Chemistry Oxidation number and types of redox reactions

### Topics to be covered

=> Oxidation number
=> Types of redox reactions

### OXIDATION NUMBER

color{green}("A less obvious example of electron transfer is realised when hydrogen combines with oxygen to form water by the reaction:")

color{red}(2H_2(g) + O_2 (g) → 2H_2O (l)) ................ (8.18)

Though not simple in its approach, yet we can visualise the color{red}(H) atom as going from a neutral (zero) state in color{red}(H_2) to a positive state in color{red}((H_2O), the color{red}(O) atom goes from a zero state in color{red}(O_2) to a dinegative state in color{red}(H_2O). It is assumed that there is an electron transfer from color{red}(H) to color{red}(O) and consequently color{red}(H_2) is oxidised and color{red}(O_2) is reduced.

=> However, as we shall see later, the charge transfer is only partial and is perhaps better described as an electron shift rather than a complete loss of electron by color{red}(H) and gain by color{red}(O). What has been said here with respect to equation (8.18) may be true for a good number of other reactions involving covalent compounds.
color{green}("Two such examples of this class of the reactions are:")

color{red}(H_2(s) +Cl_2(g) → 2HCl (g)) .................(8.19)

color{red}(CH_4(g) +4Cl_2(g) → C Cl_4 (l) +4HCl (g))  ...............(8.20)

=> In order to keep track of electron shifts in chemical reactions involving formation of covalent compounds, a more practical method of using oxidation number has been developed.

=> In this method, it is always assumed that there is a complete transfer of electron from a less electronegative atom to a more electonegative atom. For example, we rewrite equations (8.18 to 8.20) to show charge on each of the atoms forming part of the reaction :

color{red}(overset(0)(2H_2(g)) +overset(0)(O_2(g)) → overset(+1)(2H_2) overset(-2)(O )(l)) .................(8.21)

color{red}(overset(0)(H_2(s)) +overset(0)(Cl_2 (g)) → 2H overset(+1 \ \ -1)(Cl)(g)) ..............................(8.22)

color{red}(overset(-4 )(C)overset(+1)(H_4)(g) + overset(0)(4Cl_2(g)) → overset(+4)(C) overset(-1)(Cl_4)(l) + overset(+1)(H) overset(-1)(Cl)(g)) ....................(8.23)

=> Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on electron pair in a covalent bond belongs entirely to more electronegative element.The basis that electron in a covalent bond belongs entirely to more electronegative element.

=> It is not always possible to remember or make out easily in a compound/ion, which element is more electronegative than the other.

=> Therefore, a set of rules has been formulated to determine the oxidation number of an element in a compound/ion.

=> If two or more than two atoms of an element are present in the molecule/ion such as color{red}(Na_2S_2O_3//Cr_2O_7^(2–)), the oxidation number of the atom of that element will then be the average of the oxidation number of all the atoms of that element. We may at this stage, state the rules for the calculation of oxidation number. These rules are:

1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. Evidently each atom in color{red}(H_2, O_2, Cl_2, O_3, P_4, S_8, Na, Mg, Al) has the oxidation number zero.

2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Thus (Na^+) ion has an oxidation number of color{red}(+1, Mg^(2+)) ion, color{red}(+2, Fe^(3+)) ion, color{red}(+3, Cl^–) ion, color{red}(–1, O^(2–)) ion, color{red}(–2;) and so on. In their compounds all alkali metals have oxidation number of color{red}(+1), and all alkaline earth metals have an oxidation number of color{red}(+2). Aluminium is regarded to have an oxidation number of color{red}(+3) in all its compounds.

3. The oxidation number of oxygen in most compounds is color{red}(–2). However, we come across two kinds of exceptions here. One arises in the case of peroxides and superoxides, the compounds of oxygen in which oxygen atoms are directly linked to each other. While in peroxides (e.g., color{red}(H_2O_2, Na_2O_2)), each oxygen atom is assigned an oxidation number of color{red}(–1,) in superoxides (e.g., color{red}(KO_2, RbO_2)) each oxygen atom is assigned an oxidation number of color{red}(–(½)). The second exception appears rarely, i.e. when oxygen is bonded to fluorine. In such compounds e.g., oxygen difluoride color{red}((OF_2)) and dioxygen difluoride color{red}((O_2F_2)), the oxygen is assigned an oxidation number of color{red}(+2) and color{red}(+1), respectively. The number assigned to oxygen will depend upon the bonding state of oxygen but this number would now be a positive figure only.

4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compounds containing two elements). For example, in color{red}(LiH, NaH), and color{red}(CaH_2), its oxidation number is color{red}(–1.)

5. In all its compounds, fluorine has an oxidation number of color{red}(–1). Other halogens (color{red}(Cl, Br), and color{red}(I)) also have an oxidation number of color{red}(–1), when they occur as halide ions in their compounds. Chlorine, bromine and iodine when combined with oxygen, for example in oxoacids and oxoanions, have positive oxidation numbers.

6. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, the sum of oxidation number of three oxygen atoms and one carbon atom in the carbonate ion, color{red}((CO_3)^(2–)) must equal color{red}(–2).

=> By the application of above rules, we can find out the oxidation number of the desired element in a molecule or in an ion.
• It is clear that the metallic elements have positive oxidation number and nonmetallic elements have positive or negative oxidation number.
•The atoms of transition elements usually display several positive oxidation states. The highest oxidation number of a representative element is the group number for the first two groups and the group number minus 10 (following the long form of periodic table) for the other groups.
• Thus, it implies that the highest value of oxidation number exhibited by an atom of an element generally increases across the period in the periodic table.
• In the third period, the highest value of oxidation number changes from 1 to 7 as indicated below in the compounds of the elements.

=> A term that is often used interchangeably with the oxidation number is the oxidation state. Thus in color{red}(CO_2), the oxidation state of carbon is +4, that is also its oxidation number and similarly the oxidation state as well as oxidation number of oxygen is – 2. This implies that the oxidation number denotes the oxidation state of an element in a compound.

=> The oxidation number/state of a metal in a compound is sometimes presented according to the notation given by German chemist, Alfred Stock. It is popularly known as Stock notation.
•According to this, the oxidation number is expressed by putting a Roman numeral representing the oxidation number in parenthesis after the symbol of the metal in the molecular formula.
•Thus aurous chloride and auric chloride are written as color{red}(Au(I)Cl) and color{red}(Au(III)Cl_3).
•Similarly, stannous chloride and stannic chloride are written as color{red}(Sn(II)Cl_2) and color{red}(Sn(IV)Cl_4).
• This change in oxidation number implies change in oxidation state, which in turn helps to identify whether the species is present in oxidised form or reduced form.
• Thus, color{red}(Hg_2(I)Cl_2) is the reduced form of color{red}(Hg(II) Cl_2).

=> The idea of oxidation number has been invariably applied to define oxidation, reduction, oxidising agent (oxidant), reducing agent (reductant) and the redox reaction.

color { maroon} ® color{maroon} ul (" REMEMBER")

color{green}("To summarise, we may say that::")

•color{green}("Oxidation:") An increase in the oxidation number of the element in the given substance.

•color{green}("Reduction:") A decrease in the oxidation number of the element in the given substance.

• color{green}("Oxidising agent:") A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also.

• color{green}("Reducing agent:") A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.

• color{green}("Redox reactions:") Reactions which involve change in oxidation number of the interacting species.

Q 3161880725

Using Stock notation, represent the following compounds : HAuCl_4, Tl_2O, FeO, Fe_2O_3, CuI, CuO, MnO and MnO_2.

Solution:

By applying various rules of calculating the oxidation number of the desired element in a compound, the oxidation number of each metallic element in its compound is as follows:

HAuCl_4 → Au has 3

Tl_2O → Tl has 1

FeO → Fe has 2

Fe_2O_3 → Fe has 3

CuI → Cu has 1

CuO → Cu has 2

MnO → Mn has 2

MnO_2 → Mn has 4

Therefore, these compounds may be represented as:
HAu(III)Cl_4, Tl_2(I)O, Fe(II)O, Fe_2(III)O_3, Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O_2.
Q 3171880726

Justify that the reaction: 2Cu_2O(s) + Cu_2S(s) → 6Cu(s) + SO_2(g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.

Solution:

Let us assign oxidation number to each of the species in the reaction under examination. This results into:

overset(+1 qquad -2)(2Cu_2O(s)) +overset(+1 qquad -2)(Cu_2S(s)) → overset(0)(6Cu(s))+overset(+4 qquad -2)(SO_2)

We therefore, conclude that in this reaction copper is reduced from +1 state to zero oxidation state and sulphur is oxidised from –2 state to +4 state. The above reaction is thus a redox reaction. Further, Cu_2O helps sulphur in Cu_2S to increase its oxidation number, therefore, Cu(I) is an oxidant; and sulphur of Cu_2S helps copper both in Cu_2S itself and Cu_2O to decrease its oxidation number; therefore, sulphur of Cu_2S is reductant

### Types of Redox Reactions

1. color{green}("Combination reactions :")

color{green}("A combination reaction may be denoted in the manner:")

color{red}(A+B → C)

Either color{red}(A) and color{red}(B) or both color{red}(A) and color{red}(B) must be in the elemental form for such a reaction to be a redox reaction. All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions. Some important examples of this category are:

color{red}(overset(0)(C(s)) +overset(0)(O_2(g)) overset(Delta)→ overset(+4 )(C ) overset(-2)(O_2)(g)) .............(8.24)

color{red}(overset(0)(3Mg(s)) +overset(0)(N_2(g)) overset(Delta)→ overset(+2 )(Mg_3) overset(-3)(N_2)(s)) ...........(8.25)

color{red}(overset(-4 )(C) overset(+1)(H_4)(g) + overset(0)(2O_2(g)) overset(Delta)→ overset(+4 \ \ -2)(CO_2(g))+2H_2O(l))

2. color{green}("Decomposition reactions :")

Decomposition reactions are the opposite of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state. Examples of this class of reactions are:

color{red}(overset(+1 \ \ -2)(2H_2O(l)) overset(Delta)→ overset(0)(2H_2(g))+overset(0)(O_2(g))) .................(8.26)

color{red}(overset(+1 \ \ -1)(2NaH(s)) overset(Delta)→ overset(0)(2Na(s)) +overset(0)(H_2(g))) ...........(8.27)

color{red}(overset(+1 \ \ +5 \ \ -2)(2KClO_2(s)) overset(Delta)→ overset(+1 \ \ -1)(2KCl(s)) +overset(0)(3O_2(g))) .................(8.28)

It may carefully be noted that there is no change in the oxidation number of hydrogen in methane under combination reactions and that of potassium in potassium chlorate in reaction (8.28). This may also be noted here that all decomposition reactions are not redox reactions. For example, decomposition of calcium carbonate is not a redox reaction.

color{red}(overset(+2 \ \ +4 \ \ -2)(CaCO_3(s)) overset(Delta)→ overset(+2 \ \ -2)(CaO(s)) + overset(+4 \ \ -2)(CO_2(g)))

3. color{green}("Displacement reactions : ")

In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as:

color{red}(X + YZ → XZ + Y)

color{green}("Displacement reactions fit into two categories:") metal displacement and non-metal displacement

(a) color{green}("Metal displacement:") A metal in a compound can be displaced by another metal in the uncombined state. We have already discussed about this class of the reactions under section 8.2.1. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores. A few such examples are:

color{red}(overset(+2 \ \ +6 \ \ -2) (CuSO_4(aq)) + overset(0)(Zn(s)) → overset(0)(Cu(s)) + overset(+2 \ \ +6 \ \ -2)(ZnSO_4(aq))) ..........(8.29)

color{red}(overset(+5 \ \ -2)(V_2O_5 (s))+ overset(0)(5Ca(s)) overset(Delta)→ overset(0)(2V(s)) +overset(+2 \ \ -2)(5CaO(s))) ............(8.30)

color{red}(overset(+4 \ \ -1)(TiCl_4(l))+overset(0)(2Mg(s)) overset(Delta)→ overset(0)(Ti(s)) + overset(+2 \ \ -1)(2MgCl_2(s))) ..................(8.31)

color{red}(overset(+3 \ \ -2)(Cr_2O_2(s)) +overset(0)(2Al(s)) overset(Delta)→ overset(+3 \ \ -2)(Al_2O_3(s)) +overset(0)(2Cr(s))) ...................(8.32)

In each case, the reducing metal is a better reducing agent than the one that is being reduced which evidently shows more capability to lose electrons as compared to the one that is reduced.

b) color{green}("Non-metal displacement:") The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement. All alkali metals and some alkaline earth metals color{red}("(Ca, Sr, and Ba)") which are very good reductants, will displace hydrogen from cold water.

color{red}(overset(0)(2Na(s)) + overset(+1 \ \ -2)(2H_2O(l)) → overset(+1 \ \ -2 \ \ +1)(2NaOH(aq))+overset(0)(H_2(g))) .........................(8.33)

color{red}(overset(0)(Ca(s)) + overset(+1 \ \ -2)(2H_2O(l)) → overset(+2 \ \ -2 \ \ +1)(Ca(OH)_2(aq)) +overset(0)(H_2(g))) ......................(8.34)

color{green}("Less active metals such as magnesium and iron react with steam to produce dihydrogen gas:")

color{red}(overset(0)(Mg(s)) + overset(+1 \ \ -2)(2H_2O(l)) overset(Delta)→ overset(+2 \ \ -2 \ \ +1)(Mg(OH)_2(s))+ overset(0)(H_2(g))) ..........(8.35)

color{red}(overset(0)(2Fe(s)) +overset(+1 \ \ -2)(3H_2O(I)) overset(Delta)→ overset(+3 \ \ -2)(Fe_2O_3(s)) +overset(0)(3H_2(g))) ............................(8.36)

Many metals, including those which do not react with cold water, are capable of displacing hydrogen from acids. Dihydrogen from acids may even be produced by such metals which do not react with steam. Cadmium and tin are the examples of such metals. A few examples for the displacement of hydrogen from acids are:

color{red}(overset(0)(Zn(s)) + overset(+1)(2H) overset(-1)(Cl)(aq) → overset(+2)(Zn) overset(-1)(Cl_2)(aq) +overset(0)(H_2(g))) ..................(8.37)

color{red}(overset(0)(Mg(s)) + overset(+1)(2H) overset(-1)(Cl)(aq) → overset(+2)(Mg) overset(-1)(Cl_2)(aq) +overset(0)(H_2)(g)) ..............(8.38)

color{red}(overset(0)(Fe(s)) + overset(+1)(2H) overset(-1)(Cl)(aq) → overset(+2)(Fe) overset(-1)(Cl_2)(aq) +overset(0)(H_2(g))) ........................(8.39)

=> Reactions (8.37 to 8.39) are used to prepare dihydrogen gas in the laboratory. Here, the reactivity of metals is reflected in the rate of hydrogen gas evolution, which is the slowest for the least active metal color{red}(Fe), and the fastest for the most reactive metal, color{red}(Mg). Very less active metals, which may occur in the native state such as silver color{red}((Ag)), and gold color{red}((Au)) do not react even with hydrochloric acid.

=> Like metals, activity series also exists for the halogens. The power of these elements as oxidising agents decreases as we move down from fluorine to iodine in group 17 of the periodic table. This implies that fluorine is so reactive that it can replace chloride, bromide and iodide ions in solution. In fact, fluorine is so reactive that it attacks water and displaces the oxygen of water :

color{red}(overset(+1)(2H_2) overset(-2)(O)(l) + overset(0)(2F_2)(g) → overset(+1)(4H) overset(-1)(F)(aq) +overset(0)(O_2)(g)) .............(8.40)

It is for this reason that the displacement reactions of chlorine, bromine and iodine using fluorine are not generally carried out in aqueous solution. On the other hand, chlorine can displace bromide and iodide ions in an aqueous solution as shown below:

color{red}(overset(0)(Cl_2)(g) +overset(+1)(2K)overset(-1)(Br) (aq) → overset(+1)(2K) overset(-1)(Cl)(aq) + overset(0)(Br_2)(I)) ............(8.41)

color{red}(overset(0)(Cl_2(g)) + overset(+1)(2K)overset(-1)(l)(aq) →overset(+1)(2K)overset(-1)(Cl)(aq) +overset(0)(I_2)(s)) ...............(8.42)

As color{red}(Br_2) and color{red}(I_2) are coloured and dissolve in color{red}(C Cl_4), can easily be identified from the colour of the solution. The above reactions can be written in ionic form as:

color{red}(overset(0)(Cl_2(g))+overset(-1)(2Br^(-))(aq) → overset(-1)(2Cl^(-))(aq) +overset(0)(Br_2)(I)) .................(8.41a)

color{red}(overset(0)(Cl_2(g))+overset(-1)(2I^(-))(aq) → overset(-1)(2Cl^(-)) (aq) +overset(0)(I_2)(s)) .......................(8.42b)

Reactions (8.41) and (8.42) form the basis of identifying color{red}(Br^–) and color{red}(I^–) in the laboratory through the test popularly known as ‘Layer Test’. It may not be out of place to mention here that bromine likewise can displace iodide ion in solution:

color{red}(overset(0)(Br_2)(l) +overset(-1)(2I^(-)) (aq) → overset(-1)(2Br^(-)) (aq) +overset(0)(I_2)(s)) ....................(8.43)

The halogen displacement reactions have a direct industrial application. The recovery of halogens from their halides requires an oxidation process, which is represented by:

color{red}(2X^(-) → X_2 + 2e^(-)) ...................(8.44)

here color{red}(X) denotes a halogen element. Whereas chemical means are available to oxidise color{red}(Cl^–),
color{red}(Br^–) and color{red}(I^–), as fluorine is the strongest oxidising agent; there is no way to convert color{red}(F^–) ions to color{red}(F_2) by chemical means. The only way to achieve color{red}(F_2) from color{red}(F^–) is to oxidise electrolytically, the details of which you will study at a later stage.

4. color{green}("Disproportionation reactions :")

•Disproportionation reactions are a special type of redox reactions.
• In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced.
•One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states.
•The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction.
•The decomposition of hydrogen peroxide is a familiar example of the reaction, where oxygen experiences disproportionation.

color{red}(overset(+)(2H_2) overset(-1)(O_2)(aq) → overset(+1)(2H_2)overset(-2)(O)(l) + overset(0)(O_2)(g)) ....................(8.45)

Here the oxygen of peroxide, which is present in –1 state, is converted to zero oxidation state in color{red}(O_2) and decreases to –2 oxidation state in color{red}(H_2O).

color{green}("Phosphorous, sulphur and chlorine undergo disproportionation in the alkaline medium as shown below :")

color{red}(overset(0)(P_4(s)) +3OH^(-) (aq) + 3H_2O(l) → overset(-3)(PH_3)(g) +3H_2 overset(-1)(PO_2^(-))(aq)) .......................(8.46)

color{red}(overset(0)(S_8(s)) +12 OH^(-) (aq) → overset(-2)(4S^(2-) (aq)) + overset(+2)(2S_2)O_3^(2-)(aq) + 6H_2O(l)) ....................(8.47)

color{red}(overset(0)(Cl_2)(g) +2OH^(-) (aq) → overset(+1)(ClO^(-))(aq) + overset(-1)(Cl^(-))(aq) +H_2O(l)) .....................(8.48)

The reaction (8.48) describes the formation of household bleaching agents. The hypochlorite ion (color{red}(ClO^–)) formed in the reaction oxidises the colour-bearing stains of the substances to colourless compounds.

It is of interest to mention here that whereas bromine and iodine follow the same trend as exhibited by chlorine in reaction (8.48), fluorine shows deviation from this behaviour when it reacts with alkali. The reaction that takes place in the case of fluorine is as follows:

color{red}(2F_2(g) +2OH^(-)(aq) → 2F^(-)(aq) +OF_2(g) +H_2O(l)) ..................(8.49)

(It is to be noted with care that fluorine in reaction (8.49) will undoubtedly attack water to produce some oxygen also). This departure shown by fluorine is not surprising for us as we know the limitation of fluorine that, being the most electronegative element, it cannot exhibit any positive oxidation state. This means that among halogens, fluorine does not show a disproportionation tendency.
Q 3181880727

Which of the following species, do not show disproportionation reaction and why ?

ClO^(-) , ClO_2^(-) , ClO_3^(-) and ClO_4^(-)

Also write reaction for each of the species that disproportionates.

Solution:

Among the oxoanions of chlorine listed above, ClO_4^( –) does not disproportionate because in this oxoanion chlorine is present in its highest oxidation state that is, +7. The disproportionation reactions for the other three oxoanions of chlorine are as follows:

overset(+1)(3ClO^(-)) → overset(-1)(2Cl^(-))+overset(+5) (ClO_3^(-))

overset(+3)(6ClO_2^(-)) overset(hv)→ overset(+5)(4ClO_3^(-)) + overset(-1)(2Cl^(-))

overset(+5) (4ClO_3^(-))→ overset(-1) (Cl^(-)) +overset(+7)(3ClO_4^(-))
Q 3101880728

Suggest a scheme of classification of the following redox reactions

(a) N_2 (g) +O_2 (g) → 2NO (g)

(b) 2Pb (NO_3)_2(s) → 2PbO(s) +2NO_2 (g) +1/2 O_2 (g)

(c) NaH(s) +H_2O(l) → NaOH(aq) +H_2 (g)

(d) 2NO_2(g) +2OH^(-)(aq) → NO_2^(-)(aq) + NO_3^(-) (aq) +H_2O(l)

Solution:

In reaction (a), the compound nitric oxide is formed by the combination of the elemental substances, nitrogen and oxygen; therefore, this is an example of combination redox reactions. The reaction (b) involves the breaking down of lead nitrate into three components; therefore, this is categorised under decomposition redox reaction. In reaction (c), hydrogen of water has been displaced by hydride ion into dihydrogen gas. Therefore, this may be called as displacement redox reaction. The reaction (d) involves disproportionation of NO_2 (+4 state) into NO_2^(–) (+3 state) and NO_3^(–) (+5 state). Therefore reaction (d) is an example of disproportionation redox reaction
Q 3111880729

Why do the following reactions proceed differently ?
Pb_3O_4 + 8HCl → 3PbCl_2 + Cl_2 + 4H_2O and Pb_3O_4 + 4HNO_3 → 2Pb(NO_3)_2 + PbO_2 + 2H_2O

Solution:

Pb_3O_4 is actually a stoichiometric mixture of 2 mol of PbO and 1 mol of PbO_2. In PbO_2, lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO_2 thus can act as an oxidant (oxidising agent) and, therefore, can oxidise Cl^– ion of HCl into chlorine. We may also keep in mind that PbO is a basic oxide. Therefore, the reaction

Pb_3O_4 + 8HCl → 3PbCl_2 +Cl_2 +4H_2O

can be splitted into two reactions namely:

2PbO +4HCl → 2PbCl_2+2H_2O (acid-base reaction)

overset(+4)(PbO_2) + overset(-1)(4HCl) → overset(+2)(PbCl_2) +overset(0)(Cl_2) +2H_2O (redox reaction)

Since HNO_3 itself is an oxidising agent therefore, it is unlikely that the reaction may occur between PbO_2 and HNO_3. However, the acid-base reaction occurs between PbO and HNO_3 as:

2PbO +4 HNO_3 → 2Pb(NO_3)_2 +2H_2O

It is the passive nature of PbO_2 against HNO_3 that makes the reaction different from the one that follows with HCl.