Balance the redox reaction Zn+NO3^-=Zn^²+ +NH3
No, ZnS + NO₃⁻ → Zn²⁺ + S + NO is not the same as
Zn + NO3^- => Zn^²+ + NH3
4Zn + 9H^+ + NO3^- ---> 4Zn^2+ + 3H2O + NH3
The equation given in question is as follow,
ZnS + NO₃⁻ → Zn²⁺ + S + NO
Oxidation state of elements in reactant side are as follow,
Zn = +2
S = -2
N = +5
O = -2
Oxidation state of elements in product side are as follow,
Zn = +2
S = 0
N = +2
O = -2
So, In this reaction Nitrogen is Reduced while Sulfur is Oxidized.
Now Split the reaction into two half cell reactions,
Reduction Reaction,
ZnS → S + 2e⁻
Oxidation Reaction,
NO₃⁻ + 3e⁻ → NO
As the oxygen atoms are not balance, So, in acidic medium add H⁺ on the side having greater number of Oxygen atom and H₂O on the side having less number of Oxygen atoms,
Hence,
H⁺ + NO₃⁻ + 3e⁻ → NO + H₂O
Now Balance the reaction,
4H⁺ + NO₃⁻ + 3e⁻ → NO + 2H₂O
So, write both half cell equations as,
ZnS → S + 2e⁻ --------- (1)
4H⁺ + NO₃⁻ + 3e⁻ → NO + 2H₂O -------(2)
Multiply eq. 1 with 3 and eq. 2 with 2 to equalize the electron.
So,
3ZnS → 3S + 6e⁻
8H⁺ + 2NO₃⁻ + 6e⁻ → 2NO + 4H₂O
_____________________________________(e⁻ cancelled)
3ZnS + 2NO₃⁻ + 8H⁺ → 3Zn + 3S + 2NO + 4H₂O
To balance the redox reaction: Zn + NO3- = Zn2+ + NH3, you need to follow these steps:
Step 1: Identify the oxidation states of each element in the reaction.
Zn (elemental form) has an oxidation state of 0.
NO3- (nitrate ion) has an oxidation state of -1.
Zn2+ has an oxidation state of +2.
NH3 (ammonia) has an oxidation state of -3.
Step 2: Determine which elements are being oxidized and which are being reduced.
In this reaction, Zn is being oxidized from an oxidation state of 0 to +2, and NO3- is being reduced from an oxidation state of -1 to -3.
Step 3: Write the half-reactions for oxidation and reduction.
Oxidation half-reaction:
Zn → Zn2+ + 2e-
Reduction half-reaction:
NO3- + 3H2O + 2e- → NH3 + 3O2-
Step 4: Balance the atoms in each half-reaction.
Oxidation half-reaction:
Zn → Zn2+ + 2e-
Reduction half-reaction:
4NO3- + 12H2O + 10e- → 4NH3 + 12O2-
Step 5: Balance the charges in each half-reaction.
Oxidation half-reaction:
Zn → Zn2+ + 2e-
Reduction half-reaction:
10NO3- + 30H2O + 20e- → 10NH3 + 30O2-
Step 6: Balance the electrons by multiplying the half-reactions.
Oxidation half-reaction (multiplied by 10):
10Zn → 10Zn2+ + 20e-
Reduction half-reaction (multiplied by 2):
20NO3- + 60H2O + 40e- → 20NH3 + 60O2-
Step 7: Add the balanced half-reactions together and simplify.
10Zn + 20NO3- + 60H2O → 10Zn2+ + 20NH3 + 60O2-
To balance the redox reaction Zn + NO3^- = Zn^2+ + NH3, follow these steps:
1. Write down the unbalanced equation:
Zn + NO3^- = Zn^2+ + NH3
2. Identify the elements that change oxidation states.
In this reaction, Zn changes from 0 to +2, and N changes from +5 to -3.
3. Divide the reaction into two half-reactions.
The half-reactions can be written as:
Oxidation half-reaction: Zn → Zn^2+
Reduction half-reaction: NO3^- → NH3
4. Balance the atoms in each half-reaction, excluding hydrogen and oxygen.
The oxidation half-reaction is already balanced.
The reduction half-reaction needs to be balanced:
NO3^- → NH3
There are 3 oxygen atoms on the left side and none on the right side, add 3 H2O molecules to the right side:
NO3^- + 3H2O → NH3
5. Balance the oxygen atoms by adding water molecules to the side that needs more oxygen.
The oxidation half-reaction requires no further balancing.
The reduction half-reaction has 6 hydrogen atoms on the right side, add 6 H+ ions to the left side:
NO3^- + 3H2O → NH3 + 6H+
6. Balance the hydrogen atoms by adding hydrogen ions (H+) to the opposite side of the water molecules.
The oxidation half-reaction requires no further balancing.
The reduction half-reaction is now balanced:
NO3^- + 3H2O → NH3 + 6H+
7. Balance the charge by adding electrons (e^-) to the side that has a higher positive charge.
The oxidation half-reaction requires 2 electrons (2e^-) on the right side:
Zn → Zn^2+ + 2e^-
The reduction half-reaction has a net charge of -1 on the left side (NO3^-) and a charge of +1 on the right side (NH3).
To balance the charges, add 6 electrons (6e^-) to the left side:
NO3^- + 3H2O + 6e^- → NH3 + 6H+
8. Make the number of electrons transferred in the oxidation and reduction half-reactions equal.
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to make the number of electrons equal:
3Zn → 3Zn^2+ + 6e^-
2NO3^- + 6H2O + 12e^- → 2NH3 + 12H+
9. Combine the two half-reactions by adding them together.
3Zn + 2NO3^- + 6H2O → 3Zn^2+ + 2NH3 + 6H+
10. Check that the atoms, charges, and oxidation numbers are balanced.
The equation is now balanced, with the same number of atoms and charges on both sides.
Therefore, the balanced redox reaction is:
3Zn + 2NO3^- + 6H2O → 3Zn^2+ + 2NH3 + 6H+