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.

Substance oxidised	Substance reduced	Oxidising agent	Reducing agent
a) C6H6O2 (aq)	AgBr (s)	AgBr (s)	C6H6O2 (aq)
b) HCHO (aq)	[Ag(NH3)2]+	[Ag(NH3)2]+	HCHO (aq)
c) HCHO (aq)	Cu2+ (aq)	Cu2+ (aq)	HCHO (aq)
d) N2H4 (l)	H2O2 (l)	H2O2 (l)	N2H4 (l)
e) Pb(s)	PbO2 (s)	PbO2 (s)	Pb(s)

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)

Q.20 What sort of information can you draw from the following reaction?

(CN)₂(g) + 2OH^–(aq) → CN^–(aq)+CNO^–(aq)+H₂O(l)

Ans.

The reaction can be written as follows:

{\stackrel{\text{+3}}{\left(\text{CN}\right)}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{+2}}{{\text{CN}}^{\text{–}}}\left(\text{aq}\right)\text{+}\stackrel{\text{+4}}{{\text{CNO}}^{\text{–}}\left(\text{aq}\right)}\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)

The O.N. of C in (CN)₂ is +3 which increases to +4 in CNO^– and decreases to +2 in CN^–. The following information can be drawn from the above reaction:

• It is a decomposition reaction of (CN)₂ in alkaline medium to cyanide(CN^–) ion and cyanate ion(CNO^–).
• (CN)₂ is simultaneously reduced to CN^– and oxidised to CNO^– ion.
• It is a redox as well as disproportionation reaction.
• Cyanogen is pseudohalogen (behaves like halogens) and Cyanide is pseudohalides (behaves like halide ion).

Q.21 The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ion. Write a balanced ionic equation for the reaction.

Ans.

The equation can be represented as

{\text{Mn}}^{\text{3+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Mn}}^{\text{2+}}\left(\text{aq}\right)\text{+}{\text{MnO}}_{\text{2}}\left(\text{s}\right)\text{+}{\text{H}}^{\text{2}}\left(\text{aq}\right)

Oxidation half reaction:

\stackrel{+3}{{\text{Mn}}^{\text{3+}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{+4}{{\text{MnO}}_{\text{2}}}\left(\text{s}\right)

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

\stackrel{\text{+3}}{{\text{Mn}}^{\text{3+}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{+4}}{{\text{MnO}}_{\text{2}}}\left(\text{s}\right)\text{+}{\text{e}}^{\text{–}}

Balance charges by adding 4H⁺ ions on R.H.S.

\stackrel{\text{+3}}{{\text{Mn}}^{\text{3+}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{+4}}{{\text{MnO}}_{\text{2}}}\left(\text{s}\right)\text{+}{\text{4H}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}

Balance O atoms by adding 2H₂O on L.H.S:

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

Reduction half equation:

\stackrel{+3}{{\text{Mn}}^{\text{3+}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{+2}{{\text{Mn}}^{2+}}\left(\text{aq}\right)

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

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

The final balanced equation for disproportionation reaction is:

{\text{2Mn}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{MnO}}_{\text{2}}\left(\text{s}\right)\text{+}{\text{4H}}^{\text{+}}\left(\text{aq}\right){\text{+ Mn}}^{\text{2+}}\left(\text{aq}\right)

Q.22 Consider the elements:

Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only positive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Ans.

a) F is the most electronegative element and shows only -1 oxidation state.

b) Cs is alkali metal, it contains single electron in the valence shell so it exhibit an oxidation state of +1.

c) I can show both +ve and –ve oxidation states. Due to the presence of the 7 electrons in the valence shell, I shows oxidation state of +1 (in the compound of F and O which are more electronegative than I) or -1 (in compounds of I with more electropositive elements like H, Na, K). Again due to the presence of the d-orbitals, it also exhibits +ve oxidation states of +3, +5 and +7.

d) Ne is an inert gas and it has complete valence shell configuration. So it is very reluctant to lose or gain electrons. That’s why, it shows neither +ve nor –ve oxidation sates.

Q.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Ans.

The chemical equation can be represented as:

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

Reduction half equation:

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

Balance Cl atoms:

\stackrel{0}{{\text{Cl}}_{\text{2}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{–1}{2{\text{Cl}}^{\text{–}}}\left(\text{aq}\right)

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

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

…..1^st Equation

Oxidation half reaction:

\stackrel{+4}{{\text{SO}}_{\text{2}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{+6}}{{\text{SO}}_{\text{4}}^{\text{2–}}}\left(\text{aq}\right)

Add electrons on the R.H.S. to balance O.N. (oxidation number)

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

Charges are balanced by adding 4H⁺ ions on R.H.S.

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

Then, balance O atoms by adding 2H₂O on L.H.S.

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

…2^nd Equation

Add these equations, we have

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

This is the balanced redox equation.

Q.24 Refer to periodic table given in your book and now answer the following questions:

(a)Select the possible non- metals that can show disproportionation reaction.

(b)Select three metals that can show disproportionation reaction.

Ans.

The possible non-metals (which show disproportionation reaction) are P₄, Cl₂ and S₈.

{\stackrel{\text{0}}{\text{P}}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{3OH}}^{\text{–}}\left(\text{aq}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\stackrel{\text{–3}}{{\text{PH}}_{\text{3}}}\left(\text{g}\right)\text{+}\stackrel{\text{+1}}{{\text{3H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{–}}}\left(\text{aq}\right) \stackrel{\text{0}}{{\text{Cl}}_{\text{2}}}\left(\text{g}\right)\text{+}{\text{2OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{\text{cold}}{\to }\stackrel{\text{–1}}{{\text{Cl}}^{\text{–}}}\left(\text{aq}\right)\text{+}\stackrel{\text{+1}}{{\text{ClO}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)

Or

\text{3}\stackrel{\text{0}}{{\text{Cl}}_{\text{2}}}\left(\text{g}\right)\text{+}{\text{6OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{\text{Hot}}{\to }\stackrel{\text{–1}}{{\text{5Cl}}^{\text{–}}}\left(\text{aq}\right)\text{+}\stackrel{\text{+5}}{{\text{ClO}}_{\text{3}}^{\text{–}}}\left(\text{aq}\right)\text{+}{\text{3H}}_{\text{2}}\text{O}\left(\text{l}\right) \stackrel{\text{0}}{{\text{S}}_{\text{8}}}\left(\text{s}\right)\text{+}{\text{12 OH}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{–2}}{{\text{4S}}^{\text{2–}}}\left(\text{aq}\right)\text{+}\stackrel{\text{+2}}{{\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2–}}}\left(\text{aq}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}\left(\text{l}\right)

The possible metals (which show disproportionation reaction) are Cu⁺, Ga⁺, Mn³⁺.

\stackrel{\text{+1}}{{\text{2Cu}}^{\text{+}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{2+}}{{\text{Cu}}^{\text{2+}}}\left(\text{aq}\right)\text{+}\stackrel{\text{0}}{\text{Cu}\left(\text{s}\right)} \stackrel{\text{+1}}{{\text{3Ga}}^{\text{+}}}\left(\text{aq}\right)\stackrel{}{\to }\stackrel{\text{+3}}{{\text{Ga}}^{\text{3+}}}\left(\text{aq}\right)\text{+}\text{2}\stackrel{\text{0}}{\text{Ga}}\left(\text{s}\right) \stackrel{\text{+3}}{{\text{2Mn}}^{\text{3+}}}\left(\text{aq}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }\stackrel{\text{+4}}{{\text{MnO}}_{\text{2}}}\left(\text{s}\right)\text{+}\stackrel{\text{+2}}{{\text{Mn}}^{\text{2+}}}\left(\text{aq}\right)\text{+}{\text{4H}}^{\text{+}}\left(\text{aq}\right)

Q.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Ans.

The balanced chemical equation for the first step of manufacture of nitric acid is:

\underset{\text{4}\text{×}\text{17}\text{=}\text{68g}}{{\text{4NH}}_{\text{3}}\left(\text{g}\right)}\text{+}\underset{\text{5}\text{×}\text{32}\text{=}\text{160g}}{{\text{5 O}}_{\text{2}}\left(\text{g}\right)}\underset{\text{Pt}}{\overset{\text{1100K}}{\to }}\underset{\text{4}\text{×}\text{30}\text{=}\text{120g}}{\text{4 NO}\left(\text{g}\right)}\text{+}\underset{\text{6}\text{×}\text{8}\text{=}\text{108g}}{{\text{6 H}}_{\text{2}}\text{O}\left(\text{g}\right)}

Here, 68 g of NH₃ will react with 160 g of O₂ (g)

Thus 10 g of NH₃ will react with = {(160/68)x10} g =23.53 g of O₂ (g)

But the amount of oxygen available is 20 g. This value of oxygen is less than the amount required. Therefore, in this chemical reaction, O₂ is the limiting reagent.

Let us calculate it on the basis of amount of oxygen provided and not on the amount of NH₃ taken.

We can write as follows-

160 g of O₂ produces 120 g of NO.

Therefore,20 g of O₂ will produce = {(120/160) x 20}

= 15 g of NO

Q.26 Using the standard electrode potentials given below, predict if the reaction between the following is feasible or not:

\begin{array}{l}{\text{2I}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{I}}_{\text{2}}\left(\text{s}\right){\text{+2e}}^{\text{–}}\text{E°}\text{=}\text{–0}\text{.54V}\\ {\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\stackrel{}{\to }{\text{Fe}}^{\text{2+}}\text{E°}\text{=}\text{+0}\text{.77V}\\ \text{Cu}\left(\text{s}\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\text{E°}\text{=}\text{–0}\text{.34V}\\ {\text{Ag}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\stackrel{}{\to }\text{Ag}\left(\text{s}\right)\text{E°}\text{=}\text{+0}\text{.80V}\\ {\text{Br}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\stackrel{}{\to }{\text{2Br}}^{\text{–}}\left(\text{aq}\right)\text{E°}\text{=}\text{+1}\text{.09V}\end{array}

(a) Fe³⁺(aq) and I^–(aq)

(b) Ag⁺(aq) and Cu(s)

(c) Fe³⁺ (aq) and Cu(s)

(d) Ag(s) and Fe³⁺(aq)

(e) Br₂(aq) and Fe²⁺(aq)

Ans.

a) The possible reaction between Fe³⁺(aq) and I^–(aq) can be written as

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

The redox reaction can be split into two half reactions-

\begin{array}{l}\text{Oxidation:}{\text{2I}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{I}}_{\text{2}}\left(\text{s}\right)\text{+}{\text{2e}}^{\text{–}}\text{;}\text{E°}\text{=}\text{–0}\text{.54V}\left(\text{i}\right)\\ \text{Reduction:}{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\stackrel{}{\to }{\text{Fe}}^{\text{2+}}\left(\text{aq}\right)]\text{×}\text{2}\text{;}\text{E°}\text{=}\text{+0}\text{.77V}\left(\text{ii}\right)\end{array}

The sign of the electrode potential for the equation (i) (oxidation reaction) is reversed as per given standard electrode potentials. To calculate overall reaction, the number of electrons gained and lost must be canceled out. Thus, multiplying equation (ii) by 2 and adding it with equation (i) we get

{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{2I}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2Fe}}^{\text{2+}}\left(\text{aq}\right)\text{+}{\text{I}}_{\text{2}}\left(\text{s}\right)\text{;}\text{E°=+0}\text{.23V}

The emf for the overall reaction is +ve, so the reaction is feasible.

b) Reaction between Ag⁺ (aq) and Cu(s)can be written as-

\text{Cu}\left(\text{s}\right)\text{+}{\text{Ag}}^{\text{+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)\text{+Ag}\left(\text{s}\right)

The reaction can be split into two half reactions:

\begin{array}{l}\text{Oxidation:}\text{Cu}\left(\text{s}\right)\stackrel{}{\to }{\text{Cu}}^{2+}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\text{;}\text{E°}\text{=}\text{–0}\text{.34V}\\ \text{Reduction:}{\text{Ag}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\stackrel{}{\to }\text{Ag}\left(\text{s}\right)]\text{×}\text{2}\text{;}\text{E°}\text{=}\text{+0}\text{.80V}\\ \overline{\text{Overall}\text{reaction:}\text{Cu}\left(\text{s}\right)\text{+}\text{2}{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(\text{aq}\right){\text{+2Fe}}^{\text{2+}}\left(\text{aq}\right)\text{;}\text{E =}\text{+0}\text{.46V}}\end{array}

The E.M.F. of the overall reaction is +ve. Thus, the reaction is feasible.

c) The possible reaction between Fe³⁺ and Cu can occur in following two ways:

\text{Cu}\left(\text{s}\right){\text{+ 2Fe}}^{\text{3+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(\text{aq}\right){\text{+ 2Fe}}^{\text{2+}}\left(\text{aq}\right)\mathrm{\dots }\text{(i)}

The reaction (i) can be split into two half equations:

\begin{array}{l}\text{Oxidation:}\text{Cu}\left(\text{s}\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\text{;}\text{E°}\text{=}\text{–0}\text{.34V}\\ \text{Reduction:}{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\stackrel{}{\to }{\text{Fe}}^{\text{2}}\left(\text{aq}\right)]\text{×}\text{2}\text{;}\text{E°}\text{=}\text{+0}\text{.77V}\\ \overline{\text{Overall}\text{reaction:}\text{Cu}\left(\text{s}\right)\text{+}\text{2}{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(\text{aq}\right){\text{+2Fe}}^{\text{2+}}\left(\text{aq}\right)\text{;}\text{E=}\text{+0}\text{.43V}}\end{array}

E.M.F. of the overall reaction is +ve, therefore the reaction is feasible.

d) The reaction between Ag(s) and Fe³⁺ (aq)can be written as –

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

The reaction can be split into two half equations:

\begin{array}{l}\text{Oxidation:}\text{Ag}\left(\text{s}\right)\stackrel{}{\to }{\text{Ag}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\text{;}\text{E°}\text{=}\text{–0}\text{.34V}\\ \text{Reduction:}{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\stackrel{}{\to }{\text{Fe}}^{\text{2}}\left(\text{aq}\right)]\text{×}\text{2}\text{;}\text{E°}\text{=}\text{+0}\text{.77V}\\ \overline{\text{Overall}\text{reaction:}\text{Ag}\left(\text{s}\right)\text{+}{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Ag}}^{\text{+}}\left(\text{aq}\right){\text{+ Fe}}^{\text{2+}}\left(\text{aq}\right)\text{;}\text{E}\text{=}\text{–0}\text{.03V}}\end{array}

Since the E.M.F. of the overall reaction is negative, therefore this chemical reaction is not feasible.

e) The reaction between Br₂ (aq) and Fe²⁺ (aq) takes place as mentioned here:

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

Split this reaction into two half equations:

\begin{array}{l}\text{Oxidation:}{\text{Fe}}^{\text{2+}}\left(\text{s}\right)\stackrel{}{\to }{\text{Fe}}^{\text{3+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}]\text{×2}\text{;}\text{E°}\text{=}\text{–0}\text{.77V}\\ \text{Reduction:}{\text{Br}}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\stackrel{}{\to }{\text{2Br}}^{\text{–}}\left(\text{aq}\right)\text{;}\text{E°}\text{=}\text{+1}\text{.09V}\\ \overline{\text{Overall}\text{reaction:}{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right){\text{+Br}}_{\text{2}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2Fe}}^{\text{3+}}\left(\text{aq}\right){\text{+ 2Br}}^{\text{–}}\left(\text{aq}\right)\text{;}\text{E}\text{=}\text{+0}\text{.32V}}\end{array}

The E.M.F. of the overall reaction is +ve, thus the reaction is feasible.

Q.27 Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO₃ with silver electrodes.

(ii) An aqueous solution AgNO₃ with platinum electrodes.

(iii) A dilute solution of H₂SO₄ with platinum electrodes.

(iv) An aqueous solution of CuCl₂ with platinum electrodes.

Ans.

(i) In aqueous solution AgNO₃ is dissociated into Ag⁺ and NO₃^– ions.

{\text{AgNO}}_{\text{3}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Ag}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{NO}}_{\text{3}}^{\text{–}}\left(\text{aq}\right)

When the electricity is passed, the cations move towards the cathode while the anions move towards the anode. At cathode, either Ag⁺ (aq) ions or H₂O molecules may be reduced. It depends upon the electrode potential of the ions.

\begin{array}{l}{\text{Ag}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\stackrel{}{\to }\text{Ag}\left(\text{s}\right)\text{;}\text{E°}\text{=}\text{+0}\text{.80}\text{V}\\ {\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{+}{\text{2e}}^{\text{–}}\stackrel{}{\to }{\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2OH}}^{\text{–}}\left(\text{aq}\right)\text{;}\text{E°}\text{=}\text{–0}\text{.83}\text{V}\end{array}

The reduction potential of Ag⁺(aq) ion is higher than H₂O molecule. So, at cathode, Ag⁺ ions (rather than H₂O molecule) are reduced.

Similarly at anode, either Ag metal or water molecule may be oxidised. Their electrode potentials are given here-

\begin{array}{l}\text{Ag}\left(\text{s}\right)\stackrel{}{\to }{\text{Ag}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{e}}^{\text{–}}\text{;}\text{E}\text{=}\text{–0}\text{.80}\text{V}\\ {\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{O}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{4H}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{4e}}^{\text{–}}\text{;}\text{E}\text{=}\text{–1}\text{.23}\text{V}\end{array}

As oxidation potential of Ag is much higher than of H₂O molecules, at anode, Ag gets oxidised not the H₂O molecules. The oxidising potential of NO^3- ion is much lower than H₂O molecule. It is seen that more bonds break during reduction of NO^3- ions than H₂O molecule.

Thus, when an aqueous solution of AgNO₃ is electrolysed, Ag from Ag anode dissolves while Ag⁺(aq) ions present in the solution get reduced and deposited on the cathode.

(ii) Pt cannot be oxidised easily. Thus, electrolysis of AgNO₃ using Pt electrode instead of silver electrode, gives O₂ at anode and Ag⁺ ions from the solution get deposited at the cathode.

(iii) H₂SO₄ ionises in aqueous solutions in the following manner:

{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{aq}\right)\stackrel{}{\to }{\text{2H}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{SO}}_{\text{4}}^{\text{2–}}\left(\text{aq}\right)

On electrolysis, either of H⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of H⁺ ions is higher than that of H₂O molecules.

\begin{array}{l}{\text{2H}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\stackrel{}{\to }{\text{H}}_{\text{2}}\left(\text{g}\right)\text{;}\text{E° = 0}\text{.0}\text{V}\\ {\text{2H}}_{\text{2}}\text{O}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\stackrel{}{\to }{\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2OH}}^{\text{–}}\left(\text{aq}\right)\text{;}\text{E°}\text{=}\text{–0}\text{.83}\text{V}\end{array}

Hence, at the cathode, H⁺ ions get reduced to liberate H₂gas.

On the other hand, at anode, either SO₄^2– ion or H₂O molecule get oxidised. But oxidation of SO₄^2– ion involves the cleavage of many bonds as compared to H₂O molecules. Hence, SO₄^2– ions have a lower oxidation potential than H₂O. Thus, H₂O get oxidised and liberate O₂ molecules at anode.

(iii) In aqueous solutions, CuCl₂ ionises to give Cu²⁺and Cl^– ions as:

{\text{CuCl}}_{\text{2}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(\text{aq}\right){\text{+ 2Cl}}^{\text{–}}\left(\text{aq}\right)

On electrolysis, either of Cu²⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of Cu²⁺is more than that of H₂O molecules.

\begin{array}{l}{\text{Cu}}^{\text{2+}}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\stackrel{}{\to }\text{Cu}\left(\text{s}\right)\text{;}\text{E°}\text{=}\text{+0}\text{.34V}\\ {\text{2H}}_{\text{2}}\text{O}\left(\text{aq}\right)\text{+}{\text{2e}}^{\text{–}}\stackrel{}{\to }{\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2OH}}^{\text{–}}\left(\text{aq}\right)\text{;}\text{E°}\text{=}\text{–0}\text{.83V}\end{array}

Hence, Cu²⁺ ions are reduced and get deposited at cathode.

At anode, either of Cl^– or H₂O is oxidised. The oxidation potential of H₂O is higher than that of Cl^–.

\begin{array}{l}{\text{2Cl}}^{\text{–}}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cl}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{2e}}^{\text{–}}\text{;}\text{E°}\text{=}\text{–1}\text{.36V}\\ {\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\to }{\text{O}}_{\text{2}}\left(\text{g}\right){\text{+ 4H}}^{\text{+}}\left(\text{aq}\right)\text{+}{\text{4e}}^{\text{–}}\text{;}\text{E°}\text{=}\text{–1}\text{.23V}\end{array}

But oxidation of H₂O molecules occurs at a lower electrode potential than that of Cl^– ions due to the over- voltage (extra voltage required to liberate gas). As a result, Cl^– ions are oxidised and liberate Cl₂ gas at anode.

Q.28 Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn

Ans.

The electrode potential values are given below:

\begin{array}{l}\text{E°}{}_{{\text{Al}}^{\text{3+}}\text{/Al}}\text{=}\text{–1}\text{.66V}{\text{E°}}_{{\text{Zn}}^{\text{2+}}\text{/Zn}}\text{=}\text{–0}\text{.76V}\\ {\text{E°}}_{{\text{Fe}}^{\text{2+}}\text{/Fe}}\text{=}\text{–0}\text{.44V}{\text{E°}}_{{\text{Cu}}^{\text{2+}}\text{/Cu}}\text{=}\text{+0}\text{.34V}\\ {\text{E°}}_{{\text{Mg}}^{\text{2+}}\text{/Mg}}\text{= –2}\text{.36V}\end{array}

A metal having more –ve electrode potential is stronger reducing agent than the one having less –ve or +ve electrode potential values. A metal of stronger reducing power displaces another metal of weaker reducing power from its salt solution. The order of the increasing reducing power of the given metals is also mentioned here.

Cu < Fe < Zn < Al < Mg

Thus, we can conclude Zn can displace Fe from its salt but Fe cannot displace Zn from its salt.

Thus, the order in which the given metals displace each other from the solution of their salts is as follows:

Mg>Al> Zn> Fe>Cu

Q.29 Given the standard electrode potentials,

K⁺/K = –2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V

Mg²⁺/Mg = –2.37V. Cr³⁺/Cr = –0.74V

Arrange these metals in their increasing order of reducing power.

Ans.

The lower the electrode potential, the stronger is the reducing agent. Thus, the increasing order of the reducing power of the given metals is as follows:

Ag < Hg < Cr < Mg < K

Q.30 Depict the galvanic cell in which the reaction

Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)

takes place. Further show:

(i) which of the electrode is negatively charged,

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.

Ans.

The galvanic cell corresponding to the given redox reaction can be represented as:

Zn | Zn²⁺(aq) || Ag⁺(aq) | Ag

(i)Zn electrode is negatively charged because at this

electrode, Zn oxidises to Zn²⁺ and the leaving electrons accumulate on this electrode.

(ii) Electrons move from anode to cathode in the external circuit. But the direction of flow of current is always opposite to the flow of electron. Therefore, the current will flow from silver(cathode) to zinc(anode).

(iii)The reaction taking place at Zn electrode can be represented as:

Zn(s) → Zn²⁺(aq) + 2e^–

The reaction taking place at Ag electrode can be represented as:

Ag⁺(aq) + e^– → Ag(s)