Start by assigning oxidation numbers to the atoms that take part in the reaction.

#stackrel(color(blue)(0))(Zn) + stackrel(color(blue)(+1))(H) stackrel(color(blue)(+5))(N) stackrel(color(blue)(-2))(O_3) -> stackrel(color(blue)(+2))(Zn) (stackrel(color(blue)(+5))(N) stackrel(color(blue)(-2))(O_3))_2 + stackrel(color(blue)(-3))(N) stackrel(color(blue)(+1))(H_4) stackrel(color(blue)(-5))(N) stackrel(color(blue)(-2))(O_3) + stackrel(color(blue)(+1))(H_2) stackrel(color(blue)(-2))(O)#

Notice that the oxidation state of zinc changes from zero on the reactants" side lớn +2 on the products" side, which means that it is being oxidized.

Now take a look at nitrogen"s oxidation state. Notice that some nitrogen atom keep their oxidation state constant, i.e. +5 on both sides of the equation, but that other nitrogen atoms change their oxidation state from +5 lớn -3, which implies that they are being reduced.

The oxidation và reduction half reactions look like this

Oxidation half-reaction

#stackrel(color(blue)(0))(Zn) + HNO_3 -> stackrel(color(blue)(+2))(Zn) (NO_3)_2 + 2e^(-)#

Balance the nitrogen atoms first by multiplying the nitric acid by 2.

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#stackrel(color(blue)(0))(Zn) + 2HNO_3 -> stackrel(color(blue)(+2))(Zn) (NO_3)_2 + 2e^(-)#

Since you"re in acidic solution, you can balance the hydrogen atoms by adding protons, #H^(+)#, to lớn the side that lacks hydrogen, & oxygen atoms by adding water molecules to the side that lacks oxygen.

SInce you only need lớn balance the hydrogen atoms present on the reactants" side, you have

#stackrel(color(blue)(0))(Zn) + 2HNO_3 -> stackrel(color(blue)(+2))(Zn) (NO_3)_2 + 2e^(-) + 2H^(+)#

Reduction half-reaction

#Hstackrel(color(blue)(+5))(N)O_3 + 8e^(-) -> stackrel(color(blue)(-3))(N)H_4^(+)#

Balance the oxygen atoms by adding three water molecules to lớn the products" side.

#Hstackrel(color(blue)(+5))(N)O_3 + 8e^(-) -> stackrel(color(blue)(-3))(N)H_4^(+) + 3H_2O#

Balance the hydrogen atoms by adding protons to lớn the reactants" side.

#9H^(+) + Hstackrel(color(blue)(+5))(N)O_3 + 8e^(-) -> stackrel(color(blue)(-3))(N)H_4^(+) + 3H_2O#

In any redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

Since you have 4 electrons lost & 8 gained, multiply the oxidation half-reaction by 4 to get

#(4Zn + 8HNO_3 -> 4Zn(NO_3)_2 + 8e^(-) + 8H^(+)), (9H^(+) + HNO_3 + 8e^(-) -> NH_4^(+) + 3H_2O) :#

Add these two half-reactions together lớn get

#4Zn + 8HNO_3 + 9H^(+) + HNO_3 + cancel(8e^(-)) -> 4Zn(NO_3)_2 + cancel(8e^(-)) + NH_4NO_3 + 8H^(+) + 3H_2O#

If you reduce lượt thích terms that are present on both sides of the equation, you"ll get

#4Zn + 9HNO_3 + H^(+) -> 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O#

SInce you"re in acidic solution, the proton present on the reactants" side can only come from the nitric acid, which implies that you actually have 10 molecules of acid, instead of 9.

In the oxidation number change method the underlying principle is that the gain in the oxidation number (number of electrons) in one reactant must be equal to the loss in the oxidation number of the other reactant.

Step 1. Write down the unbalanced equation ("skeleton equation") of the chemical reaction. All reactants và products must be known. For a better result write the reaction in ionic form.

Step 2. Separate the process into half reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

b) Identify và write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Write down the transfer of electrons. Carefully, insert coefficients, if necessary, khổng lồ make the numbers of oxidized và reduced atoms equal on the two sides of each redox couples.

c) Combine these redox couples into two half-reactions: one for the oxidation, & one for the reduction (see: Divide the redox reaction into two half-reactions).

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas to lớn balance the number of atoms. Never change any formulas.

a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation và products lớn the right.

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b) Balance the charge. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a H+ ion khổng lồ the side deficient in positive charge.

c) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left và right side, if they aren"t equilibrate these atoms by adding water molecules.

Balanced half-reactions are well tabulated in handbooks and on the web in a "Tables of standard electrode potentials". These tables, by convention, contain the half-cell potentials for reduction. Khổng lồ make the oxidation reaction, simply reverse the reduction reaction và change the sign on the E1/2 value.

Step 4. Make electron gain equivalent khổng lồ electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. Khổng lồ make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

Step 5. địa chỉ the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side và all of the products together on the other side.

Step 6. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest phối of integers possible.

Finally, always check to see that the equation is balanced. First, verify that the equation contains the same type và number of atoms on both sides of the equation.

Second, verify that the sum of the charges on one side of the equation is equal to the sum of the charges on the other side. It doesn"t matter what the charge is as long as it is the same on both sides.

Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms on the right side, & since the charges on both sides are equal we can write a balanced equation.

Citing this page:

Generalic, Eni. "Balancing redox reactions by oxidation number change method." Eni
G. Periodic Table of the Elements. KTF-Split, 14 Jan. 2023. Web. Date of access. .


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