 →   Ag⁺¹ + N⁺²O

^{Silver (Ag) is increasing its oxidation number 0 → +1(Oxidation occurring)}

^{Recall Increase in oxidation number is Oxidation.}

^{Nitrogen (N)decreasing it oxidation number +5  → +2 ( reduction occurring)}

^{Recall: Decrease in oxidation number is reduction.}

^{Oxygen (O) is not changing its oxidation number} 

 

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)

 Ag⁰  →   Ag⁺¹ 

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

 ^{Ag0  →   Ag+1  + 1e-} 

Ag^{0  →}   Ag^{+1  +} 1e^-) x 3

3Ag⁰→ 3Ag⁺¹ + 3e^-

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

N⁺⁵O₃^-1     →   N⁺²O

There is one N at both side and 3 O at reactants and 1 N at product side. Here  N atoms are equal at both side  O is not balance . In such  case we are allowed to balance the O by adding H2O as medium is acidic. we can add 2H2O to balance the deficiency of two oxygen atoms at product side.

 

N⁺⁵O₃^-1     →   N⁺²O   +  2H₂O

Now O atoms are balanced at both side but now products have 4 H but there is no H at all at reactant side now we have option to add H+ to balance H at both side of chemical equation. Add 4H+ at reactant side.

4H⁺ +N⁺⁵O₃^-1              →   N⁺²O   +  2H₂O

Now balance number of electrons to balance charges numbers.

4H⁺ +N⁺⁵O₃^-1    + 3e^-          →   N⁺²O   +  2H₂O

There is  gain of 3 electrons by N atom - as oxidation number is decreasing +5 to+2 - but in oxidation half reaction there is loss of 1 electron (look at oxidation half reaction). There must be loss of 3 electrons by Ag in oxidation half reaction so N can gain 3 electrons to get  reduce. so balance the number of electron in oxidation half reaction by multiplying with 3 ( see above) in oxidation half reaction. 

Step 5:Now add both oxidation half and reduction half reactions.

3Ag⁰→ 3Ag⁺¹ + 3e^-

4H⁺ +N⁺⁵ O₃^-1    + 3e^-   →   N⁺² O   +  2H₂O

Sum Equation:

3Ag⁰ +N⁺⁵O₃^-1 +4H^+  →3Ag⁺¹ + N⁺²O   +  2H₂O

Now check number of atoms and  charges at both side of chemical equation.

Final Balanced Equation

3Ag +NO₃^-1 +4H^+  →3Ag⁺¹ + NO   +  2H₂O

 

charge on left side = +4(H) -1( NO3)= +3

charge on right side= +1 0n 3 Ag =+3