Maharashtra Board Solutions Class 11-Chemistry: Chapter 6- Redox Reactions

Class 11: Chemistry Chapter 6 solutions. Complete Class 11 Chemistry Chapter 6 Notes.

Maharashtra Board Solutions Class 11-Chemistry: Chapter 6- Redox Reactions

Maharashtra Board 11th Chemistry Chapter 6, Class 11 Chemistry Chapter 6 solutions

1. Choose the most correct option

Question A.
Oxidation numbers of Cl atoms marked as Cl^a and Cl^b in CaOCl₂ (bleaching powder) are

a. zero in each
b. -1 in Cl^a and +1 in Cl^b
c. +1 in Cl^a and -1 in Cl^b
d. 1 in each
Answer:
b. -1 in Cl^a and +1 in Cl^b

Question B.
Which of the following is not an example of redox reacton ?
a. CuO + H₂ → Cu + H₂O
b. Fe₂O₃ + 3CO₂ → 2Fe + 3CO₂
c. 2K + F₂ → 2KF
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
Answer:
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Question C.
A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by
a. A₂(BC₃)₂
b. A₃(BC₄)₂
c. A₃(B₄C)₂
d. ABC₂
Answer:
b. A₃(BC₄)₂

Question D.
The coefficients p, q, r, s in the reaction
pCr2O2−7 + q Fe^2⊕ → r Cr^3⊕ + s Fe^3⊕ + H₂O respectively are :
a. 1, 2, 6, 6
b. 6, 1, 2, 4
c. 1, 6, 2, 6
d. 1, 2, 4, 6
Answer:
c. 1, 6, 2, 6

Question E.
For the following redox reactions, find the correct statement.
Sn^2⊕ + 2Fe^3⊕ → Sn^4⊕ + 2Fe^2⊕
a. Sn^2⊕ is undergoing oxidation
b. Fe^3⊕ is undergoing oxidation
c. It is not a redox reaction
d. Both Sn^2⊕ and Fe^3⊕ are oxidised
Answer:
a. Sn^2⊕ is undergoing oxidation

Question F.
Oxidation number of carbon in H₂CO₃ is
a. +1
b. +2
c. +3
d. +4
Answer:
d. +4

Question G.
Which is the correct stock notation for magenese dioxide ?
a. Mn(I)O₂
b. Mn(II)O₂
c. Mn(III)O₂
d. Mn(IV)O₂
Answer:
d. Mn(IV)O₂

Question I.
Oxidation number of oxygen in superoxide is
a. -2
b. -1
c. –12
d. 0
Answer:
c. –12

Question J.
Which of the following halogens does always show oxidation state -1 ?
a. F
b. Cl
c. Br
d. I
Answer:
a. F

Question K.
The process SO₂ → S₂Cl₂ is
a. Reduction
b. Oxidation
c. Neither oxidation nor reduction
d. Oxidation and reduction.
Answer:
a. Reduction

2. Write the formula for the following compounds :
A. Mercury(II) chloride
B. Thallium(I) sulphate
C. Tin(IV) oxide
D. Chromium(III) oxide
Answer:
i. HgCl₂
ii. Tl₂SO₄
iii. SnO₂
iv. Cr₂O₃

3. Answer the following questions

Question A.
In which chemical reaction does carbon exibit variation of oxidation state from -4 to +4 ? Write balanced chemical reaction.
Answer:
In combustion of methane, carbon exhibits variation from -4 to +4. The reaction is as follows:
CH₄ + 2O₂ → CO₂ + 2H₂O
In CH₄, the oxidation state of carbon is -4 while in CO₂, the oxidation state of carbon is +4.

Question B.
In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5 ?
Answer:

C. Calculate the oxidation number of underlined atoms.
a. H₂SO₄
b. HNO₃
c. H₃PO₃
d. K₂C₂O₄
e. H₂S₄O₆
f. Cr₂O₇^2-
g. NaH₂PO₄
Answer:
i. H₂SO₄
Oxidation number of H = +1
Oxidation number of O = -2
H₂SO₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0
∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0
∴ Oxidation number of S + 2 – 8 = 0
∴ Oxidation number of S in H₂SO₄ = +6

ii. HNO₃
Oxidation number of H = +1
Oxidation number of O = -2
HNO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0
∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0
∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0
∴ Oxidation number of N + 1 – 6 = 0
∴ Oxidation number of N in HNO₃ = +5

iii. H₃PO₃
Oxidation number of O = -2
Oxidation number of H = +1
H₃PO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0
∴ Oxidation number of P + 3 – 6 = 0
Oxidation number of P is H₃PO₃ = +3

iv. K₂C₂O₄
Oxidation number of K = +1
Oxidation number of O = -2
K₂C₂O₄ is a neutral molecule.
∴ Sum of the oxidation number of all atoms = 0
∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0
∴ 2 × (Oxidation number of C) + 2 – 8 = 0
∴ 2 × (Oxidation number of C) = + 6
∴ Oxidation number of C = +62
∴ Oxidation number of C in K₂C₂O₄ = +3

v. H₂S₄O₆
Oxidation number of H = +1
Oxidation number of O = -2
H₂S₄O₆ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0
∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0
∴ 4 × (Oxidation number of S) + 2 – 12 = 0
∴ 4 × (Oxidation number of S) = + 10
∴ Oxidation number of S = +104
∴ Oxidation number of S in H₂S₄O₆ = +2.5

vi. Cr₂O₇^2-
Oxidation of O = -2
Cr₂O₇^2- is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 2
∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2
∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2
∴ 2 × (Oxidation number of Cr) – 14 = – 2
∴ 2 × (Oxidation number of Cr) = – 2 + 14
∴ Oxidation number of Cr = +122
∴ Oxidation number of Cr in Cr₂O₇^2- = +6

vii. NaH₂PO₄
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
NaH₂PO₄ is a neutral molecule
Sum of the oxidation numbers of all atoms = 0
(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0
(+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0
(Oxidation number of P) + 3 – 8 = 0
Oxidation number of P in NaH₂PO₄ = +5

Question D.
Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
b. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
c. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
Answer:
i. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from -2 to +4 and that of Cu decreases from +1 to 0. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by gain of electrons and therefore, Cu is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agents (Reduced species): Cu₂O/ Cu₂S
3. Reductant/reducing agent (Oxidised species): Cu₂S

[Note: Cu in both Cu₂O and Cu₂S undergoes reduction. Hence, both Cu₂O and Cu₂S can be termed as oxidising agents in the given reaction.]

ii. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction. Result:
The given reaction is NOT a redox reaction.

iii. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and itself is oxidised. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent (Reduced species): I₂
3. Reductant/reducing agent (Oxidised species): S₂O₃^2-

Question E.
What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg²⁺
b. Cu / Cu²⁺
c. O₂ / O^2-
d. Cl₂ / Cl^–
Answer:
a. Mg / Mg²⁺
Here, Mg loses two electrons to form Mg²⁺ ion.
Mg(s)⟶Mg2+(aq)+2e−
Hence, Mg / Mg²⁺ is an oxidized state.

b. Cu/Cu²⁺
Here, Cu loses two electrons to form Cu²⁺ ion.
Cu(s)⟶Cu2+(aq)+2e−
Hence, Cu/Cu²⁺ is in an oxidized state.

c. O₂ / O^2-
Here, each O gains two electrons to form O^2- ion.
O2( g)+4e−⟶2O2−(aq)
Hence, O₂ / O^2- is in a reduced state.

d. Cl₂ / Cl^–
Here, each Cl gains one electron to form Cl^– ion.
Cl2( g)+2e−⟶2Cl−(aq)
Hence, Cl₂ / Cl^– is in a reduced state.

Question F.
Justify the following reaction as redox reaction.
2 Na_2(s) + S_(s) → Na₂S_(s)
Find out the oxidizing and reducing agents.
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + S_(s) → 2Na⁺ + S^2-
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and sulphur atom gains two electrons to form S^2-. This can be represented as follows:

iii. When Na is oxidised to Na₂S, the neutral Na atom loses electrons to form Na⁺ in Na₂S while the elemental sulphur gains electrons and forms S^2- in Na₂S.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.
vi. Oxidising agent is an electron acceptor and hence, S is an oxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.

Question G.
Provide the stock notation for the following compounds : HAuCl₄, Tl₂O, FeO, Fe₂O₃, MnO and CuO.
Answer:

Question H.
Assign oxidation number to each atom in the following species.
a. Cr(OH)₄^–
b. Na₂S₂O₃
c. H₃BO₃
Answer:
i. Cr(OH)₄^–
Oxidation number of O = -2
Oxidation number of H = +1
Cr(OH)₄^– is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1
∴ Oxidation number of Cr + 4 × (-2) + 4 × (+1) = – 1
∴ Oxidation number of Cr – 8 + 4 = – 1
∴ Oxidation number of Cr – 4 = – 1 –
∴ Oxidation number of Cr = – 1 + 4
∴ Oxidation number of Cr in Cr(OH)₄^– = +3

ii. Na₂S₂O₃
Oxidation number of Na = +1
Oxidation number of O = -2
Na₂S₂O₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Na) + 2 × (Oxidation number of S) + 3 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of S) + 3 × (-2) = 0
∴ 2 × (Oxidation number of S) + 2 – 6 = 0
∴ 2 × (Oxidation number of S) = + 4
∴ Oxidation number of S = +42
∴ Oxidation number of S in Na₂S₂O₃ = +2

iii. H₃BO₃
Oxidation number of H = +1
Oxidation number of O = -2
H₃BO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of B) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of B) + 3 × (-2) = 0
∴ Oxidation number of B + 3 – 6 = 0
∴ Oxidation number of B in H₃BO₃ = +3

Question I.
Which of the following redox couple is stronger oxidizing agent ?
a. Cl₂ (E⁰ = 1.36 V) and Br₂ (E⁰ = 1.09 V)
b. MnOΘ4 (E0 = 1.51 V) and Cr2O2Θ7 (E⁰ = 1.33 V)
Answer:
a. Cl₂ has a larger positive value of E⁰ than Br₂. Thus, Cl₂ is a stronger oxidizing agent than Br₂.
b. MnOΘ4 has larger positive value of E⁰ than Cr2O2Θ7. Thus, MnOΘ4 is stronger oxidizing agent than Cr2O2Θ7

Question J.
Which of the following redox couple is stronger reducing agent ?
a. Li (E⁰ = – 3.05 V) and Mg(E⁰ = – 2.36 V)
b. Zn(E⁰ = – 0.76 V) and Fe(E⁰ = – 0.44 V)
Answer:
a. Li has a larger negative value of E⁰ than Mg. Thus, Li is a stronger reducing agent than Mg.
b. Zn has a larger negative value of E⁰ than Fe. Thus, Zn is a stronger reducing agent than Fe.

4. Balance the reactions/equations :

Question A.
Balance the following reactions by oxidation number method

Answer:
i. Cr2O2−7(aq)+SO2−3(aq)⟶Cr3+(aq)+SO2−4(aq)( acidic )
Step 1: Write skeletal equation and balance the elements other than O and H.
Cr2O2−7(aq)+SO2−3(a)⟶2Cr3+(aq)+SO2−4(aq)
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)
Step 3: Balance ‘O’ atoms by adding 4H₂O to the right-hand side.
Cr2O2−7(aq)+3SO2−3(aq)⟶2Cr3+(aq)+3SO2−4(aq)+4H2O(l)
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H on the left-hand side.
Cr2O2−7(aq)+3SO2−3(aq)+8H+(aq)⟶2Cr3+(aq)+3SO2−4(aq)+4H2O(l)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:

ii. MnO−4(aq)+Br−(aq)⟶MnO2( s)+BrO−3(aq) (basic) 
Step 1: Write skeletal equation and balance the elements other than O and H.
MnO−4(aq)+Br−(aq)⟶MnO2( s)+BrO−3(aq)
Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of Mn.
2MnO−4(aq)+Br−(aq)⟶2MnO2( s)+BrO−3(aq)
Step 3: Balance ‘O’ atoms by adding H₂O to the right-hand side.
2MnO−4(aq)+Br−(2q)⟶2MnO2( s)+BrO−3 (aq) +H2O(l)
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.

iii. H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l) (acidic)
Step 1: Write skeletal equation and balance the elements other than O and H.
H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l)
Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of S.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)
Step 3: Balance ‘O’ atoms by adding H2O to the right-hand side.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l) + H₂O_(l)
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
2H₂SO_4(aq) + C_(s) → CO₂ + 2SO_2(g) + H₂O_(l)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: 2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)

iv. Bi(OH)3( s)+Sn(OH)−3(aq)⟶Bi(s)+Sn(OH)2−6(aq) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H.
Bi(OH)3( s)+Sn(OH)−3(aq)⟶Bi(s)+Sn(OH)2−6(aq)
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.

Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3W on the right-hand side.

Question B.
Balance the following redox equation by half reaction method

Answer:
i. H₂C₂O_4(aq) + MnO−4(aq) → CO_2(g) + Mn2+(aq)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 8H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

ii. Bi(OH)3( s)+SnO2−2(aq)⟶SnO2−3(aq)+Bi(s)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance half equations for O atoms by adding H₂O to the side with less O atoms. Add 1H₂O to left side of oxidation half equation and 3H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 3H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

Reaction occurs in basic medium. However, H⁺ ions cancel out and the reaction is balanced. Hence, no need to add OH^– ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:

5. Complete the following table :

Assign oxidation number to the underlined species and write Stock notation of compound

Compound	Oxidation number	Stock notation
AuCl3	……………..	……………..
SnCl2	……………..	……………..
V––2O4−7	……………..	……………..
Pt–––Cl2−6	……………..	……………..
H3AsO3	……………..	……………..

Answer:

Compound	Oxidation number	Stock notation
AuCl3	+3	Au(III)Cl3
SnCl2	+2	Sn(II)Cl2
V––2O4−7	+5	V2(V)O4−7
Pt–––Cl2−6	+4	Pt(IV)Cl2−6
H3AsO3	+3	H3As(III)O3

Can you tell? (Textbook Page No. 81)

Question i.
Why does cut apple turn brown when exposed to air?
Answer:
Cut apple turns brown when exposed to air because polyphenols are released. These polyphenols undergo oxidation in the presence of air and impart brown colour.

Question ii.
Why does old car bumper change colour?
Answer:
Car bumper is made of iron which undergoes rusting over a period of time. Hence, old car bumper changes colour.

Question iii.
Why do new batteries become useless after some days?
Answer:
Batteries generate electricity by redox reactions. Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. Hence, new batteries become useless after some days.

Can you recall? (Textbook Page No. 81)

Question i.
What is combustion reaction?
Answer:
Combustion is a process in which a substance combines with oxygen.

Question ii.
Write an equation for combustion of methane.
Answer:
Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O + Heat + Light

Question iii.
What is the driving force behind reactions of elements?
Answer:
The ability of element to combine with other element or the ability of element to replace other element in compound is the driving force behind the reactions. This may involve formation of precipitates, formation of water, release of gas, etc.

Try this. (Textbook Page No. 82)

Question 1.
Complete the following table of displacement reactions. Identify oxidising and reducing agents involved.

Reactants	Products
Zn(s) + ————(aq)	————-(aq) + Cu(s)
Cu(s) + 2Ag+(aq)	—————– + ————–
———– + ————-	Co2+(aq) + Ni(s)

Answer:

Try this (Textbook Page No. 88)

Question 1.
Classify the following unbalanced half equations as oxidation and reduction.

Answer:

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Maharashtra Board Solutions Class 11-Chemistry: Chapter 6- Redox Reactions

Download PDF: Maharashtra Board Solutions Class 11-Chemistry: Chapter 6- Redox Reactions PDF

Chapterwise Maharashtra Board Solutions Class 11 Chemistry :

• Chapter 1- Some Basic Concepts of Chemistry
• Chapter 2- Introduction to Analytical Chemistry
• Chapter 3- Basic Analytical Techniques
• Chapter 4- Structure of Atom
• Chapter 5- Chemical Bonding
• Chapter 6- Redox Reactions
• Chapter 7- Modern Periodic Table
• Chapter 8- Elements of Group 1 and 2
• Chapter 9- Elements of Group 13, 14 and 15
• Chapter 10- States of Matter
• Chapter 11- Adsorption and Colloids
• Chapter 12- Chemical Equilibrium
• Chapter 13- Nuclear Chemistry and Radioactivity
• Chapter 14- Basic Principles of Organic Chemistry
• Chapter 15- Hydrocarbons
• Chapter 16- Chemistry in Everyday Life

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