NCERT Solutions for Class 11 Chemistry Chapter 8

Q.1 Assign oxidation number to the underlined elements in each of the following species:

(a) NaH₂PO₄ (b) NaHSO₄ (c) H₄P₂O₇ (d) K₂MnO₄

(e) CaO₂ (f) NaBH₄ (g) H₂S₂O₇ (h) KAl(SO₄)₂.12H₂O

Ans.

a) Let the oxidation number of P be x. Write the oxidation number of each atom above it.

+1 +1 x –2

Na H₂ P O₄

Thus, the sum of the oxidation number in NaH₂PO₄ = 1(+1) + 2(+1) + x + 4(-2) = x – 5

An atom as a whole is neutral. So sum of the oxidation numbers should be zero.

x – 5 = 0

or, x = +5

Thus the oxidation number of P in NaH₂PO₄ = +5

b) +1 +1 x –2

Na H S O₄

1(+1) + 1(+1) + x + 4(-2) = 0

or, x – 6 = 0

or, x = +6

Thus, the oxidation number of S in NaHSO₄ is +6.

c) +1 x –2

H₄ P₂ O₇

4(+1) + 2x + 7(-2) = 0

or, 2x – 10 = 0

or, x = +5

Thus, the oxidation number of P in H₄P₂O₇ is +5.

d) +1 x –2

K₂ Mn O₄

2(+1) + x + 4(-2) = 0

or, x – 6 = 0

or, x = +6

Thus, the oxidation number of Mn in K₂MnO₄ is +6.

e) +2 x

Ca O₂

1(+2) + 2x = 0

or, x = -1

Thus, the oxidation number of O in CaO₂ is -1.

f) In NaBH₄, H is present as hydride ion. Therefore, the oxidation state of H^– = -1.

+1 x –1

Na B H₄

1(+1) + x + 4(-1) = 0

or, x – 3 = 0

or, x = +3

Thus, the oxidation number of B in NaBH₄ is +3.

g)

+1 x –2

H₂ S₂ O₇

2(+1) + 2x + 7(-2) = 0

or, 2x – 12 = 0

or, x = +6

Thus, the oxidation number of S in H₂S₂O₇ is +6.

h)

+1 +3 x –2 +1 –2

K Al P (SO₄)₂ • 12H₂O

+1 + 3 + 2x + 8(-2) + 12 (2 X 1 – 2) = 0

or, 1 + 3 + 2x – 16 = 0

or, x = +6

Thus, the oxidation number of S in KAl(SO₄)₂.12H₂O is +6.

Q.2 What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your results?

(a) KI₃ (b) H₂S₄O₆ (c) Fe₃O₄ (d) CH₃CH₂OH (e) CH₃COOH

Ans.

a) In KI₃, the oxidation number of K is +1, based on that oxidation number of I = –1/3. But the oxidation number can’t be fractional. The oxidation number of iodine can be explained by considering the structure of KI₃.

K⁺[I — I ← I]^–

Thus, a coordinate bond is formed between I₂ molecule and I^– ion. The oxidation number of I₂ molecule is zero. The O.N. of iodine forming the coordinate bond is –1. Therefore the O.N. of iodine in KI₃ is –1.

b) Considering the structure of tetrathionate (H₂S₄O₆)

The oxidation number of the each S atom covalently bonded with each other in the middle is zero. But the oxidation number of each terminal S atom is +5 because these are attached with 3 oxygen atoms and one hydrogen atom.

c) If we consider the stoichiometry of Fe₃O₄, the chemical formula becomes as FeO.Fe₂O₃.

+2 –2 +3 –2

Fe O • Fe₂ O₃

Thus the oxidation number of Fe in Fe₃O₄ = +2 and +3.

d) By conventional method, let the oxidation number of C in CH₃CH₂OH or C₂H₆O = x

x +1 –2

C₂ H₆ O

or, 2x + 6(+1) + 1(-2) = 0

or, x= -2

By chemical bonding, second (marked by 2) carbon is attached with 3 hydrogen (H) atoms (less electronegative than carbon) and 1 CH₂OH group (more electronegative than carbon).

Therefore O.N. of second carbon = 3(+1) + x + 1(-1) = 0

or, x = -2

First carbon (marked by 1) is attached with one OH (O.N. = -1), two H and one CH₃ (O.N. = +1) group.

Thus O.N. of first carbon = 1 + 2(+1) + x + 1(-1) =0

or, x = -2

e) By conventional method,

x +1 –2

C₂ H₄ O₂

or, 2x + 4 + 2(-2) = 0

or, x = 0

By chemical bonding method, second (marked by 2) carbon is attached with three hydrogen (H) atoms (less electronegative than C) and one COOH group (more electronegative than C).

O.N. of second carbon = 3(+1) + x + 1(-1) = 0

or, x = -2

First carbon is attached with one double bonded oxygen atom, one OH (O.N.= -1) and one CH₃ (O.N = +1).

O.N. of first carbon = 1(-2) + 1(-1) + x + 1 = 0

or, x = +2

Q.3 Justify that the following reactions are redox reactions:

(a) CuO (s) + H₂ (g) → Cu (s) + H₂O (g)

(b) Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g)

(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆ (g) + 3LiCl (s) + 3 AlCl₃ (s)

(d) 2K (s) + F₂ (g) → 2K⁺F^– (s)

(e) 4 NH₃ (g) + 5 O₂ (g) → 4NO (g) + 6H₂O (g)

Ans.

(a)

\stackrel{\text{+2}}{\text{C}}\text{u}\stackrel{\text{–2}}{\text{O}}\text{(s)}\text{+}{\stackrel{\text{0}}{\text{H}}}_{\text{2}}\text{(g)}\to \stackrel{\text{0}}{\text{C}}\text{u(s)}\text{+}\stackrel{\text{+1}\text{–2}}{{\text{H}}_{\text{2}}\text{O}}\text{(g)}

CuO is reduced to Cu by removing oxygen and on the other hand H₂ is oxidised to H₂O by adding oxygen atom.

Again O.N. of Cu decreases from +2 to 0 and O.N. of H increases from 0 to +1. Thus, it is a redox reaction.

b)

{\stackrel{\text{+3}}{\text{Fe}}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s) + 3}\stackrel{\text{+2}}{\text{C}}\text{O(g)}\to \text{2}\stackrel{\text{0}}{\text{F}}\text{e(s) + 3}\stackrel{\text{+4}}{\text{C}}{\text{O}}_{\text{2}}\text{(g)}

In the above reaction, oxygen is removed from Fe₂O₃ and added to CO. The O.N. of Fe is decreases from +3 to 0 while O.N. of C increases from +2 to +4. Fe₂O₃ is reduced to Fe and CO is oxidised to CO₂. Thus, it is a redox reaction.

c)

\stackrel{\text{+3}}{\text{4B}}{\stackrel{\text{–1}}{\text{Cl}}}_{\text{3}}\text{(g) +}\stackrel{\text{+1}}{\text{3L}}\stackrel{\text{+3}}{\text{iA}}{\stackrel{\text{–1}}{\text{lH}}}_{\text{4}}\text{(s)}\to \stackrel{\text{–3}\text{+}\text{1}}{{\text{2B}}_{\text{2}}{\text{H}}_{\text{6}}}\text{(g) +}\stackrel{\text{+1}\text{–1}}{\text{3LiCl}}\text{(s) +}{\stackrel{\text{+3}\text{–1}}{\text{3AlCl}}}_{\text{3}}\text{(s)}

O.N. of B decreases from +3 in BCl₃ to -3 in B₂H₆ while that of H increases from -1 in LiAlH₄ to +1 in B₂H₆. Thus,BCl₃ is reduced and LiAlH₄ is oxidised. Thus, it is a redox reaction.

d)

\text{2}\stackrel{\text{0}}{\text{K}}\text{(s) +}{\stackrel{\text{0}}{\text{F}}}_{\text{2}}\text{(g)}\to \text{2}\stackrel{+1–1}{{\text{K}}^{\text{+}}{\text{F}}^{–}}\text{(s)}

K atom loses one electron and gets oxidised to K⁺ and F₂ gains one electron and is reduced to F^–. Thus, it is a redox reaction.

e)

{\stackrel{–3+1}{\text{4NH}}}_{\text{3}}\text{(g) + 5}{\stackrel{0}{\text{O}}}_2(\text{g})\to \stackrel{\text{+2}\text{–2}}{\text{4NO(}}\text{g) +}\stackrel{\text{+1}\text{–2}}{{\text{6H}}_{\text{2}}\text{O}}\text{(g)}

The O.N. of N increases from -3 in NH₃ to +2 in NO. Thus NH₃ is oxidised while O.N. of O decreases from 0 in O₂ to -2 in NO or H₂O. Thus O₂ is reduced. Thus, it is a redox reaction.

Q.4 Fluorine reacts with ice and results in the change: H₂O(s) + F₂ (g) → HF (g) + HOF (g)

Justify that this reaction is a redox reaction.

Ans.

Let us write the O.N. of each atom in above chemical reaction:

\stackrel{\text{+1}\text{–2}}{{\text{H}}_{\text{2}}\text{O}}\text{+}{\stackrel{\text{0}}{\text{F}}}_{\text{2}}\to \stackrel{\text{+1}\text{–1}}{\text{HF}}\text{+}\stackrel{\text{+1}\text{–2}\text{+1}}{\text{HOF}}

The O.N. of F decreases from 0 in F₂ to -1 in HF and increases from 0 in F₂ to +1 in HOF. Thus, F₂ simultaneously gets reduced and oxidised. Thus, it is a redox reaction. This type of reaction is also called as disproportionation reaction.

Q.5 Calculate the number of sulphur and nitrogen in H₂SO₅ and NO₃^–. Suggest structure of these compounds. Count for the fallacy.

Ans.

The structure of H₂SO₅ is shown here. Let us calculate O.N. of S by conventional method.

By conventional method, O.N. of S in H₂SO₅ is:

2(+1) + x + 5(-2) = 0

or, x = +8

But the maximum O.N. of S can’t exceed 6 as S has only six electrons in its valence shell. This fallacy is overcome if we calculate it by chemical bonding method.

Hence, 2(+1) +3(-2) + x + 2(-1) = 0

(For H) (For O) (For S) (For O-O)

or, x = +6

Let us calculate O.N. of NO₃^– ion by conventional method.

x + 3(-2) = -1

or, x= +5

According to chemical bonding, structure of NO₃^–ion is:

Let us calculate O.N. of NO₃^– ion by chemical bonding method.

x + 1(-1) + 1(-2) + 1(-2) =0

(For N) (For O-) (For =O) (For → O)

or x = +5

Thus, there is no fallacy about the O.N. of N in NO₃^–.

Q.6 Write formulas for the following compounds:

(a) Mercury (II) chloride (b) Nickel (II) sulphate

(c) Tin (IV) oxide (d) Thallium (I) sulphate

(e) Iron (III) sulphate (f) Chromium (III) oxide

Ans.

a) Hg (II)Cl₂ b) Ni(II)SO₄ c) Sn(IV)O₂

d) Tl₂(I)SO₄ e) Fe₂(III)(SO₄)₃ f) Cr₂(III)O₃

Q.7 Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

Ans.

Q.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

Ans.

In SO₂, the O.N. of S,

x + 2(–2) = 0

or, x = +4

Sulphur shows variable oxidation state. Oxidation number of S varies from -2 to +6. Thus, S in SO₂ can increase or decrease its oxidation number. So we can state that it can act as oxidising as well as reducing agent.

In H₂O₂, the O.N. of O is –1 (calculated by chemical bonding method). The range of O.N. of O may be from -2 to 0. So we find that O in H₂O₂ can give as well as gain electrons. It can act as oxidising as well as reducing agent.

In O₃, the O.N. of O is zero. It can only decrease its O.N. from O to –2 but can’t increase it. That’s why O₃ can act only as oxidising agent.

In HNO₃, O.N. of N,

1(+1) + x + 3(–2) = 0

or, x = +5

The range of O.N. of N is from +5 to -3. N can show maximum O.N. as +5. Therefore, it can only decrease its O.N. Hence, it acts only as oxidant.

Q.9

Consider the reactions:

(a)6 CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(aq) + 6O₂(g)

(b) O₃(g) + H₂O₂(l) → H₂O(l) + 2O₂(g)

Why it is more appropriate to write these reactions as:

(a) 6CO₂(g) + 12H₂O(l) → C₆H₁₂O₆(aq) + 6H₂O(l) + 6O₂(g)

(b) O₃(g) + H₂O₂ (l) → H₂O(l) + O₂(g) + O₂(g)

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Ans.

a) The 1^st reaction is reaction taking place during photosynthesis. It occurs in two steps. In 1^st step, H₂O decomposes to H₂ and O₂ and in 2^nd step, the produced H₂ reduces CO₂ to C₆H₁₂O₆.

\begin{array}{l}{\text{12H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{12H}}_{\text{2}}\left(\text{g}\right){\text{+ 6O}}_{\text{2}}\text{(g)}\\ {\text{6CO}}_{\text{2}}\left(\text{g}\right){\text{+ 12H}}_{\text{2}}\left(\text{g}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{+ 6H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ \overline{{\text{6CO}}_{\text{2}}\left(\text{g}\right){\text{+ 12H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right){\text{+ 6H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ 6O}}_{\text{2}}\left(\text{g}\right)}\end{array}

Therefore, the most appropriate form of the photosynthesis reaction is the last one, because it shows that 12 H₂O are used per molecule of carbohydrate formed and 6 H₂O molecules are produced during this process.

b) In 2^nd reaction, the purpose of writing the O₂ molecule two times suggests that oxygen molecule is produced from each of the two reactants.

\begin{array}{l}{\text{O}}_3\left(\text{g}\right)\to {\text{O}}_{\text{2}}\left(\text{g}\right)\text{+ O(g)}\\ {\text{H}}_2{\text{O}}_{\text{2}}\left(\text{l}\right)\text{+ O}\left(\text{g}\right)\to {\text{H}}_{\text{2}}{\text{O(l) + O}}_2\left(\text{g}\right)\\ \overline{{\text{O}}_{\text{3}}\left(\text{g}\right){\text{+ H}}_{\text{2}}\text{O}{}_2\left(\text{l}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ O}}_2\left(\text{g}\right){\text{+ O}}_{\text{2}}\left(\text{g}\right)}\end{array}

The path of the two reactions (a) and (b) can be determined by using the isotopes of hydrogen (i.e. deuterium) and oxygen (i.e. O¹⁸).

Q.10 The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?

Ans.

The electronic configuration of Ag is [³⁶Kr] 3d¹⁰4s¹. It shows that its stable oxidation state is +1, not +2. In AgF₂, Ag has +2 O.N. which represents a highly unstable oxidation state. So it quickly accepts an electron to form more stable +1 oxidation state. Therefore, AgF₂ acts as strong oxidising agent.

Ag²⁺ + e^– → Ag⁺

Q.11 Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Ans.

i) Let us consider the following two chemical reactions.

\begin{array}{l}\underset{\left(Excess\right)}{{\text{2C(s) + O}}_{\text{2}}}\left(\text{g}\right)\to \stackrel{\text{+2}}{\text{2CO}}\left(\text{g}\right)\\ \end{array}

…1st equation

\begin{array}{l}\underset{\left(\text{Excess}\right)}{\text{C}\left(\text{s}\right)\text{+}{\text{O}}_{\text{2}}\left(\text{g}\right)}\to \stackrel{\text{+4}}{{\text{CO}}_{\text{2}}}\left(\text{g}\right)\\ \end{array}

…2nd equation

Here, C acts as a reducing agent and O₂ acts as an oxidising agent. If excess amount of carbon is burnt with limited supply of oxygen, CO is formed. In this chemical reaction (1^st), oxidation state of C is +2. If excess amount of O₂ is used, then initially formed CO gets oxidised to CO₂. In this chemical reaction (2^nd) the oxidation state of C is +4.

ii) Let us consider next chemical reaction.

\begin{array}{l}\underset{\left(\text{Excess}\right)}{{\text{P}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{6Cl}}_{\text{2}}}\left(\text{g}\right)\to \stackrel{\text{+3}}{{\text{4PCl}}_{\text{3}}}\\ \end{array}

…1st equation

\begin{array}{l}\underset{\left(\text{Excess}\right)}{{\text{P}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{10Cl}}_{\text{2}}}\left(\text{g}\right)\to \stackrel{\text{+5}}{{\text{4PCl}}_{\text{2}}}\\ \end{array}

…2nd equation

In 1^st chemical reaction, P₄ is a reducing agent and Cl₂ is an oxidising agent. When P is in excess, PCl₃ is formed in which oxidation state of P is +3. When Cl₂ is present in excess, then initially formed PCl₃ reacts further to form PCl₅ in which the oxidation state of P is +5.

iii) Let us consider another chemical reaction.

\begin{array}{l}\underset{\left(\text{Excess}\right)}{\text{4Na}\left(\text{s}\right)\text{+}{\text{O}}_{\text{2}}}\left(\text{g}\right)\to \stackrel{\text{–2}}{{\text{2Na}}_2\text{O(s)}}\\ \end{array}

…1st equation

\begin{array}{l}\underset{\left(\text{Excess}\right)}{\text{2Na}\left(\text{s}\right)\text{+}{\text{O}}_{\text{2}}}\left(\text{g}\right)\to \stackrel{\text{–1}}{{\text{Na}}_2\text{O(s)}}\\ \end{array}

…2nd equation

In 1^st chemical reaction, Na is a reducing agent and O₂ is an oxidising agent. When excess of Na is used, Na₂O is formed in which O has oxidation state of –2. When oxygen is used in excess, Na₂O₂ is formed in which the oxidation state of O is –1 which is higher than –2.

Q.12 How do you count for the following observations?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?

Ans.

KMnO₄ is oxidising agent and it oxidises toluene to benzoic acid in acidic, basic and neutral medium.

(i) Redox reaction in acidic medium:

ii) In basic and neutral medium:

Alcoholic KMnO₄ is used as an oxidant in the manufacture of benzoic acid. It is so because organic reactions must be performed in a medium of organic solvent (alcohol) in which the oxidising agent (inorganic salt) can be dissolved.

b) When conc. H₂SO₄ is added to chloride salt, a pungent smell is produced due to formation of HCl. Stronger acid replaces weaker acid from its salt. Since HCl is weak reducing agent, it can’t reduce H₂SO₄ to SO₂. So, HCl is not oxidised to Cl₂.

2NaCl+ 2H₂SO₄ → 2NaHSO₄ + 2HCl

Stronger acid Weaker acid

{\text{2HCl + H}}_{\text{2}}{\text{SO}}_{\text{4}}\xcancel{\stackrel{}{\to }}{\text{Cl}}_{\text{2}}{\text{+SO}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}

When it is the salt containing bromine ion, HBr is produced. HBr is stronger reducing agent than HCl so it reduces H₂SO₄ to SO₂ and itself oxidises to give red vapour of Br₂.

NaBr + dil H₂SO₄ → NaHSO₄ + HBr

2HBr + dil H₂SO₄ → Br₂ + SO₂ + 2H₂O

Q.13 Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

(a) 2AgBr (s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr (aq) + C₆H₄O₂(aq)

(b) HCHO (l) + 2[Ag(NH₃)₂]⁺(aq) + 3OH^–(aq) → 2Ag(s) + HCOO^–(aq) + 4NH₃(aq) + 2H₂O(l)

(c) HCHO (l) + 2 Cu²⁺(aq) + 5 OH^–(aq) → Cu₂O(s) + HCOO^–(aq) + 3H₂O(l)

(d) N₂H₄(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l)

(e) Pb(s) + PbO₂(s) + 2 H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

Ans.

Q.14 Consider the reactions:

2 S₂O₃^2–(aq) + I₂(s) → S₄O₆^2–(aq) + 2I^–(aq)

S₂O₃ ^2–(aq) + 2Br₂(l) + 5 H₂O(l) → 2SO₄^2–(aq) + 4Br^–(aq) + 10H⁺(aq)

Why does the same reactant thiosulphate react differently with iodine and bromine?

Ans.

O.N. of S in S₂O₃^2–

or, 2x + 3(-2) = -2

or, x = +2

O.N. of S in S₄O₆^2–

or,4x + 6(-2) = -2

or, x = +2.5

O.N. of S in SO₄^2–

or x + 4(-2) = -2

or, x = +6

Since Br₂ is stronger oxidising agent than I₂, it oxidises S of S₂O₃^2– to a higher oxidation state of +6 in SO₄^2-. I₂ is weaker oxidising agent, it oxidises S to a lower oxidation of +2.5 in S₄O₆^2–. For this reason, thiosulphate reacts differently towards Br₂ and I₂.

Q.15 Justify given reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Ans.

Halogens are strong oxidising agents because they quickly accept electrons. Their oxidising power can be measured by their electrode potentials. According to the electrode potential, the order of oxidising power is as follows:

F₂ (+2.8V) > Cl₂ (+1.36V) > Br₂ (+1.09V)> I₂ (+0.54V)

From these electrode potentials it is clear that, F₂ can oxidise Cl^– to Cl₂, Br^– to Br₂ and I^– to I₂. Cl₂ can oxidise Br^– to Br₂ and I^– to I₂ but not F^– to F₂. Br₂ can oxidise I^– to I₂ but not Cl^– to Cl₂ and Br^– to Br₂.

\begin{array}{l}{\text{F}}_{\text{2}}\left(\text{g}\right){\text{+ 2Cl}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2F}}^{\text{–}}\left(\text{aq}\right){\text{+ Cl}}_{\text{2}}\left(\text{g}\right)\\ {\text{F}}_{\text{2}}\left(\text{g}\right){\text{+ 2Br}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2F}}^{\text{–}}\left(\text{aq}\right){\text{+ Br}}_{\text{2}}\left(\text{l}\right)\\ {\text{F}}_{\text{2}}\left(\text{g}\right){\text{+ 2I}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2F}}^{\text{–}}\left(\text{aq}\right){\text{+ I}}_{\text{2}}\left(\text{s}\right)\\ {\text{Cl}}_{\text{2}}\left(\text{g}\right){\text{+ 2Br}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2Cl}}^{–}\left(\text{aq}\right){\text{+ Br}}_{\text{2}}\left(\text{l}\right)\\ {\text{Cl}}_{\text{2}}\left(\text{g}\right){\text{+ 2I}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2Cl}}^{\text{–}}\left(\text{aq}\right){\text{+ I}}_{\text{2}}\left(\text{l}\right)\\ {\text{Br}}_{\text{2}}\left(\text{l}\right){\text{+ 2I}}^{\text{–}}\stackrel{}{\to }{\text{2Br}}^{\text{–}}\left(\text{aq}\right){\text{+ I}}_{\text{2}}\left(\text{s}\right)\end{array}

Thus, F₂ is the strongest oxidising agent among the others.

The halide ions have the tendency to lose electrons hence, can act as reducing agents. Since the electrode potentials of the halide ions decreases in the order,

I^–(-0.54V)>Br^–(-1.09)>Cl^–(-1.36V)>F^–(-2.87V)

the reducing power of the corresponding hydrohalic acids follows the same order.

HI>HBr>HCl>HF

HI is best reducing agent. HI and HBr can reduce H₂SO₄ to SO₂ while HCl and HF can’t.

\begin{array}{l}{\text{2HBr + H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{Br}}_{\text{2}}{\text{+ SO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}\\ {\text{2HI + H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{I}}_{\text{2}}{\text{+ SO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}\\ {\text{2Cu}}^{\text{2+}}\left(\text{aq}\right){\text{+ 4I}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cu}}_{\text{2}}{\text{I}}_{\text{2}}\left(\text{s}\right){\text{+ I}}_{\text{2}}\left(\text{s}\right)\end{array}

I^– reduces Cu²⁺ to Cu⁺ but Br^– can’t. So HI is stronger reducing agent than HBr. If we compare HF and HCl, HCl is stronger reducing agent than HF because HCl can reduce MnO₂ to Mn²⁺ but HF can’t.

{\text{MnO}}_{\text{2}}\left(\text{s}\right)\text{+ 4HCl}\left(\text{aq}\right)\stackrel{}{\to }{\text{MnCl}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{Cl}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2H}}_{\text{2}}\text{O} \begin{array}{l}{\text{MnO}}_{\text{2}}\left(\text{s}\right)\text{+ 4HCl}\left(\text{l}\right)\underset{Noreaction}{\xcancel{\stackrel{}{\to }}}{F}_2\\ \end{array}

Q.16 Why does the following reaction occur?

XeO₆^4– (aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g)+ F₂(g) + 3H₂O(l)

What conclusion about the compound Na₄XeO₆ (of which XeO₆^4– is a part) can be drawn from the reaction?

Ans.

The above reaction can be written as:

\stackrel{\text{+8}}{{\text{XeO}}_{\text{6}}^{\text{4–}}}\left(\text{aq}\right)\text{+}\stackrel{\text{–1}}{{\text{2F}}^{\text{–}}}\left(\text{aq}\right){\text{+ 6H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{+6}}{\text{Xe}}{\text{O}}_{\text{3}}\left(\text{s}\right)\text{+}{\stackrel{\text{0}}{\text{F}}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)

The O.N. of Xe in XeO₆^4– is +8 which decreases to +6 in XeO₃ while in F atom O.N. increases from –1 to 0. So it is a redox reaction where XeO₆^4– is reduced and F^– is oxidised.

The compound Na₄XeO₆ (of which XeO₆^4– is a part) is stronger oxidant than F₂.

Q.17 Consider the reactions:

(a) H₃PO₂(aq) + 4 AgNO₃(aq) + 2 H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)

(b) H₃PO₂(aq) + 2CuSO₄(aq) + 2 H₂O(l) → H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)

(c) C₆H₅CHO(l) + 2[Ag (NH₃)₂]⁺(aq) + 3OH^–(aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃ (aq) + 2 H₂O(l)

(d) C₆H₅CHO (l) + 2Cu²⁺(aq) + 5OH^–(aq) → No change observed.

What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?

Ans.

The reactions (a) and (b) indicate that H₃PO₂ (hypophosphorous acid) acts as a reducing agent and thus AgNO₃ and CuSO₄ get reduced by the hypophosphorous acid. Conversely, AgNO₃ and CuSO₄ act as oxidising agent, H₃PO₂ oxidises to H₃PO₄.

In the third reaction (c) Ag⁺ oxidises C₆H₅CHO to C₆H₅COO^– ion but in the reaction (d) Cu²⁺ can’t oxidise C₆H₅CHO to C₆H₅COO^–. Thus, Ag⁺ ion is stronger oxidising agent than Cu²⁺ ion.

Q.18 Balance the following redox reactions by ion – electron method:

(a) MnO₄^– (aq) + I^– (aq) → MnO₂ (s) + I₂(s) (in basic medium)

(b) MnO₄^– (aq) + SO₂ (g) → Mn²⁺ (aq) + HSO₄^– (aq) (in acidic solution)

(c) H₂O₂ (aq) + Fe²⁺ (aq) → Fe³⁺ (aq) + H₂O (l) (in acidic solution)

(d) Cr₂O₇ ^2– + SO₂ (g) → Cr³⁺ (aq) + SO₄^2– (aq) (in acidic solution)

Ans.

a) Oxidation half equation:

\stackrel{\text{–1}}{{\text{2I}}^{\text{–}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{0}}{{\text{I}}_{\text{2}}}\left(\text{s}\right)\text{+}{\text{2e}}^{\text{–}}

….1^st equation

Reduction half equation:

\stackrel{\text{+7}}{{\text{MnO}}_4^{-}}\left(\text{aq}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ 3e}}^{\text{–}}\stackrel{}{\to }\stackrel{\text{+4}}{{\text{MnO}}_{\text{2}}}\left(\text{s}\right)\text{+}{\text{4OH}}^{\text{–}}\left(\text{aq}\right)

…2^nd equation

Now multiply 1^st equation by 3 and 2^nd equation by 2, then add them to get the balanced chemical equation:

{\text{2MnO}}_{\text{4}}^{\text{–}}\left(\text{aq}\right){\text{+ 4H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ 6I}}^{\text{–}}\stackrel{}{\to }{\text{2MnO}}_{\text{2}}\left(\text{s}\right){\text{+ 8OH}}^{\text{–}}\left(\text{aq}\right){\text{+ 3I}}_{\text{2}}\left(\text{s}\right)

b) Oxidation half equation:

\stackrel{\text{+4}}{{\text{SO}}_{\text{2}}}\left(\text{g}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\stackrel{\text{+6}}{{\text{HSO}}_{\text{4}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{3H}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}

….1^st equation

Reduction half equation:

\begin{array}{l}\stackrel{\text{+7}}{{\text{MnO}}_{\text{4}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{8H}}^{\text{+}}\left(\text{aq}\right){\text{+ 5e}}^{\text{–}}\stackrel{}{\to }\stackrel{\text{+2}}{{\text{Mn}}^{\text{2+}}}{\text{+ 4H}}_{\text{2}}\text{O}\left(\text{l}\right)\\ \end{array}

…2^nd equation

Multiply 1^st equation by 5 and 2^nd equation by 2 and then add these chemical equations. Finally we get:

{\text{2MnO}}_{\text{4}}^{\text{–}}\left(\text{aq}\right){\text{+ 5SO}}_{\text{2}}\left(\text{g}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ H}}^{+}\left(\text{aq}\right)\stackrel{}{\to }{\text{2Mn}}^{\text{2+}}\left(\text{aq}\right){\text{+ 5HSO}}_{\text{4}}^{\text{–}}\left(\text{aq}\right)

c) Oxidation half equation:

{\text{Fe}}^{\text{2+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}

….1^st equation

Reduction half equation:

\stackrel{\text{–1}}{{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}}\left(\text{aq}\right){\text{+ 2H}}^{\text{+}}\left(\text{aq}\right){\text{+ 2e}}^{\text{–}}\stackrel{}{\to }\stackrel{\text{–2}}{{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)}

…2^nd equation

Multiply 1^st equation by 2 and add it to 2^nd equation. We get,

{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right){\text{+ 2H}}^{\text{+}}\left(\text{aq}\right){\text{+ 2Fe}}^{\text{2+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+ 2Fe}}^{\text{3+}}\left(\text{aq}\right)

d) Oxidation half equation:

\stackrel{\text{+4}}{{\text{SO}}_{\text{2}}}\left(\text{g}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\stackrel{\text{+6}}{{\text{SO}}_{\text{4}}^{\text{2–}}}\left(\text{aq}\right){\text{+ 4H}}^{\text{+}}\left(\text{aq}\right){\text{+ 2e}}^{\text{–}}

….1^st equation

Reduction half equation:

\stackrel{\text{+6}}{{\text{Cr}}_{\text{2}}}{\text{O}}_{\text{7}}^{\text{2–}}\left(\text{aq}\right){\text{+ 14H}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{6e}}^{\text{–}}\stackrel{}{\to }\stackrel{\text{+3}}{{\text{2Cr}}_{}^{\text{3+}}}\left(\text{aq}\right){\text{+ 7H}}_2\text{O}\left(\text{l}\right)

…2^nd equation

Now multiply 1^st equation by 3 and then add it to 2^nd equation. Final balanced equation is written here.

{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2–}}\left(\text{aq}\right){\text{+ 3SO}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2H}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2Cr}}_{}^{\text{3+}}\left(\text{aq}\right){\text{+ 3SO}}_{\text{4}}^{\text{2–}}\left(\text{aq}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)

Q.19 Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

(a) P₄(s) + OH^–(aq) → PH₃(g) + H₂PO₂^– (aq)

(b) N₂H₄(l) + ClO₃^–(aq) → NO(g) + Cl^–(g)

(c) Cl₂O₇ (g) + H₂O₂(aq) → ClO₂^–(aq) + O₂(g) + H⁺

Ans.

P₄ acts as oxidising as well as reducing agent.

Oxidation number method:

Total increase in O.N. of P in H₂PO₂^– = 1 x 4 = 4

Total decrease in O.N. of P in PH₃ = 3 x 4 = 12

To balance increase/decrease in O.N., multiply PH₃ by 1 and H₂PO₂^– by 3, we get

{\text{P}}_{\text{4}}\left(\text{s}\right){\text{+ OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{PH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{3H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)

Now balance O atoms, multiply OH^– by 6, we have

{\text{P}}_{\text{4}}\left(\text{s}\right){\text{+ 6OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{PH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{3H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)

To balance H atoms, add 3 H₂O to L.H.S. and 3 OH^– to R.H.S. we have

{\text{P}}_{\text{4}}\left(\text{s}\right){\text{+ 6OH}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{PH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{3H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{3OH}}^{\text{–}}\left(\text{aq}\right) {\text{P}}_{\text{4}}\left(\text{s}\right){\text{+ 3OH}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{PH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{3H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)

By Ion-electron method:

Oxidation half reaction:

{\text{P}}_{\text{4}}\left(\text{s}\right)\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)

Balance P atoms, we have

{\stackrel{0}{\text{P}}}_{\text{4}}\left(\text{s}\right)\stackrel{}{\to }\stackrel{+1}{4{\text{H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)}

Balance of O.N. by adding electrons,

{\stackrel{\text{0}}{\text{P}}}_{\text{4}}\left(\text{s}\right)\stackrel{}{\to }\stackrel{\text{+1}}{{\text{4H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)}\text{+}{\text{4e}}^{\text{–}}

Balance charge by adding 8 OH^– ions,

{\text{P}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{8OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{4H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{4e}}^{\text{–}}

….1^st Equation

Reduction half reaction:

{\stackrel{0}{\text{P}}}_{\text{4}}\left(\text{s}\right)\stackrel{}{\to }\stackrel{–3}{{\text{PH}}_3\left(\text{g}\right)}

Balancing of P atoms:

{\stackrel{0}{\text{P}}}_{\text{4}}\left(\text{s}\right)\stackrel{}{\to }\stackrel{–3}{{\text{4PH}}_3\left(\text{g}\right)}

Balance of O.N. by adding 12 electrons on L.H.S.:

{\stackrel{\text{0}}{\text{P}}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{12e}}^{\text{–}}\stackrel{}{\to }{\text{4PH}}_{\text{3}}\left(\text{g}\right)

Balance charge by adding 12 OH^– ions on R.H.S.:

{\stackrel{\text{0}}{\text{P}}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{12e}}^{\text{–}}\stackrel{}{\to }{\text{4PH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{12OH}}^{\text{–}}\left(\text{aq}\right)

Balance O atoms by adding 12 H₂O to L.H.S. of chemical equation:

{\stackrel{\text{0}}{\text{P}}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{12H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{12e}}^{\text{–}}\stackrel{}{\to }{\text{4PH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{12OH}}^{\text{–}}\left(\text{aq}\right)

…2^nd Equation

Multiply equation 2 by 3.

After canceling out the electrons, the ultimate equation becomes:

{\text{4P}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{12H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{24OH}}^{\text{–}}\stackrel{}{\to }{\text{4PH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{12OH}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{12H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)

Divide the whole chemical equation by 4 and balance OH^– ions. Finally, we get

{\text{P}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{3OH}}^{\text{–}}\stackrel{}{\to }{\text{PH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{3H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}\left(\text{aq}\right)

b)

ClO₃^–acts as the oxidising agent and N₂H₄ acts as reducing agent.

Oxidation number method:

Total increase in O.N. of N = 2 x 4 = 8

Total decrease in O.N. of Cl = 1 x 6 = 6

To balance increase/decrease in O.N., multiply N₂H₄ by 3 and ClO₃^– by 4:

{\text{3N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{4ClO}}_{\text{3}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }\text{NO}\left(\text{g}\right)\text{+}{\text{Cl}}^{\text{–}}\left(\text{aq}\right)

To balance N and Cl atoms, multiply NO by 6 and Cl^– by 4, we have

{\text{3N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{4ClO}}_{\text{3}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }\text{6NO}\left(\text{g}\right)\text{+}{\text{Cl}}^{\text{–}}\left(\text{aq}\right)

Let us balance O atoms by adding 6H₂O,

{\text{3N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{4ClO}}_{\text{3}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }\text{6NO}\left(\text{g}\right)\text{+}{\text{Cl}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}\left(\text{l}\right)

Ion-electron method:

Oxidation half reaction:

\stackrel{–2}{{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\left(\text{l}\right)\stackrel{}{\to }\stackrel{+2}{\text{NO}}\left(\text{g}\right)

Balancing of N atoms:

\stackrel{–2}{{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\left(\text{l}\right)\stackrel{}{\to }\stackrel{+2}{\text{2NO}}\left(\text{g}\right)

Balance O.N. by adding 8 electrons on R.H.S.

{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\stackrel{}{\to }\text{2NO}\left(\text{g}\right)\text{+}{\text{8e}}^{\text{–}}

Balance charges by adding OH^– ions on L.H.S.

{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{8OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }\text{2NO}\left(\text{g}\right)\text{+}{\text{8e}}^{\text{–}}

Finally, balance O atoms by adding 6 H₂O on R.H.S.

{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\text{+}{\text{8OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }\text{2NO}\left(\text{g}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{8e}}^{\text{–}}

….1^st Equation

Reduction half reaction:

\stackrel{\text{+5}}{{\text{ClO}}_{\text{3}}^{\text{–}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{–1}}{{\text{Cl}}^{\text{–}}}\left(\text{aq}\right)

Balance O.N. by adding 6 electrons on L.H.S.

\stackrel{}{{\text{ClO}}_{\text{3}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{6e}}^{\text{–}}\stackrel{}{\to }\stackrel{}{{\text{Cl}}^{\text{–}}}\left(\text{aq}\right)

Balance charges by adding 6 OH^– ion on R.H.S.

\stackrel{}{{\text{ClO}}_{\text{3}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{6e}}^{\text{–}}\stackrel{}{\to }\stackrel{}{{\text{Cl}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{6OH}}^{\text{–}}\left(\text{aq}\right)

Balance O atoms by adding 3 H₂O on L.H.S.

\stackrel{}{{\text{ClO}}_{\text{3}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{6e}}^{\text{–}}\stackrel{}{\to }\stackrel{}{{\text{Cl}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{6OH}}^{\text{–}}\left(\text{aq}\right)

Equalize the electron transfer between oxidation and reduction half-equations , the final equation we get:

\text{4}\stackrel{}{{\text{ClO}}_{\text{3}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{3N}}_{\text{2}}{\text{H}}_{\text{4}}\left(\text{l}\right)\stackrel{}{\to }\stackrel{}{{\text{4Cl}}^{\text{–}}}\left(\text{aq}\right)\text{+}\text{6NO}\left(\text{g}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}\left(\text{l}\right)

c)

Cl₂O₇ acts as an oxidising agent while H₂O₂ acts as reducing agent.

Oxidation number method:

To balance increase/decrease in O.N., multiply H₂O₂ and O₂ by 4, we have

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}^{}}\left(\text{g}\right)\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{}{{\text{ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{4O}}_{\text{2}}\left(\text{g}\right)

To balance Cl atoms, multiply ClO₂^– by 2 on R.H.S. we get,

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}^{}}\left(\text{g}\right)\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{4O}}_{\text{2}}\left(\text{g}\right)

To balance O atoms, add 3H₂O to R.H.S. we have

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}^{}}\left(\text{g}\right)\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{4O}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)

To balance H atoms, add 2H₂O to R.H.S. and 2 OH^– to L.H.S. We get

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}^{}}\left(\text{g}\right)\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{2OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{4O}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{5H}}_{\text{2}}\text{O}\left(\text{l}\right)

Ion-electron method:

Oxidation half reaction:

\stackrel{\text{–1}}{{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)}\stackrel{}{\to }{\stackrel{\text{0}}{\text{O}}}_{\text{2}}\left(\text{g}\right)

Balance O.N. by adding 2 electrons on R.H.S.

\stackrel{}{{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)}\stackrel{}{\to }{\stackrel{}{\text{O}}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2e}}^{\text{–}}

Balance charges by adding 2OH^– ions on L.H.S.

\stackrel{}{{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)}\text{+}{\text{2OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\stackrel{}{\text{O}}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2e}}^{\text{–}}

Balance O atoms by adding 2H₂O on R.H.S.

\stackrel{}{{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)}\text{+}{\text{2OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\stackrel{}{\text{O}}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{2e}}^{\text{–}}

…1^st Equation

Reduction half reaction:

\stackrel{\text{+7}}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{g}\right)}\stackrel{}{\to }\stackrel{\text{+3}}{{\text{ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)

Balance Cl atoms:

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{g}\right)}\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)

Balance O.N. by adding 8 electrons on L.H.S.:

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{g}\right)}+8e^{–}\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)

Adding of 6 OH^– ions on R.H.S to balance charges:

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{g}\right)}\text{+}{\text{8e}}^{\text{–}}\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{6OH}}^{\text{–}}

Balance O atoms by adding 3H₂O to L.H.S. We have,

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{g}\right)}\text{+}{\text{3H}}_{\text{2}}\text{O}\text{+}{\text{8e}}^{\text{–}}\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{6OH}}^{\text{–}}\left(\text{aq}\right)

..2^nd Equation

Multiply equation 1 by 2 and then add both the equations.

After cancelling electrons in both chemical equations we get final chemical equation as –

\stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{g}\right)}\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{8OH}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{6OH}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{4O}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{8H}}_{\text{2}}\text{O}\left(\text{l}\right) \stackrel{}{{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{g}\right)}\text{+}{\text{4H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{2OH}}^{–}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{}{{\text{2ClO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{4O}}_2\left(\text{g}\right)\text{+}{\text{5H}}_{\text{2}}\text{O}\left(\text{l}\right)