Question

Be sure to answer all parts. Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identifies of the oxidizing and reducing agents.

Please show work. Thank you.

Answer 1

the given reaction is

Zn(s)   + N03- ---> Zn(OH)42- + NH3

consider the oxidation reaction :

Zn ----> Zn(OH)42-

the steps to balance a redox reaction in a basic medium are as follows

1) first balance the Zinc atoms

Zn ---> Zn(OH)42-

2) now balance the oxygen atoms using H20

Zn + 4H20 ----> Zn(OH)42-

3) now balance the hydrogen atoms using H+

Zn + 4H20 ---> Zn(OH)42- + 4H+

4) now add equal number of OH- as H+ on both sides

Zn + 4H20 + 4OH- ---> Zn(OH)42- + 4H+ + 4OH-

5) now combine OH- and H+ to from H20

Zn + 4H20 + 4 OH- ---> Zn(OH)42- + 4H20

6) now balance the H20

Zn + 4OH- ---> Zn(OH)42-

7) now balacne the charge using electrons

Zn + 4OH- ---> Zn(OH)42- + 2e-

now consider the reduction reaction :

N03- --> NH3

follow the same steps as above

1) N03- ---> NH3

2) N03- ----> NH3 + 3H20

3) N03- + 9H+ ---> NH3 + 3H20

4) N03- + 9H+ + 9OH- ---> Nh3 + 3H20 + 9OH-

5) N03- + 9H20 ---> NH3 + 3H20 + 9OH-

6) N03- + 6H20 ---> NH3 + 9OH-

7) N03- + 6H20 + 8e- ---> NH3 + 9OH-

now

equate the number of electrons in both oxidation and reduction reactions using common multiples

4Zn + 16OH- ---> 4Zn(OH)42- + 8e-

N03- + 6H20 + 8e- ---> NH3 + 9OH-

cancel out the electrons and OH-

now final balanced reaction is

4 Zn (s) + N03- (aq) + 7OH- (aq) + 6H20 (l) ---> 4 Zn(OH)42- (aq) + NH3 (g)

now

oxidzing reagent is the one which undergoes reduction

so from the above reaction

the one that undergoes reduction is N03-

so

N03- is the oxidizing reagent

Similarly

the one which undergoes oxidation is the reducing agent

so

Zn is the reducing agent

Answer 2

To balance a redox equation you can do the following steps:

STEP 1. Assign oxidation numbers to identify your elements that oxidized and reduced.

+I +5 1a t

from this we can see that Zn's oxidation number went from 0 to 2+.

This means Zn is oxidized, so NO₃^- is oxidizing agent

on the other hand, N's oxidation state changed from +5 to -3

This means N is reduced, so Zn is the reducing agent

STEP 2. Split the reaction into its oxidation half and reduction half.

OXIDATION REDUCTION NO, (a) NHg)

STEP 3. Balance the atoms in each half. To balance O, add a molecule of H₂O for every O that needs to be balanced. To balance H, add an H⁺ for every H that needs to be balanced.

4H20(aq) tZa(s)→Zn(OH)42-(al ) + 4 H+ 9H'(aqjt NO3-(aq)-NH3(g)+ 3H2。

STEP 4. Balance the charge. Consider the total charge on each side (product and reactant side) and balance them by adding electrons. Remember that an electron has -1 charge.

0 o- +4

STEP 5. After balancing the charge on each half reaction, this time we will make sure that the number of electrons lost is equal to the number of electrons gained. We do this by multiplying a factor that will make them equal.

16 H20(I) + 4 Zn(s)→ 4 Zn(OH)42-(aq) + 16 H+(aq) + 8 e.

STEP 6. Combine the two half reactions. Cancel terms that appear on both reactant and product side.

13 oidation 16 HO) +4 Zn(s)> 4 Zn(OH(a) +16 H(a)se 13H20(I) + 4 Zn(s) + NO3(аф-) 4 Zn(OH)42-(aq) + NH3(g)+7 H"(аф

STEP 7. Normally we would end in STEP 6 if this was in acidic conditions. However, your problem says [basic] that means we need to make the reaction in basic medium.

To convert to basic medium, we add an OH- for every H+ in the reaction in STEP 6. H+ and OH- will combine to form H₂O and we further cancel terms to get the final balanced reaction.

+7 OH 7-H20(1)

SO our final balanced reaction in basic medium is:

7 OH-(аф+6 H2O(I) + 4 Zn(s) + NO3(aq)→ 4 Zn(OH)42-(aq) + ŅH3(g)

I hope this helps!