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Chemical Reactions & Equations

Chemical Reactions & Equations CHEMICAL REACTIONS

Section - 1

There are infinite chemical processes taking place in our daily lives where the properties of a substance change. However, during these processes atoms of an element do not transform into atoms of another element. Also, there is no gain or loss of atoms during a chemical reaction. Some examples of chemical change observed in our daily lives are:

1.



A freshly cut apple turns brown after some time.



Milk left for some time during summer turns sour.



Iron rusts on being exposed to humid conditions.

Chemical Reactions : During this process, chemical substances, with inherent properties, interact with each other and form completely new chemical substances with different properties. The substances that undergo reaction are called reactants and the substances thus formed as a result of evolution or absorption of heat & energy etc. are called products. During a chemical reaction, a few bonds between the atoms of reactants break. Then, the atoms rearrange themselves to form the products.

2.

Characteristics of Chemical Reactions : The important characteristics of chemical reactions are: I.

Evolution of gas : Some chemical reactions are characterized by the evolution of a gas. For Example :

II.

(a)

When zinc granules are dipped in with dilute sulphuric acid, bubbles of hydrogen gas are produced.

(b)

The chemical reaction between sodium carbonate and dilute hydrochloric acid is characteristed by the evolution of CO2 gas.

Formation of Precipitate :  A precipitate is a solid product which separates out from the solution during a chemical reaction.  Some chemical reactions are characterized by the formation of a precipitate. For Example : (a)

The chemical reaction between potassium iodide and lead nitrate is characterized by the formation of a yellow precipitate of lead iodide.

(b)

The chemical reaction between sulphuric acid and barium chloride solution is characterised by the formation of a white precipitate of baruim sulphate.

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Section 1

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Chemical Reactions & Equations III.

Change in Colour : Some chemical reactions are characterized by a change in colour. For Example : (a) The chemical reaction between citric acid and purple coloured potassium permanganate solution is characterized by a change in colour from purple to colourless. (b) The chemical reaction between sulphur dioxide gas and acidified potassium dichromate solution is characterized by a change in colour from orange to green.

IV.

Change in Temperature : Some chemical reactions are characterized by a change in temperature.

V.

For Example : (a) The chemical reaction between quicklime and water to form slaked lime is characterized by a change in temperature (rise in temperature). (b) The chemical reaction between zinc granules and dilute sulphuric acid is also characterized by a change in temperature (rise in temperature). Change in state : Some chemical reactions are characterized by a change in state. For Example : (a) The combustion reaction of candle wax is characterized by a change in state from solid to liquid and gas.

Illustrating the concept : When a magnesium ribbon rubbed with a sand paper (required to remove the layer of magnesium oxide formed over it’s surface) is burnt using a spirit lamp, it burns with a dazzling white light and a white-coloured powder is obtained. This powder is magnesium oxide that is formed as a result of a chemical reaction between magnesium and oxygen present in the air.

Magnesium  Oxygen   Magnesium oxide Ribbon (solid)

3.

(white powder)

Endothermic Reactions & Exothermic Reactions : Exothermic Reaction : A chemical reaction which produces (releases) energy in the form of heat and light is called a exothermic reaction. Examples : (i)

CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O () + Energy (heat)

(ii)

2Al (s) + Fe2O3(s)   Al2O3 (s) + 2Fe (s) + Energy (heat)

Endothermic Reactions : A chemical reaction which absorbs energy in the form of heat and light is called a endothermic reaction. Example :  Ba(SCN) (aq) + 2NH (aq) + 10 H O () Ba(OH)2 . 8 H2O(s) + 2NH4SCN(s)  2 3 2

In this, the reaction mixture becomes so cold that moisture from the air forms a layer of frost. 2

Section 1


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CHEMICAL EQUATIONS

1.

Section - 2

Chemical Equations : This is the short form of writing down chemical reactions. Every chemical species can be represented in the form of certain symbols, and these symbols are used to write a chemical reaction. In a chemical equation, the reactants are written on the left-hand side and products on the right- hand side, separated by an arrow. Physical states of the reactant and products are expressed within a bracket put before the corresponding chemical species. Following short forms are used for different physical states: (s)

for solid state

(l) for liquid state

(aq) for solution in water

(g) for gaseous state

Different physical conditions, such as temperature and pressure can be shown over the separating arrow.

2. Balancing Chemical Equations : The balancing of chemical equation means to equalise the number of atoms of each element on the both sides of the equation. Hit and Trial Method : A chemical equation can be balanced by making trials using the smallest whole number coefficients. Steps to be followed are given below. Step 1 : Write the skeleton equation with the correct formulae of reactants and products. Step 2 : Count the number of atoms of various elements on both sides of the equation and make them equal to multiplying the formulae by suitable numbers. Step 3 : To equate the number of atoms on both sides of equation, select the formula containing maximum number of atoms and start the process of balancing. Step 4 : Then pick the second element to balance the partly balanced equation. Similarly, pick the third and fourth, and so on. Step 5 : Atoms of elementary gases such as hydrogen, oxygen, nitrogen, etc. are balanced last of all. Sometimes, these are first changed to atomic state. After balancing, these are changed back to molecular form. Step 6 : In ionic equations, the charges must also be balanced in addition to balancing the atoms. This means that there should be same total charge on both sides of the equation.


Section 2

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Illustration - 1

Let us try balancing the following chemical equation: Fe + H2O  Fe3O4 + H2

SOLUTION : Step 1: Count the number of atoms of various elements on both sides of the equation

Step 2: Start balancing with the compound which contains maximum number of atoms. It may be a reactant or a product. In that compound select the element which has maximum number of atoms. According to this rule, select Fe3O4 and the element oxygen in it. There are four O atoms on the RHS and only one O atom on the LHS. To balance, multiply H2O by 4 so that there are four atoms of O on LHS also.

The partly balanced equation becomes: Fe + 4H2O  Fe3O4 + H2 Step 3: In above equation Fe and H are still not balanced. Pick any of these atoms to proceed further, atoms in the above partly balance the number of H atoms, make the number of atoms of hydrogen as 8 or molecules of hydrogen 4 on R.H.S.

Fe + 4H2O  Fe3O4 + 4H2 Step 4: Examine the above equation and pick up the other atom which is not balanced. It is clear that only one atom iron is left unbalanced. To equalise Fe atoms, put coefficient 3 before Fe.

Thus, the equation becomes 3Fe + 4H2O  Fe3O4 + 4H2 4

Section 2


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Step 5: Finally check, the correctness of the balanced equation by counting the number of atoms on both sides of the equation 3Fe + 4 H2O  Fe3O4 + 4 H2

Illustrating the concept : Zn + H 2SO4  ZnSO 4 + H 2 zinc sulphuric zinc sulphate hydrogen acid

In this equation, both the sides have: 

1 atom of zinc



2 atoms of hydrogen



1 atom of sulphur



4 atoms of oxygen

So, this is a balanced chemical equation. Now, let us try to balance a chemical equation.

Illustration - 2

Glucose oxidises during the following reaction C6 H12O6  O2   CO2  H 2O

SOLUTION : Step 1: All the elements present in the equation are written down. Then, the number of atoms present of these elements, both on the reactants side and on the products side are counted and written down before the corresponding element.

Step 2: The number of atoms of each element on each side are compared. If they are not equal for any atom, they are multiplied by a number to make them equal. Generally, comparison is started with the element having maximum number of atoms. Here it is hydrogen. It has 12 atoms on the reactants’ side and just 2 atoms on the products’ side. So, to equalise the number of atoms on the products’ side, H 2O is multiplied by 6.

The equation becomes: C6 H12O6  O2   CO 2  6H 2O Self Study Course for IITJEE with Online Support

Section 2

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Step 3: Since O atoms occur in maximum entities, it should be balanced last of all. Balancing of carbon atoms : 6 atoms of carbon on LHS, 1 atom of carbon on RHS. So, to equalise the number of atoms on the products’ side, CO2 is multiplied by 6.

 6CO 2  6H 2O The equation becomes: C6 H12O6  O2  Step 4: This leads to a change in the number of atoms of oxygen on the products’ side.

To make the number of atoms of oxygen on the reactants’ side equal to 18, dioxygen on the reactants’ side is multiplied by 6 to make the number of oxygen atoms 12 on this side. The other reactant glucose already contains 6 oxygen atoms. So, the total number of oxygen atoms on the reactants’ side becomes 12 + 6 = 18, i.e., equal to the no. of oxygen atoms present on the products’ side. Step 5: The resultant equation is: C6H12O6 + 6O2  6CO2 + 6H2O This has the same number of atoms of all the elements on both sides; so now, it is a balanced chemical equation.

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Section 2


Vidyamandir Classes


IN-CHAPTER EXERCISE - A 1.

When sulphuric acid is poured in a flask containing zinc granules, hydrogen gas is emitted. Is this an example of a physical change or a chemical reaction.

2.

Ice contained in a beaker becomes water due to the sun’s heat. Is this an example of a physical change or a chemical reaction.

3.

When hydrogen is passed over copper oxide, copper and steam are formed. Write a balanced equation for this reaction and state which of the chemicals are: (i) Reactants (ii) Products (iii) Metals (iv) Non-metals.

4.

Magnesium carbonate reacts with hydrochloric acid to produce magnesium chloride, carbon dioxide and water. Write a balanced equation for this reaction.

5.

Which of the following equations are balanced. (a)

 CO2 + H2O CH4 + 2O2 

(b)

2HgO   2Hg + O2

6.

Translate the following statements into chemical equations and then balance these: (a) Aluminium metal replaces iron from ferric oxide (Fe2O3) giving aluminium oxide and iron. (b) Phosphorus burns in oxygen to form phosphorus pentoxide.

7.

Balance the following equations: (i)

 KCl + O2 KClO3 

(ii)

Al + H2SO4   Al2(SO4)3 + H2


Section 2

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TYPES OF CHEMICAL REACTIONS

Section - 3

Chemical reactions can be classified into the following categories:

1.

1.

Combination reactions.

2.

Decomposition reactions.

3.

Displacement reactions.

4.

Double displacement reactions.

5.

Oxidation and reduction reactions.

6.

Effects of Oxidation Reactions in Everyday Life (Rusting and Rancidity).

Combination Reactions : During combination reactions, two or more chemical substances combine to form a single product. Combination reactions are also called synthesis reactions. Such reactions play an important role in inorganic chem -istry. Combination reactions may occur between: (i) Two elements (ii) An element and a compound (iii) Two compounds

Illustrating the concept :

2.



When hydrogen molecules combine with oxygen, water is formed. 2 H 2 (g) + O2 (g)  2 H 2O(l )



Coal burns in the presence of oxygen to form carbon dioxide. C(s) + O 2 (g)  CO 2 (g)



Sodium oxide and water make sodium hydroxide. Na2O + H2O  2 NaOH

Decomposition Reaction : During a decomposition reaction, a compound splits into elements or smaller compounds. It may be consid -ered as the opposite of combination reaction. The chances of a decomposition reaction occuring increase under the influence of : (a) Heat (b) Radiation (c) Humidity (d) Acids Decomposition reactions are used in analytical methods. Analytical methods that use decomposition reactions are :  Mass spectrometry  Thermogravimetric analysis  Gravimetric analysis Endothermic Reactions : Those reactions in which heat is absorbed is called endothermic reactions. heat

2Pb(NO3)2  2PbO(s) + 4NO2(g) + O2(g) Thus we observe that decomposition reactions required energy either in form of heat, light or electricity of breaking down reactants. 8

Section 3


Vidyamandir Classes Illustration - -33 Illustration


When limestone is heated, it gets broken down into quicklime and carbon dioxide.

SOLUTION : heat

 

CaCO3 (s) Limestone (Calcium carbonate)

 CO 2 (g)

CaO(s) Quicklime (Calcium oxide)

Illustration - 4 is used to manufacture cement. Quicklime

Illustration - 4 When white silver chloride is exposed to sunlight, it decomposes to give grey-coloured silver and chlorine gas. SOLUTION : sunlight 2AgCl( s )   2Ag( s )  Cl ( g ) Silver chloride (white)

3.

Silver (grey)

2

Chlorine

Displacement Reaction : In these reactions, a more reactive element displaces a less reactive element from a compound. Such reactions are also called displacement reactions.

Illustrating the concept : If an iron nail is dipped in blue-coloured copper sulphate solution, the iron displaces the copper from copper sulphate and forms iron sulphate. Iron sulphate is light green in colour. Therefore, the blue colour of the solution fades. Due to the deposition of copper on iron, the nail turns brown. Fe (s) + CuSO4 (aq)  FeSO4 (aq) + Cu (s)

4.

Double Displacement Reaction : In a double displacement reaction, the reacting substances exchange their ions or bonds and form new compounds.

Illustrating the concept : When barium chloride is added to sodium sulphate, exchange of ions takes place and SO 24  (sulphate ions) go to barium chloride. The sulphate ions displace Cl– (chloride ions) from barium chloride to form insoluble barium sulphate. Barium sulphate appears as a white precipitate. The chloride ions from barium chloride go to sodium sulphate, displace it’s SO 24  ions and form sodium chloride. Na2SO4 (aq) + BaCl2 (aq)  2 NaCl (aq) + BaSO4 (s) This is an example of a double displacement reaction that is also a precipitation reaction. Self Study Course for IITJEE with Online Support

Section 3

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5.

Oxidation and Reduction Reactions : In order to understand oxidation and reduction reactions, the concept of oxidation and reduction should be first explained. Oxidation is addition of oxygen to a substance. The oxygen donor is called an oxidising agent and the recipient of oxygen is oxidised. For Example : When copper is heated, it reacts with oxygen to form black-coloured copper oxide as shown below :

2Cu + O2   2CuO Here, copper is oxidised because it has accepted oxygen. For Example : Ammonia gas on combustion gives nitrogen gas and water as shown below : 4 NH3 + 3O2   2 N 2 + 6 H 2O In this reaction NH3 is oxidised because it loses hydrogen (and we can also say that oxygen is reduced as it gets associated with the hydrogen). Reduction is removal of oxygen (or addition of hydrogen) from a substance. The substance that removes oxygen (or adds hydrogen) is called reducing agent, and the substance that loses oxygen (or gains hydrogen) is said to be reduced. For Example : When hydrogen gas is passed over black-coloured copper oxide, it loses oxygen as shown below : heat

C uO + H 2   Cu + H 2O Here, CuO is an oxidising agent because it donates oxygen. In the process, CuO gets reduced and H 2 gains oxygen. So, it is the reducing agent that gets oxidised. For Example : When carbon monoxide reacts with hydrogen, methanol is formed as shown below :

CO + H 2   CH3OH In this reaction CO is reduced because it gains hydrogen (and we can also say that hydrogen is oxidized as it gets associated with the oxygen). Such reactions are also called redox reactions because of simultaneously occurring reduction and oxidation reactions.

NOW ATTEMPT IN-CHAPTER EXERCISE-B BEFORE PROCEEDING AHEAD IN THIS EBOOK

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Section 3


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OXIDATION NUMBER CONCEPT

Section - 4

Oxidation Number : Oxidation number of an element in a particular compound represents the number of elections lost or gained by an element. OIL – Oxidation is loss, RIG – Reduction is gain Oxidation State : Oxidation state of an atom is defined as oxidation number per atom. For Example : In K 2 MnO 4 Oxidation number of Mn = +6 ; Oxidation state of Mn = Mn 6+

The Rules for Deriving Oxidation Number : (i) (ii)

In uncombined state or free state, oxidation number of an element is zero. O is 2; in peroxides ( O  O ) it is 1.

(iii) (iv) (v) (vi) (vii) (viii)

H is +1 ; in ionic hydrides it is 1. Halogen in halides is always 1. Sulphur in sulphides is always 2. Alkali metal (Li, Na, K, Rb, Cs, Fr) is in +1. Alkaline earth metal (Be, Mg, Ca, Sr, Ba, Ra) is in always +2. The algebraic sum of all the oxidation number of elements in a compounds X is equal to zero or net charge respectively. (ix) Some variable oxidation state of elements

Fe(2, 3), Cu (1&  2); Mn(7, 6, 5, 4, 3, 2, 1); Sn(2, 4) For Example : K 2 MnO 4

2  (oxidation number of K) + (Oxidation number of Mn) + 4  (Oxidation number of O) = 0 For Example : MnO 4 1  (oxidation number of Mn) + 4  (Oxidation number of O) = 1

Maximum oxidation number of an element (except O and F) = Group number. Minimum oxidation number of an element (except metal) = Group number 8.


Section 4

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For Example : In compounds of carbon, the oxidation number varies from 4 to +4 but valency of carbon always 4. Carbon Compound

CH 4

Oxidation number of C

4

CH3Cl

CH 2Cl2 0

2

CHCl3 +2

CCl4 +4

Oxidation State/Oxidation number in Complex Molecules : (i)

Carbon in Glucose (C6 H12O6 ) : Let the oxidation number of carbon be x, hydrogen +1 and oxygen 2 Sum of oxidation number of various atoms in C6H12O6 is

6 x  12(1)  6(2)  0 6 x  12  12  0 6 x  0, x = 0 In glucose the oxidation state of carbon is zero.

(ii)

Ni in [Ni(CO)4 ] : The oxidation state of CO is zero, hence the oxidation state of nickel will also zero.

(iii) Sulphur in (CH3 )2 SO i.e., dimethyl sulphoxide : Let the oxidation state of S be x, oxygen 2, and each methyl group is +1 Sum of oxidation numbers of various atoms in (CH3 )2 SO is 2  x  2  0 x=0

Thus the oxidation state of sulphur in (CH3 )2 SO is zero. (iv) Boron in LiBH 4 : In metal hydrides, the oxidation state of hydrogen is 1 and the metal Lithium is +1 Sum of oxidation number of various atom in LiBH 4 is, x = +3 1  x  4  0 Thus the oxidation state of B in LiBH 4 is +3

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Illustration - 5


Find out the oxidation number/oxidation state of :

(a)

S in H 2 S2O7

(b)

S in Na2 S2O3

(c)

Cr in Cr (CO )6

(d)

Fe in Fe2 (CO )9

(e)

Fe in Fe3O4

(f)

Mn in MnO4

SOLUTION : (a) Let the oxidation no. of S be x, H = +1 and  2

(d)

2(1)  2 x  7(2)  0

(b)

(c)

2  2 x  14  0 x = +6 2 x  12; Let the oxidation number of S be x, Na = +1 and O  2

(e)

O  2. 3 x  4(2)  0

2  2x  6  0 2 x  4; x  2 Let oxidation number of Cr be x, and CO = zero. x  6(0)  0; x  0

Let oxidation number of Fe be x, and CO = zero. 2 x  9(0)  0 2x = 0 x=0 Let oxidation number of Fe be x, and

3x  8  0

8 3 Let the oxidation number of Mn be x and x

(f)

O = 2. x  4(2)  1 x  7

Balancing of redox Equations : The chemical equations which involve oxidation and reduction (redox equation) are balanced with the help of the following methods : 

Oxidation number method.



Iron electron method (or Half reaction method)

Let us try to balance a few redox equations with the help of these methods.

Balancing by Oxidation Number Method : The various steps involved in balancing a redox equation by oxidation number method are : 

Write the skeleton equation.



Indicate the oxidation numbers of all the atoms involved in the equation above their symbols.



Identify the elements which undergo change in oxidation number.



Calculate the increase and decrease in oxidation number per atom with respect to the reactants. If more than one atom is involved, then multiply with the number of the atom undergoing the change to calculate the total change in oxidation number.


Section 4

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Chemical Reactions & Equations    

Equate the increase and decrease in oxidation number on the reactant side by multiplying the formulae of the oxidising and reducing agents suitably. Balance the equation with respect to all the atoms excepts hydrogen and oxygen. Finally balance hydrogen and oxygen atoms also. In the reactions taking place in the acidic medium, balance the O atoms by adding required number



of H 2O molecules to the side deficient in O atoms. Then balance the H atoms by adding H + to the side deficient in H atoms. In the basic medium, first balance the number of negative charges by adding required number of OH  ions to the side deficient in the magnitude of the charges. Then add H 2O molecules on the

other side in order to balance the OH  ions added. Example : CuO + NH3   Cu + N 2 + H 2O

The balancing is done in the following steps : 1.

Write the O.N. of each atom in the skeleton equation +2 2

3 +1

0

0

+1 2

Cu O + N H3   Cu + N 2 + H 2 O +2

2.

3

0

0

3.

Identify the atoms which undergo change in O.N. Cu O + N H3   Cu + N 2 + H 2O Calculate the increase and decrease in O. N. w.r.t. reactant atoms

4.

Equate the increase and decrease in O.N. on the reactant side.

3CuO  2 NH 3   Cu  N 2  H 2O 5.

Balance the number of Cu and N atoms on both sides of the equation.

3CuO  2 NH 3   3Cu  N 2  H 2O 6.

Now balance H and O atoms by hit and trial method

3CuO + 2 NH3   3Cu + N 2 + 3H 2O

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Illustration - 6


Balance the following equations by oxidation number method

(a)

Cu  NO3   NO2  Cu 2

(Acidic Medium)

(b)

[Cr (OH )4 ]  H 2O2   CrO42   H 2O

(Basic Medium)

SOLUTION : (a)

5 2

0

4 2

2

1.

Write the O.N. of each atom in the skeleton equation. Cu  ( N O )    N O2  (Cu ) 2  3

2.

 N O2  (Cu )2  Identify the atoms which undergo change in O.N. Cu  ( N O3 ) 

3.

Calculate the increase and decrease in O.N. w.r.t to reactant atoms

4.

Equate the increase and decrease in O.N. on the reactant side. Cu  2 NO3   NO2  Cu 2 

5.

Balance the number of Cu and N atoms on both sides of the equation.

0

5

4

2

Cu  2 NO3   2 NO2  Cu 2  6.

As the reaction is carried in the acidic medium balance the number of O atoms by adding two H 2O molecules on the product side. Cu  2 NO3   2 NO2  Cu 2   2 H 2O

7.

To balance the number of H atoms, add 4H  on the reactant side

Cu  2 NO3  4 H    2 NO2  Cu 2   2 H 2O. The final equation is balanced w.r. to charge also. (b)

[Cr (OH )4 ]  H 2O2   CrO42   H 2O

(Basic Medium)

The balancing is done in the following steps : 1.

Write the O.N. of each atom in the skeleton equation :

2.

Identify the atoms which undergo change in O.N. 3

1

6

2

[Cr (OH )4 ]  H 2 O 2  (Cr O4 )2   H 2 O Self Study Course for IITJEE with Online Support

Section 4

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Chemical Reactions & Equations 3.

Calculate the increase and decrease in O.N. w.r. to reactant atoms

4.

Equate the increase and decrease in O.N. in the reactant side.

2[Cr (OH )4 ]  3H 2O2   2(CrO4 )2   H 2O 5.

Balance the number of Cr atoms in the equation.

2[Cr(OH)4 ] + 3H 2O2   2(CrO 4 )2 + H 2O 6.

As the reaction occurs in the basic medium, and the ionic charges are not equal on both side, add 2 OH  ions on reactant side to make ionic charges equal.

2[Cr(OH)4 ] + 3H 2O2 + 2OH    2CrO 42  + H 2O 7.

Finally, count hydrogen atoms and add appropriate number of (H 2O) molecules on right side to achieve balanced redox change.

Balancing by Ion-Election Method (or Half Reaction Method) Balancing can also be done by another method known as ion-electron method. It is based on the principle that the electrons lost during oxidation half reaction in a particular redox reaction is equal to the electrons gained in the reduction half reaction. The method is, therefore, called half reaction method. The balancing is completed in the following steps : (i)

Write the redox reaction in ionic form. Total electrons lost by oxidant = Total electrons gained by reductant

(ii)

Find out species which are getting oxidised and also which are getting reduced.

(iii) Split the whole equation into two half reactions i.e. oxidation half reaction and reduction half reaction. (iv) While balancing each half reaction, add electrons for the number of atoms of each element. (v)

In the acidic medium, and neutral medium add water molecules to the side deficient in O and H + to the side deficient in hydrogen.

(vi) For the reaction in basic medium, first balance the atoms as done in acidic medium. Then for each H + ion, add an equal number of OH  ions to both sides of the equation. Where H + and OH  appear on the same side of the equation, combine these to give H 2O.

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(vii) (viii)


 Multiply one or both half reactions by suitable number so that the number of e s become equal in both the equation.

Add the two balanced half reactions and cancel any term common to both sides.

Illustration - 7

Balance the following chemical equation by ion-electron method.

(a)

Cr2O72   Fe2   H    Cr 3  Fe3  H 2O

(b)

MnO4 (aq)  I  (aq )   MnO2 ( s )  I 2 ( s )

SOLUTION : Step 1 : Separation of the equation in two half reactions (i)

Write the O.N. of all the atoms involved in the skeleton equation 6 2

2

1

3

3

1 2

(Cr2 O7 )2   ( Fe)2   ( H )   (Cr )3  ( Fe)3  H 2 O (ii)

Identify the atoms which undergo change in O.N. 6

2

1

3

3

(Cr2 O7 )2   ( Fe)2   ( H )  (Cr )3  ( Fe)3  H 2O (iii)

Find out the species involved in the oxidation and reduction half reactions.

Thus, the two half reactions are : Oxidation half reaction : Fe2    Fe3 ,

Reduction half reaction : (Cr2O7 )2    Cr 3

Step 2 : Balancing of oxidation half reaction The oxidation half reaction is : Fe2    Fe3 (i)

As the increase in O.N. as a result of oxidation is 1, add one e on the product side to balance 2  Fe3  e change in O.N. Fe 

(ii)

The charge is already balanced, and the thus the equation is also balanced

Fe2    Fe3  e


….(i)

Section 4

17

Vidyamandir Classes

Chemical Reactions & Equations Step 3 : Balancing of reduction half reaction 6

3

The reduction half reaction is : (Cr2 O7 ) 2   (Cr )3 (i)

The decrease in O.N. per Cr atom is 3 and the total decrease in O.N. for two Cr atoms is 6. 2   Cr 3 Therefore, add 6e  on the reactant side (Cr2O7 )  6e 

(ii)

 2Cr 3 Balance Cr atoms on both sides of the equation (Cr2O7 ) 2  6e 

(iii) In order to balance O atoms add seven H 2O molecules on the product side and then to balance H atoms add 14 H + on the reactant side.

(Cr2O7 )2   6e  14 H    2Cr 3  7 H 2O

….(ii)

Step 4 : Adding the two half reactions In order to equate the electrons, multiply the equation (i) by 6 and then add to equation (ii) in order to get the final equation.

[b]

MnO4 (aq)  I  (aq )   MnO2 ( s )  I 2 ( s )

Step 1 : First we write the skeletal ionic equation which is

MnO4 (aq)  I  (aq )   MnO2 ( s )  I 2 ( s ) 1 

0

Step 2 : The two half-reactions are: Oxidation half : I (aq )   I 2 ( s ), Reduction half : 7

4

Mn O4 (aq )   Mn O2 (s ) Step 3 : To balance the I atoms in the oxidation half reaction, we rewrite it as : 2 I  (aq )   I 2 ( s) Step 4 : To balance the O atoms in the reduction half reaction, we add two water molecules on the right :

MnO4 (aq)   MnO2 ( s )  2 H 2O(l ) To balance the H atoms, we add four H  ions on the left:

MnO4 (aq)  4 H  (aq )   MnO2 (s )  2 H 2O (l ) As the reaction takes place in a basic solution, therefore, for H  ions, we add four OH  ions to both sides of the equation : 18

Section 4


Vidyamandir Classes


Mn O4 (aq)  4 H  (aq )  4OH  (aq )   MnO2 ( s)  2 H 2O(l )  4OH  (aq) Replacing the H  and OH  ions with water, the resultant equation is :

MnO4 (aq)  2 H 2O(l )   MnO2 (s )  4OH  (aq) Step 5 : In this step we balance the charges of the two half-reactions in the manner depicted as :

2 I  (aq )   I 2 ( s )  2e , Mn O4 (aq )  2 H 2O(l )3e   MnO2 (s )  4OH  (aq) Step 6 : Add two half-reactions to obtain net reactions after cancelling electrons on both sides.

6 I  (aq )  2MnO4 (aq )  4 H 2O(l )   3I 2 ( s )  2MnO2 ( s )  8OH  (aq) Step 7: A final verification shows that the equation is balanced in respect of the number of atoms and charge on both sides.

IN-CHAPTER EXERCISE-C A.

Assign oxidation number to the underlined elements in each of the following species : (i)

NaH 2 PO 4

(ii)

NaHSO 4

(iii)

H 2 P2O7

(iv) K 2 MnO 4

(v)

CaO 2

(vi)

NaBH 4

(vii)

H 2S2O7

(viii) KAl(SO)2 12 H 2O

(ix)

KI3

(x)

H 2S4O6

(xi)

Fe3O 4

(xii) C2 H5OH

(xv)

 

(xvi) NH 2 - NH 2

(xiii) C2 H 4O2

B.

(xiv) N  N

Balance the following equations in Acidic medium by both oxidation number and ion electron methods and identify the oxidants and reductants (i)

  MnO4(  Mn(2aq aq.)  C2 H 2O( aq.)  .)  CO2 g   H 2O(l )

(ii)

2 Cr2O7(  Cr 3 ( aq.)  C2 H 4O2( aq.) aq.)  C2 H 4O( aq.) 

(iii)

 MnO4 (aq.)  Br(aq.)   Mn(2aq .)  Br( aq.)

(iv)

   Cu( s )  NO3(  Cu(2aq aq.)  H (aq.)  .)  NO( g )  H 2O( l )

(v)

2 3  3 Cr2O7(  Cr(3aq aq.)  Fe( aq.)  .)  Fe( aq.)

 2  3 (vi) MnO4(  Mn(2aq aq.)  Fe( aq.)  .)  Fe( aq.)


Section 4

19

Vidyamandir Classes

Chemical Reactions & Equations C.

Balance the following equation in Basic medium by both oxidation number and ion electron methods and identify the oxidants and reductants. (i)

P4  NaOH  H 2O   PH 3  3NaH 2 PO2

(ii)

 N 2 H 4( g )  ClO3(  NO( g )  Cl  (aq.) aq.) 

 (iii) Cl2O7( g )  H 2O2( aq.)   ClO2( aq.)  O2( g )

(iv)

   Zn(s)  NO3(  Zn(2aq aq.)  .)  NH 4( aq.)

Effects of Oxidation Reactions in Everyday Life : Corrosion : One of the effects of oxidation reactions that we see around us is corrosion. Corrosion is the chemical reaction during which the characteristic properties of a material are adversely affected due to oxidation reactions with its surroundings. Corroded materials losses electrons while reacting with water and oxygen. Corrosion of iron leads to weakening of a iron structure. This process is called rusting. A film of reddish iron oxide forms on the surface of the corroded iron. Exposure to moisture present in the atmosphere is sufficient to rust iron. Corrosion is represented by the following chemical reaction :

4 Fe  3O2  xH 2O   2 Fe2O3 . xH 2O Painting iron prevents it from rusting. The layer of paint forms a protective coating that does not allow moist air to corrode the iron’s surface. Other way to prevent rusting are Oiling or Greasing, Galvanisation, coating by chromium. (Chrome plating) Rancidity : Oxidation has damaging effect on foods containing fats and oils. When the fats and oil present in food materials get oxidised by the oxygen, their oxidation products have unpleasant smell and taste. The condition produced by aerial oxidation of fats and oils in food marked by unpleasant smell and taste is called Rancidity. We can prevent rancidity by : (i)

Adding anti-oxidants to food containing fats and oils.

(ii)

Packaging fat and oil containing foods in nitrogen gas.

(iii) Keeping food in a refrigerator (reducing the temperature). (iv) Storing food in air-tight containers. (v)

20

Storing food away from light.

Section 4


Vidyamandir Classes


IN-CHAPTER EXERCISE-D 1.

What happens when zinc removes copper from a copper sulphate solution. Write a balanced chemical equation for this reaction.

2.

Explain double displacement reaction with the help of an example.

3.

Explain what happens when manganese oxide reacts with hydrochloric acid, and identify the oxidising and reducing agents.

4.

How will you prevent oils from turning rancid ?

5.

Explain the process of rusting of iron with the relevant chemical reaction.

SUBJECTIVE SOLVED EXAMPLES Example - 1

Why should a magnesium ribbon be cleaned before burning in air ?

SOLUTION : When magnesium ribbon remains exposed to moist air, a white layer of magnesium oxide is formed on its surface. This hinders the burning of magnesium. Hence, this layer is first removed by rubbing with sand paper before burning.

Example - 2

Write the balanced equation for the following chemical reactions :

(i)

Hydrogen + Chlorine  Hydrogen chloride

(ii)

Barium chloride + Aluminium sulphate  Barium sulphate + Aluminium chloride

SOLUTION : (i)

The given word equation is : Hydrogen + Chlorine  Hydrogen chloride Step 1:Writing the skeletal chemical equation. H2 + Cl2  HCl Step 2:Balancing of different elements. To balance H atoms on the two sides, multiply HCl on RHS by 2. We get : H2 + Cl2  2HCl Chlorine atoms are automatically balanced. Hence, the above equation is the balanced equation.

(ii)

Step 1: Writing the word equation. The given word equation is : Barium chloride + Aluminium sulphate  Barium sulphate + Aluminium chloride


Subjective Solved Example

21

Vidyamandir Classes

Chemical Reactions & Equations Step 2: Writing the skeletal chemical equation. In terms of formulae, the skeletal chemical equation will be BaCl2 + Al2 (SO4)3  BaSO4 + AlCl3

Step 3: Selecting the biggest formula and balancing of its different elements. Starting balancing of elements of the compound with the biggest formula. i.e., Al2(SO4)3, we balance elements one by one as follows : Al atoms on LHS = 2. Al atoms on RHS = 1. To balance Al atoms, multiply AlCl3 on RHS by 2. We get : BaCl 2  Al2 (SO4 )3   BaSO4  2AlCl3

S atoms on LHS = 3. S atoms on RHS = 1. To balance S atoms, multiply BaSO4 on RHS by 3. We get : BaCl 2  Al2 (SO4 )3   3BaSO4  2AlCl3

O - atoms are automatically balanced Ba atoms on LHS = 1. Ba atoms on RHS = 3. To balanced Ba atom, multiply BaCl2 on LHS by 3. We get : 3BaCl 2  Al2 (SO4 )3   3BaSO4  2AlCl3

Cl - atoms are automatically balanced. Step 3: Checking the correctness of the balanced chemical equation

Thus, the above equation is the required balanced chemical equation.

22



Vidyamandir Classes

Example - 3


Write a balanced chemical equation with state symbols for the following reactions :

(i)

Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.

(ii)

Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride and water.

SOLUTION : (i)

Step 1: Writing the word equation : The word equation for the given reaction is Barium chloride  Sodium sulphate   Barium sulphate  Sodium chloride

Solution 

Solution 

 Insoluble solid 

Solution 

Step 2: Writing the skeletal chemical equation BaCl2  aq   Na 2SO4  aq    BaSO 4  2 NaCl Step 3: Balancing of different elements Ba atoms are already balanced Cl atoms on LHS = 2. Cl atoms on RHS = 1. To balance Cl atoms, multiply NaCl on RHS by 2, We get : BaCl2  Na 2SO 4   BaSO 4  2 NaCl Na atoms are automatically balanced S atoms are already balanced O atoms are already balanced Step 4: Checking the correctness of the balanced equation

Thus, the number of atoms of each element is equal on the two sides. Hence, the equation is balanced. Step 5: Writing the equation with state symbols. We have BaCl2 (aq)  Na 2SO 4 (aq)   BaSO4 (s)  2NaCl (aq)



23

Vidyamandir Classes

Chemical Reactions & Equations (ii)

Step 1: Writing the word equation The word equation for the given reaction is : Sodium hydroxide + Hydrochloric acid  Sodium Chloride + Water Step 2: Writing the skeletal chemical equation NaOH + HCl   NaCl  H 2O Step 3: Balancing of different elements The number of atoms of each element is equal on the two sides, therefore, this equation is already balanced. Step 4: Writing the equation with state symbols, we have NaOH (aq) + HCl (aq)   NaCl (aq)  H 2O ()

Example - 4

Translate the following statements into chemical equation and then balance them :

(i)

Nitrogen gas and hydrogen gas react to form ammonia gas.

(ii)

Hydrogen sulphide gas burns in air to give water and sulphur dioxide.

(iii)

Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

SOLUTION : (i)

Step 1: Writing the chemical equation in the word form Nitrogen + Hydrogen  Ammonia Step 2: Writing the sekeletal chemical equation N2 + H2  NH3 Step 3: Balancing of atoms on both sides. There are 2 N atoms on LHS and 1 N atoms on RHS. To balance N atoms, multiply NH3 on RHS by 2, we get : N2 + H2  2NH3 Now, there are 6 H - atoms on RHS and 2 H atoms on LHS To balance H - atoms, multiply H2 on LHS by 3, We get : N2 + 3H2  2NH3 Step 4: To check the correctness of the balanced equation

(ii)

To translate the given statement into chemical equation The word equation for the given statement will be : Hydrogen sulphide + Oxygen   Water  Sulphur dioxide

 from air  24



Vidyamandir Classes


Hence, the skeletal chemical equation will be H 2S  O2  H 2O  SO2 Balancing of the solve equation Step 1: Writing the skeletal chemical equation H 2S  O2  H 2O + SO2 Step 2: Balancing the element which occurs at minimum number of places. S and H occur in two compounds whereas O occurs in three compounds. We, therefore, first balance S and H. S atoms on LHS = 1, S atoms on RHS = 1 H atoms on LHS = 2, H atoms on RHS = 2 Thus, S and H atoms are already balanced. Step 3: To balance O-atoms. O-atoms on LHS = 2, O-atoms on RHS = 3(1 in H2O, 2 in SO2) To balance O-atoms, multiply O2 on LHS by 3, and on RHS H2O and SO2 by 2 each. We get : H 2S  3O 2   2H 2 O  2SO 2

Step 4: To rebalance H and S atoms. Now. H-atoms in LHS = 2. H-atoms on RHs = 4. To balance H atoms, multiply H2S on LHS by 2. We get : 2H 2S  3O2   2H 2 O  2SO2 Now, S-atoms on LHS = 2, S - atoms on RHS = 2  S-atoms are balanced.

Step 5: To check that the equation is balanced

Thus, the above equation is the required balanced chemical equation. (iii) To translate the given statement into chemical equation. The given word equation is : Potassium + Water  Potassium hydroxide + Hydrogen

 KOH + H 2 In term of formulae, the skeletal chemical equation will be K + H 2O  Balancing of the above equation



25

Vidyamandir Classes


H atoms occur in maximum compounds. Hence, they should be balanced last of all. Balancing of K atoms. Already balanced (1 on each side) Balancing of O atoms. Already balanced (1 on each side) Balancing of H atoms. If we multiply KOH on RHS by 2, number of H-atoms on RHS becomes = 4. To balance H atoms, multiply H2O on LHS by 2. We get : K + 2H 2O   2KOH + H 2 To re-balance K atoms, multiply K on LHS by 2. We get : 2K + 2H 2O   2KOH + H 2 This is required balanced equation. Alternatively, sometimes, in such equations, it is convenient to write the equation first in the atomic form and then make it molecular. This is explained below : Writing the equation in the atomic form, we have K + H 2O   KOH + H In this equation, the number of atoms of each element are equal on the two sides. Hence, it is a balanced chemical equation in the atomic form. To make it molecular, multiply throughout by 2 and write 2 H as H2. We get : 2K + 2H 2O   2KOH + H 2

EXERCISE BASED ON CBSE [SUBJECTIVE] SHORT ANSWER TYPE QUESTIONS 1.

2. 3. 4. 5. 6.

26

Supporse you burn a magnesium ribbon of about 2 cm after cleaning it by sand paper and collecting it in a watch glass. Now answer the following : (i) What is the colour of the flame ? (ii) What is the colour and name of the compound that your have collected in the watch glass ? Why is it necessary to balance an equation ? Name two methods to balance a equation ? How will you know that a chemical reaction has occurred ? What information does an equation give us ? Write the balanced equation for the following chemical reactions :

CBSE Exercise


Vidyamandir Classes


(i)

7.

8. 9. 10.

Hydrogen + Chlorine   Hydrogen Chloride (ii) Barium chloride + Aluminium Sulphate Barium sulphate + Aluminium chloride (iii) Sodium + Water Sodium Hydroxide + Hydrogen A solution of a substance ‘x’ is used for white washing. (i) Name the substance ‘x’ and write its formula. (ii) Write the reaction of the substance ‘x’ named in (a) above wit water. Why does the colour of copper sulphate solution change when an iron nail is dipped in it. Give an example of a double displacement reaction. Identify the substances that are oxidised and the substances that are reduced in the following reactions : (i)

4 Na(s) + O 2 (g)   2 Na 2O(s)

(ii)

CuO(s) + H 2 (g)   Cu(s) + H 2O(l)

LONG TYPE QUESTIONS 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

Distinguish between exothermic and endothermic reactions. Give examples. Why is respiration an exothermic reaction. Explain. Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity. Distinguish between displacement and double displacement reactions. Write equations for these reactions. (a) Why do we apply paint on iron articles. (b) Oil and fat containing food items are flushed with nitrogen Why? Explain the following terms with one example each : (i) Corrosion (ii) Rancidity. What are the drawbacks of the chemical equation? Explain it with an example Illustrate a combination reaction with the help of an experiment. Can a displacement reaction be a redox reaction. Explain with the help of an example. What happens when a piece of (a) zinc metal is added to copper sulphate solution? (b) aluminium metal is added to dilute hydrochloric acid? (c) silver metal is added to copper sulphate solution ? NOW ATTEMPT EXERCISE BASED ON CBSE [OBJECTIVE] IN THIS EBOOK NOW ATTEMPT EXERCISE BASED ON NTSE IN THIS EBOOK

NOW ATTEMPT EXERCISE BASED ON JEE FOUNDATION IN THIS EBOOK


CBSE Exercise

27

Vidyamandir Classes


SOLUTIONS - IN-CHAPTER EXERCISES-A H 2SO4  Zn   ZnSO 4  H 2 

1.

dilute

It is a chemical reaction. 2.

Since ice and water can be interconverted into each other by changing the condition therefore it a physical change.

3.

 Cu(s) + H2O(g) CuO(s) + H2(g) 

(i)

CuO (s) and H2(g)

(ii)

Cu(s) and H2O (g)

(iii)

Cu(s)

(iv)

H2(g), and H2O(g)

4.

 MgCl2 (aq) + CO2(g) + H2O() MgCO3(s) + 2HCl(aq) 

5.

In question (a) the no. of oxygen atoms is not balanced.

6.

(a) (b)

7.

 2Fe(s) + Al2O3(s) 2Al(s) + Fe2O3(s)   2P2O5 P4 + 5O2 

(i)

 2KCl + 3O2 2KClO3 

(ii)

 Al2(SO4)3 + 3H2 2Al + 3H2SO4 

SOLUTIONS - IN-CHAPTER EXERCISES-C A.

Let the oxidation state of the underlined atom be x. (i)

NaH 2 PO 4  1  (2  1)  x  (4) ( 2)  0

(ii)

NaHSO 4  1  1  x  (4) (  2)  0 x   6

(iii)

H 2 P2 O 72 

(2  1)  2x  (7) (  2)  2

x

(iv)

K 2MnO 4

(2  1)  x  (4) (  2)  0

x=6

(v)

CaO2

2  2x  0

x  1

(vi)

NaBH 4

 1  x  (4) (  1)  0

x=3



x 5

10 5 2

(Here the 0.S of H is –1) (vii) 28

H2S2 O7

Solutions

(2  1)  2x  (7) (2)  0

x=6


Vidyamandir Classes (viii)


KAl(SO4 )2 . 12H 2O 1  3  2x  (8) (  2)  0

x=6

(ix)

KI3

1  3x  0

x

(x)

H 2S4O6

(2  1)  (4  x)  (6) ( 2)  0

x

(xi)

Fe3O4

3x  (4) (  2)  0

x

(xii)

C2 H 2OH

2x  (5  1)  (2)  1  0

x 2

(xiii)

C2 H 4 O 2

2x  (4  1)  (2) (  2)  0

x=0

1 3 10 5  4 2

8 3

(The oxidation state calculated here is the average of sthe oxidation states of the individual C atoms) present in the molecule (xiv) N  N x=0 2x  0

B.

C.

(xv)

N 2O5

2x  (5) (  2)  0

x=5

(xvi)

NH 2  NH 2

2x  (4  1)  0

x

(i)

2MnO4  5C2 H 2O 4  6H    2Mn 2   10CO2  8H 2O

(ii)

CrO27   3C2 H 4O  8H    2Cr 3  3C 2 H 4O 2  4H 2O

(iii)

2MnO 4  10Br   16H    2Mn 2   5Br2  8H 2O

(iv)

3Cu  2NO3  8H    3Cu 2   2NO  4H 2O

(v)

Cr2O72   6Fe 2   14H    6Fe3  2Cr3  7H 2O

(vi)

MnO 4  5Fe 2   8H    Mn 2   5Fe3  4H 2O

(i)

P4  3H2O  3NaOH   PH3  3NaH 2 PO2

(ii)

3N 2 H 2  4ClO3   6NO  4Cl  6H 2O

(iii)

Cl2O7  4H 2O 2  2OH    2ClO 2  4O 2  5H 2O

(iv)

4Zn  NO3  7H 2O   4Zn 2   4NH 4  10OH 


4  2 2

Solutions

29

Vidyamandir Classes


SOLUTIONS - IN-CHAPTER EXERCISES-D 1.

Zn(s)  CuSO 4 (aq)  ZnSO4 (aq)  Cu(s)

2.

When Zn removes copper from copper sulphate solution, the blue color of the solution disappears. Refer module

3.

MnO2  2HCl   MnCl2  H2O  Cl2  MnO 2 is oxidising agent & HCl is reducing agent

4. 5.

Refer Module Corrosion of iron leads to weakening of an iron structure. This process is called rusting. A film of reddish iron oxide forms on the surface of the corroded iron. Exposure to moisture present in the atmosphere is sufficient to rust iron. Corrosion is represented by the following chemical reaction: 4 Fe + 3 O2 + x H2O  2Fe2O3.xH2O ; Painting iron prevents it from rusting.

SOLUTIONS - EXERCISE BASED ON CBSE [SUBJECTIVE] EXERCISE SHORT ANSWER TYPE 1.

Magnesium ribbon + oxygen   Magnesium oxide (White powder) (a) It burns with a dazzling white light. (b) The white powder that has been collected on the watch glass is magnesium oxide.

2.

5.

There is no gain or loss of atoms during a chemical reacion. The balancing of chemical equation means to equalise the number of atoms of each element as well as the net charge on both sides of the equation. (i) Oxidation number method (ii) Ion electron method. During this process, chemical substances with inherent properties interact with each other and form com pletely new chemical substances with different properties. The important characteristics of chemical reactions An equation tells us about the reactants, reagents & the product and their states.

6.

(i)

H 2  Cl2   2HCl

(ii)

3BaCl2  Al2 (SO4 )3   2AlCl3  3BaSO4

(iii)

2Na  2H 2O   2NaOH  H 2

7.

(i)

‘x’ is CaO used for white washing.

8.

Fe(s)  CuSO 4 (aq)   FeSO4 (aq)  Cu(s)

3. 4.

Blue

30

Solutions

(ii)

CaO  H 2O   Ca(OH)2

Green


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9.

2NaCl  ZnSO4   Na 2SO4  ZnCl2

10.

(i)

0

1 2

0

(ii)

4 Na(s)  O 2 (g)   2 Na 2 O (s)

2 2

0

0

1 2

Cu O 2 (s)  H 2 (g)   Cu  H 2 O ( l)

oxidised   Na

Oxidised   H2

Reduced   O2

Re duced   CuO2

EXERCISE LONG ANSWER TYPE 11.

Exothermic Reaction : A chemical reaction which produces (releases) energy in the form of heat and light is called exothermic reaction. Examples : 2Al (s) + Fe2O3(s)   Al2O3 (s) + 2Fe (l) + Energy (heat) Endothermic Reactions : A chemical reaction which absorbs energy in the form of heat and light is called endothermic reaction. Example : Ba(OH)2 . 8 H2O(s) + 2NH4SCN(s)   Ba(SCN)2 (aq) + 2NH3 (aq) + 10 H2O (l)

12.

Respiration is a metabolic process by which an organism obtains energy (in form of ATP) by oxidizing nutrients and releasing waste products.

13.

CaCO3 (s)   CaO(s)  CO 2 (g) ; AgCl  2Ag  Cl2 ; 2H 2O   2H 2  O2

14.

2Na(s)  ZnCl2 (aq)   2NaCl(aq)  Zn(s) . It is an example of displacement reaction



sunlight

electricity

2AgNO3  ZnCl2   2AgCl  Zn(NO3 )2 15. 16. 17.

It is an example of double displacement reaction. (a) To prevent them from rusting. (b) Refer Module

To prevent rancidity

A chemical equation does not provide information about the spontaneity & the rate of reaction or how fast the reaction is occuring.

Cu  ZnSO4   CuSO4  Zn is not a spontaneous reaction whereas Zn  CuSO4   ZnSO 4  Cu is a spontaneous reaction. 18.

Refer Module

19.

Yes.

0

2

2

0

Zn  Cu SO4   Zn SO 4  Cu

It is a displacement as well as a redox reaction.


Solutions

31

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Chemical Reactions & Equations 20.

(a)

Zn  CuSO4   ZnSO4  Cu

(b)

2Al  6HCl(dil.)   2AlCl3  3H2 (g)

(c)

Ag(s)  CuSO 4 (aq)   No reaction

ANSWERS TO IN-CHAPTER EXERCISES 1. Chemical reaction

A

2. Physical change

(i) CuO (s) and H2(g) (ii) Cu(s) and H2O (g) (iii) Cu(s) (iv) CuO(s) H2(g), and H2O(g)  MgCl2 (aq) + CO2(g) + H2O() 4. MgCO3(s) + 2HCl(aq)  5. (b)

 2Fe(s) + Al2O3(s) 6.(a) 2Al(s) + Fe2O3(s) 

 2KCl + 3O2 7.(i) 2KClO3 

32

 Cu(s) + H2O(g) 3. CuO(s) + H2(g) 

Answers

 2P2O5 (b) P4 + 5O2 

 Al2(SO4)3 + 3H2 (ii) 2Al + 3H2SO4 


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My Chapter Notes


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Illustration - 1


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