JEE Exam  >  JEE Questions  >  Balance the following ionic equation by oxida... Start Learning for Free
Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+?
Verified Answer
Balance the following ionic equation by oxidation number method :-zn+N...
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 →→      Zn2+  + 2e-
Reduction half reaction:   NO3- + e-→→    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 →→  Zn2+  + 2e-   
Reduction half reaction:   
Add 3 molecules of H2O  on RHS
NO3- + e - →→     NH4+ + 3H2O
 
Step 4:  Balance H atoms by adding appropriate number of H+ on the other side.
Oxidation half reaction remains as such.  Zn  →→    Zn2+  + 2e- 
Reduction half reaction:  Add 10H+ to LHS
NO3- + 10H+ + e- →→     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  →→   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: NO3- + 10H+ + 8e-  →→     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:NO3- + 10H2O  + 8e-  →→     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 →→     4Zn2+  + 8e- 
 Reduction half reaction:      NO3- + 10H2O  + 8e- →→     NH4+ + 3H2O + 10 OH-
 
Step 8: Add the two reactions.
Cancel common terms on the two sides.
Combined equation:   4 Zn  +   NO3- + 10H2O + 8e -  →→    4Zn2+  + 8e- + NH4+ + 3H2O  + 10OH-
8 electrons and 3 molecules of H2O being common on either side get cancelled.
Final balanced equation is:  4Zn  +   NO3- + 7H2O    →→   4Zn2+  + NH4+  + 10OH-
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
Balance the following ionic equation by oxidation number method :-zn+N...
Balancing the Ionic Equation: Zn + NO3- → Zn2+ + NH4+

Step 1: Assigning oxidation numbers
- The oxidation number of zinc (Zn) is 0 because it is a pure element.
- The oxidation number of nitrate (NO3-) is -1.
- The oxidation number of zinc (Zn2+) is +2.
- The oxidation number of ammonium (NH4+) is +1.

Step 2: Identifying the oxidized and reduced species
- In this equation, zinc is oxidized from an oxidation number of 0 to +2, so it is the reducing agent.
- Nitrate is reduced from an oxidation number of -1 to 0, so it is the oxidizing agent.

Step 3: Writing the half-reactions
- Half-reaction for oxidation: Zn → Zn2+
- Half-reaction for reduction: NO3- → NH4+

Step 4: Balancing the half-reactions
- Balancing the oxidation half-reaction: Since zinc is being oxidized, we need to add two electrons (e-) to the left side to balance the charge. The balanced half-reaction is: Zn → Zn2+ + 2e-
- Balancing the reduction half-reaction: The nitrogen atom in nitrate (NO3-) has an oxidation number of +5, while the nitrogen atom in ammonium (NH4+) has an oxidation number of -3. To balance the nitrogen atoms, we need to add three electrons (3e-) to the right side. The balanced half-reaction is: NO3- + 3e- → NH4+

Step 5: Balancing the number of electrons transferred
- To equalize the number of electrons in both half-reactions, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. This will result in 6 electrons in both half-reactions.
Balanced oxidation half-reaction: 3Zn → 3Zn2+ + 6e-
Balanced reduction half-reaction: 2NO3- + 6e- → 2NH4+

Step 6: Combining the balanced half-reactions
- Now that the number of electrons is balanced, we can combine the two half-reactions into the overall balanced ionic equation:
3Zn + 2NO3- → 3Zn2+ + 2NH4+

Step 7: Final check
- Verify that the equation is balanced in terms of atoms and charge. In this case, there are three zinc atoms, two nitrogen atoms, ten hydrogen atoms, and twelve oxygen atoms on both sides of the equation. The charges are also balanced, with a total charge of -6 on both sides.

Explanation:
The oxidation number method is used to balance ionic equations by assigning oxidation numbers to each element involved. The oxidation half-reaction represents the species losing electrons (oxidation) and the reduction half-reaction represents the species gaining electrons (reduction). By balancing the number of electrons transferred in each half-reaction and combining them, the overall balanced equation is obtained. In this specific equation, zinc is oxidized
Explore Courses for JEE exam

Similar JEE Doubts

Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+?
Question Description
Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+?.
Solutions for Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+? defined & explained in the simplest way possible. Besides giving the explanation of Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+?, a detailed solution for Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+? has been provided alongside types of Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+? theory, EduRev gives you an ample number of questions to practice Balance the following ionic equation by oxidation number method :-zn+NO3 gives ZN2+ + NH4+? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev