balance in basic medium Zn + NO3- gives Zn2+ +NH4+

Dear Student,
Please find below the solution to the asked query:

The following are the steps to balance a redox reaction in basic medium
 
Step1 :  Separate  the reaction into oxidation and reduction half reactions.
 
Oxidation half reaction:  Zn $\to$      Zn²⁺  + 2e^-
Reduction half reaction:   NO₃^- + e^-$\to$    NH₄⁺
 
Step2:  Balance the atoms in the two equation except for H and O

Both sides are already balanced.
 
Step 3:  Balance O atoms by adding appropriate number of H₂O molecules on the other side.
Oxidation half reaction does not contain O atoms and thus remains as such:  Zn $\to$  Zn²⁺  + 2e^-   
Reduction half reaction:  
Add 3 molecules of H₂O  on RHS
NO^3- + e ^- $\to$     NH₄⁺ + 3H₂O
 
Step 4:  Balance H atoms by adding appropriate number of H⁺ on the other side.
Oxidation half reaction remains as such.  Zn  $\to$    Zn²⁺  + 2e^- 
Reduction half reaction:  Add 10H⁺ to LHS
NO^3- + 10H⁺ + e^- $\to$     NH₄⁺ + 3H₂O 
 
Step5: Balance charge on both sides  of the two half reactions by adding electrons
Oxidation half reaction is balanced as LHS has not charge and RHS has +2 charge on Zn and -2 charge on electrons.
So, reactions remains the same. Zn  $\to$   Zn²⁺  + 2e^- 
 
Reduction half reaction:
Total charge on RHS = +1
Total charge on LHS = (-1) + 10 X (+1) =+9. To make the charge equivalent to RHS, we need 8 electrons.
So, the reaction is: NO^3- + 10H⁺ + 8e^-  $\to$     NH₄⁺ + 3H₂O 
 
Step 6: Reaction occurs in basic medium, so H⁺ ions need to be neutralised by addition of appropriate number of hydroxyl ions.
Oxidation half reaction has no H⁺ ion so the reaction remains the same.
Reduction half reaction has 10H⁺ ions on LHS and so 10 OH^- ions are added to both sides.  This will convert H⁺ ions into H₂O
So, the reaction is:NO^3- + 10H₂O  + 8e^-  $\to$     NH₄⁺ + 3H₂O + 10OH^-
 
Step 7: The number of electrons must be the same in the two half reactions. If not, multiply the equations with an appropriate integer.
In this case, oxidation half reaction needs to be multiplied by 4
Oxidation half reaction  x 4:   4 Zn $\to$     4Zn²⁺  + 8e^-
 Reduction half reaction:      NO^3- + 10H₂O  + 8e^- $\to$     NH₄⁺ + 3H₂O + 10 OH^-
 
Step 8: Add the two reactions.
Cancel common terms on the two sides.
Combined equation:   4 Zn  +   NO^3- + 10H₂O + 8e -  $\to$    4Zn²⁺  + 8e^- + NH₄⁺ + 3H₂O  + 10OH^-
8 electrons and 3 molecules of H2O being common on either side get cancelled.
Final balanced equation is:  4Zn  +   NO^3- + 7H₂O    $\to$   4Zn²⁺  + NH₄⁺  + 10OH^-

Hope this information will clear your doubts about balancing redox reactions.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards