Chemistry FBISE SSC-1 Keybook Chapter 7

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Example 7.1

 

Following reaction occurs when you burn Sui gas

            CH4 + 2O2       →        CO2 +2H2O + heat

            Identify the element undergoing oxidation:

Solution:

Since C in CH4 losses H-atoms and combines with oxygen atoms, thus C atoms undergo oxidation. At the same time O-atoms combine with H-atoms to form H2O, thus O-atoms undergo reduction.

 

 

Identify elements undergoing oxidation and reduction in the following reactions.

  1. i) N2 + 3H2 →        2NH3

N-atoms undergo reduction because they gain hydrogen.

H-atoms undergo oxidation.

 

  1. ii) 2H2 + O2 →        2H2O

H-atoms undergo oxidation because they gains oxygen.

O-atoms undergo reduction because they losses oxygen.

 

iii)        Fe2O3+ 3CO    →        2Fe + 3CO2

C-atoms undergo oxidation because they gains oxygen.

Fe-atoms undergo reduction because they losses oxygen.

 

  1. iv) 4 Al + 3O2 →        2Al2O3

Al-atoms under oxidation because they gains oxygen.

O-atoms undergo reduction.

 

Example 7.2

 

In the following reaction identify which element is oxidized and which element is being reduced.

 

2Ca + O2         →        2CaO

Ca belongs to Group IIA and loses two electrons, so it undergoes oxidation.

O belongs to Group VIA and gains two electrons, so it undergoes reduction.

 

In the following reactions, identify which element is oxidized and which element is reduced?

 

Mg + S            →        MgS

Mg belongs to Group IIA and loses two electrons, so it undergo oxidation. S belongs to Group VI A and gains two electrons, so it undergo reduction.

 

4Na + O2         →        2Na2O

Na belongs to Group IA and loses one electron, so it undergo oxidation.

O belongs to Group VIA and gain two electrons, so it undergo reduction.

 

2Al + 3Cl2       →        2AlCl3

Al belongs to Group III A and loses three electrons, so it undergo oxidation. Cl belongs to Group VII A and gains one electron, so it undergo reduction.

 

Mg + H2          →        MgH2

Mg belongs to Group IIA and loses two electron, so it undergo oxidation. H belongs to Group IA and gain one electron so undergo reduction.

 

Process leading to oxidation and reduction

OxidationReduction
Gain of oxygenLoss of oxygen
Loss of hydrogenGain of hydrogen
Loss of electronsGain of electrons
Increase in oxidation numberDecrease in oxidation number

 

Oxidation states of some elements in binary compounds that rarely change

ElementsOxidation state
Group – IA+1
Group – II A+2
Group – III A+3
H+1 (except in metal hydrides where it is-1)
Group-VII A-1
O-2 (except peroxides and in OF2)

 

Example 7.3

A device called Breath Alyzer is used by police to test a person’s breath for alcohol. It contains an acidic solution of potassium dichromate K2Cr2O7. It is strong oxidizing agent. Determine oxidation state of Cr in it.

Solution:

The sum of oxidation numbers must be zero

K2Cr2O7

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

2 + 2x – 14 = 0

2x-12 = 0

2x=12

x=6

The oxidation state for C in K2Cr2O7 is +6.

 

Example 7.4

Boric acid H3BO3 is used in eye wash. What is the oxidation state of B in this acid?

Solution:

The sum of oxidation numbers must be zero

H3BO3

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

3+x-6=0

x-3=0

x=3

Thus the oxidation state for B in H3BO3 is +3

 

One major problem of air pollution is the formation of acid rain. Air pollutants such as SO2 and NO2 combine with oxygen and water vapours in the air to form H2SO4 and HNO3. These acids fall to the ground with the rain, making the rain acidic. Clouds can also absorb the acids and carry them hundreds of kilometer away from where the pollutants are released. Determine the oxidation number of N is NO2 and HNO3, S is SO2 and H2SO4.

Solution:

The sum of oxidation number must be zero

Oxidation number of N in NO2:

x + 2 (-2) = 0

x – 4 = 0

x = 4

Thus the oxidation state for N in NO2 is+4

            Oxidation number of N in HNO3

1 + x + 3(-2) = 0

1 + x – 6 = 0

x – 5 = 0

x = 5

Thus the oxidation state for N is HNO3 is +5.

 

Oxidation number of S in SO2:

X + 2 (-2) = 0

X – 4 = 0

X = 4

Thus oxidation state for S in SO2 is +4

 

Oxidation number of S in H2SO4:

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

2 + x – 8 = 0

X – 6 = 0

X = 6

The oxidation state for S in H2SO4 is +6

 

  1. Consider the following reaction that takes place in the manufacture of steel.

            Fe2O3 + 3CO   →        2Fe + 3CO2

                Identify the oxidizing and reducing agents.

 

Solution:

First assign oxidation numbers to each atom.

2(+3) 3(-2)     +2 -2                    0         +4 2(-2)

Fe2O3 + 3CO   →        2Fe + 3CO2

The oxidation number of Fe decreases, Fe is reduced. So Fe2O3 is oxidizing agent.

The oxidation number of C increases, C is oxidized. So CO is reducing agent.

 

Example 7.5

Tungsten is used to make filaments for electric bulbs because it has the highest milting point and high electrical resistance. This metal is obtained from tungsten (VI) oxide, WO3 by reducing it with hydrogen gas.

            WO3 + 3H2      →        W + 3H2O.

Identify the oxidizing and reducing agents in this reaction.

 

Solution:

First assign oxidation number to each atom.

+ 3                 → + 3

Because the oxidation number of W decreases, so WO3 is an oxidizing agent. Similarly the oxidation number of H increases, therefore H2 is reducing agent.

 

  1. The torch cell discharges electricity because of an oxidation reduction reaction that takes place between zinc and manganese dioxide.

 

Zn + 2MnO2 + H2O    →        Zn (OH) 2 + Mn2O3

Identify the oxidizing and reducing agents in this reaction.

 

Solution:

First assign oxidation number to each atom.

+2+→+

The Zn is going from an oxidation state of 0 to +2 and so is being oxidized. Because it is being oxidized, it is the reducing agent.

Mn is going from +4 in MnO2 to +3 in Mn2O3, and so it is being reduced. Because it is being reduced, Mn is the oxidizing agent.

 

  1. Identify oxidizing and reducing agents in the following reactions.
  • 2S + Cl2 →        S2 Cl2

First assign oxidation number to each atom.

2 +           →

Oxidation number of Cl is decreased, Cl is reduced so Cl2 is oxidizing agent. Oxidation number of S is increased, S is oxidized. So S is reducing agent.

 

  • 2Na + Br2 →        2NaBr

First assign oxidation number to each atom.

2 +       →        2

Oxidation number of Br is decreased, Br is reduced, so Br2 is oxidizing agent.

Oxidation number of Na is increased, Na is oxidized, So Na is reducing agent.

 

  • H2 + S → H2S

First assign oxidation number to each atom.

+ →

Oxidation number of S is decreased, S is reduced. So S is oxidizing agent. Oxidation number of H is increased, H is oxidized. So H is reducing agent.

 

  1. Define Oxidation-reduction reaction?

Ans.                 The reaction which involved transfer of electron is known as oxidation-reduction reaction. These reactions are also called redox reactions. Such reactions are commercially very important. Most of the metals are recovered from their ores by redox reaction.

 

  1. What is electrolyte and non-electrolyte? Also give examples

Ans.     Electrolyte:

A substance that conducts electricity when it is dissolved in water or in the molten state is called electrolyte.

Example: NaCl, KCl, HCl, NaOH etc.

Non-Electrolyte:

A substance that cannot conduct electricity when dissolved in water or in the molten state is called non-electrolyte.

Example: Urea, glucose, sucrose, benzene etc.

 

  1. What are spontaneous and non-spontaneous processes? Give examples

Ans.     Spontaneous process:

A physical or chemical change that occurs by itself is called a spontaneous process. It does not require source of energy to make them happen.

Example:

  1. Water flows from higher level to lower level.
  2. Iron placed in moist air, rusts.

Non-spontaneous process:

A physical or chemical change that requires a source of energy to make them happen is called non-spontaneous process.

Example:

Water flows from lower to higher level by using a pump.

 

  1. Write a note on electrolytic cells?

Ans.     An electrochemical cell in which electrical energy is used to drive a chemical reaction is called an electrolytic cell.

 

An electrolytic cell consists of

  1. A vessel containing an electrolyte (MX)
  2. Two inert electrodes
  3. A battery

Electrons move from anode to cathode in the outer circuit, in the solution the cations move towards cathode and anions towards anode. At anode anions oxidize by loosing electrons. At cathode cations reduce by gaining electrons. This means oxidation occurs at anode and reduction at cathode.

At anode:

→        X + 1ē

At Cathode:

+ 1ē   →     M

 

  • Sketch an electrolytic cell for the electrolysis of fused KCl.

 

 

 

  • Electrolytic cell shown in figure is used for the electrolysis of fused sodium chloride. Indicate the direction of flow of electrons. Identify anode and cathode.

 

 

 

 

 

  1. Write a note on Daniel Cells?

Ans.     The cell which involves spontaneous redox reaction to generates electricity is called a galvanic or voltaic cell. The name Voltaic is given to this cell because Alessandro Volta discovered first such cell. The English chemist Fredrick Daniel constructed first voltaic cell using zinc (Zn) and copper (Cu) electrodes. Therefore this-cell is named as Daniel Cell.

 

A galvanic cell consists of the following parts:

  1. A zinc bar dipped into a 1M ZnSO4
  2. A copper bar dipped into a 1M CuSO4
  3. A salt bridge which is inverted U-tube containing an inert electrolyte such as KCl. Its ions do not react with the electrodes or with the ions in solution. It makes the electrical contacts between the solutions through which ions can move.
  4. A voltmeter to measure current.

 

Each compartment of cell is called a half cell. Thus a Daniel Cell consists of two half-cell joined in series. When circuit is complete, electrons begin to flow from Zn rod through the external wires to Cu rod. Thus Zn half-cell acts as anode and Cu half-cell as cathode. Note that a half-cell consists of a metal rod dipped in the solution of its salt.

 

Sketch a voltaic cell labeling the cathode; the anode and the direction of flow of the electrons. Use the following chemicals.

Silver, Zinc, Silver, Nitrate (AgNO3) and Zinc Sulphate (ZnSO4)

(Hint: Zinc is more active than Ag)

 

 

  1. Write reactions in a Daniel Cell?

Ans.     In Daniel cell, electrons flow from Zn rod. through the external wire to Cu rod. This is because Zn has more tendency to undergo oxidation than Cu. Zn atoms from the rod go into the solution as Zn+2 ions leaving electrons on the rod. These electrons flow in the external circuit. Thus, oxidation half reaction occurs at anode compartment. Cu+2 ions in copper sulphate solution capture electrons from Cu electrode and are reduced. Reduction half reaction occurs at the cathode compartment. Such oxidation and reduction reactions are called half-cell reactions.

At anode (Oxidation half reaction):

Zn(S)     → Zn+2 (aq) + 2ē

At cathode(Reduction half reaction):

Cu+2(aq) + 2ē → Cu(s)

 

Identify the half-cell in which oxidation occurs and the half cell in which reduction occurs in the following voltaic cell.

 

 

Following reactions occur in the cell

At anode:        Zn(S)   →        Zn+2 (aq) + 2ē

At cathode:     H2(g) + 2ē →   2H+ (aq)

Solution:

Oxidation half reaction occur at anode.

Reduction half reaction occur at cathode.

 

  1. Write a note on Dry Cell?

Ans.     The dry cell batteries are used to power many flashlights, toys and small appliances. The anode is the zinc metal of the container and the cathode is an inert graphite rod at the center of the container in contact with a mixture of MnO2 and carbon (charcoal).

The electrolyte is a mixture of moist NH4CI and ZnCl2. Following reactions take place in it.

 

At Anode:

Zn        →        Zn+2 + 2e

 

At Cathode:

2NH4+ + 2MnO2 +2e  → Mn2O3 + 2NH3 + H2O
This cell produces a potential of 1.5 V.

 

  1. What are functions of electrochemical industries?

Ans.     Functions of Electrochemical Industries:

Electrical energy is extensively consumed by the chemical process industries.

Electrochemical industries use electricity to bring chemical change and produce wide variety of substances. Such industries are called electrolytic industries. These industries produce substances that cannot be made economically in another way.

 

 

  1. Explain the manufacturing of sodium metal from fused sodium chloride?

Ans.     On the large scale sodium metal is produced by the electrolysis of fused NaCl. An electrolytic cell called Down’s Cell is used for this purpose as shown below.

 

The electrodes are iron cathode and graphite anode. Chlorine is obtained as by-product. In molten sodium chloride (NaCl), Na+ ions are free to move about. Under the influence of electric current, Na+ ions move towards the cathode and Cl ions towards the anode.

At Anode:       2Cl     →        Cl2 (g) +2e–           (oxidation)

At Cathode:    2Na+ + 2e       →        2Na      (reduction)

 

Molten sodium is collected into a sodium collecting ring,       from where it is periodically drained. Whereas, chlorine gas in collected into the funnel at the top of the cell.

 

  1. Explain the manufacturing of sodium hydroxide from Brine?

 

Ans.     Electrolysis of brine, a concentrated aqueous solution of sodium chloride is used for the industrial production of sodium hydroxide. Electrolysis of brine produces simultaneously three important industrial chemicals, chlorine gas, hydrogen gas and sodium hydroxide. The electrolytic cell called Nelson’s Cell

 

 

Working of Nelson’s Cell

Aqueous solution of sodium chloride consists of Na+, CI, H+ and OH ions. These ions move towards their respective electrodes and redox reactions take place at these electrodes. When electrolysis takes place Clions are discharged at anode and CI2 gas rises into the dome at the top of the cell. The H+ ions are discharged at cathode and H2 gas escapes through a pipe. The sodium hydroxide solution slowly percolates into a catch basin.

At Anode:

2Cl (aq) →        Cl2 (g) +2e

At Cathode:

                        2H2O(l) +2e→  H2 (g) +2OH(aq)

Overall reaction:

2Cl (aq)+ 2H2O(l) → Cl2 (g)+ H2 (g) +2OH(aq)

 

  1. Define electroplating. Write good conditions for electroplating.

Ans.     Electrolytic process used to deposit one metal on another metal is called electroplating.

Conditions:

  1. High current density.
  2. Low temperature
  • High concentration of metal in its electrolyte.

 

  1. Explain Zinc Plating on steel

Ans.     Zinc plating on steel is done by using zinc metal as anode. A solution of potassium zinc cyanide K2[Zn(CN)4] containing little sodium cyanide. The steel object is made cathode. During the electrolysis zinc at the anode enters the solution as Zn+2 ions, which are deposited at the cathode. The electrolyte ionizes as follows.

K2[Zn(CN)4] (aq) ⇋      2K+(aq) + [Zn(CN)4]-2(aq)

                                [Zn(CN)4]-2(aq) ⇋           Zn+2 +4CN(aq)

Following reactions occur at the electrodes:

At anode:              Zn(s)     →     Zn+2(aq) +2e-1

At cathode:        Zn+2(aq) +2e-1 →          Zn(s)

Sodium cyanide prevents the hydrolysis of the electrolyte.

 

  1. What do you know about tin plating on steel?

Ans.     Tin plating on steel is done by using tin as anode and a solution of stannous sulphate, (SnSO4) as electrolyte. Few drops of dil. H2SO4are added in the electrolyte to prevent its hydrolysis. The electrolyte ionizes as follows.

SnSO4 (aq)         ⇋      Sn+2(aq)+ (aq)

During the electrolysis following reactions occur:

At anode:              Sn(s)     →      Sn+2(aq) +2e-1

At cathode:        Sn+2(aq) +2e-1 →          Sn(s)

 

 

  1. Explain Chromium plating on steel.

Ans.     Since chromium metal does not adhere strongly to the steel therefore steel is first plated with copper or nickel and then chromium. For electroplating chromium, chromium metal is used as anode and chromium sulphate, Cr2(SO4)3 as an electrolyte. A few drops of dil. H2SO4 are added in the electrolyte to prevent its hydrolysis. The electrolyte ionizes as follows:

Cr2(SO4)3 (aq)     ⇋      2Cr+3(aq)+ (aq)

During the electrolysis following reactions occur:

At anode:              Cr(s)     →      Cr+3(aq) +3e-1

At cathode:        Cr+3(aq) +3e-1 →           Cr(s)

 

  1. Define corrosion?

Ans.     Corrosion is the process in which a metal reacts with oxygen and moisture in the atmosphere.

 

 

  1. Write ways of prevention of corrosion?

Ans.     Prevention of Corrosion

Prevention of corrosion is an important way of conserving our natural resources. Following methods have been devised to protect metals from corrosion:

  1. Coating with paint:
    Corrosion can be prevented by painting the metal, so that it does not come in contact with oxygen and moisture and other harmful agents.  Paint is cheap and easily applied. Paint is used to protect many everyday steel objects such as cars, trucks, trains, bikes, bridges etc. Paint also provides visual appeal.
  2. Alloying:
    The tendency of iron to oxidize can be greatly reduced by alloying it with other metals. For example, stainless steel is an alloy of iron chromium and nickel. It is protected from corrosion by an outer layer of Cr2O3.
  3. Coating with a thin layer of another metal:
    Metals that readily corrode can be protected by coating with a thin layer of another metal that resists corrosion. This can be done by:
  • Tinning
  • Galvanizing
  • Electroplating

 

(a) Tinning:

In the process of tin plating, clean iron sheet is dipped in a bath of molten tin. It is then passed through hot pair of rollers. Tin protects iron effectively, since, it is very stable.

(b) Galvanizing (Coating with Zinc):

The process of galvanizing consists of dipping a clean iron sheet in a hot zinc chloride bath and heating. After this sheet is rolled into zinc bath and cooled.

(c) Electroplating:

In electroplating an electrolytic process is used to deposit one metal on another metal.

 

  1. Cathodic Protection:

Cathodic protection is the process in which the metal that is to be protected from corrosion is made cathode and is connected to metals such as magnesium or aluminum. These metals are more active than iron, so they act as anode and iron as cathode. The more active metals themselves oxidize and save iron from corrosion. Cathodic protection is employed to prevent iron and steel structures such as pipes, tanks, oil rigs etc in the moist underground and marine environment.

 

Review Questions

  1. Encircle the correct answer.
  • In which of the following changes the nitrogen atom is reduced?
  1. N2 to NO
  2. N2 to NO2
  3. N2 to NH3
  4. N2 to HNO3
  • Which of the following changes reaction is an example of oxidation?
  1. Chlorine molecule to chloride ion
  2. Silver atoms to silver (I) ion
  3. Oxygen molecule to oxide ion
  4. Iron (III) ion to iron(II) ion
  • Which of the following elements in the given reaction is reduced?

ZnO + H2                                           Zn + H2O

  1. H2
  2. ZnO
  3. Zn
  4. O
  • Consider the following reaction:

H2S   +Cl2         2HCl +   S

In this reaction what does H2S behave as?

  1. Reducing agent
  2. Oxidizing agent
  3. Catalyst
  4. Electrolyte
  • The oxidation state of Cr in K2Cr2O7 is
  1. 12
  2. 6
  3. 3
  4. Zero
  • Which of the following statement is not correct about the galvanic cell,
  1. Cations are reduced at cathode
  2. Anions arc oxidized at anode
  3. Electrons flow from cathode to anode
  4. Oxidation occurs at anode
  • Which of the following is not true about the Daniel cell?
  1. Half-cell of an active metal acts as cathode.
  2. Half-cell contains an element in contact with its ions in aqueous solution.
  3. A salt bridge connects the two half cells.
  4. A spontaneous oxidation-reduction reaction generates electricity.
  • Which of the following do not involve electrolytic process?
  1. Refining of copper
  2. Manufacture of sodium from NaCl
  3. Electroplating of steel
  4. Reduction of metal oxide by a reducing agent.
  • Galvanizing is
  1. Coating with Sn
  2. Coating with Zn
  3. Coating with Cr
  4. Coating with Cu
  • Which of the following is true for the Nelson cell ?
  1. Sodium metal is produced at anode
  2. Chlorine gas is produced at anode
  3. Hydrogen gas is produced at anode
  4. Sodium ions are not reduced at cathode

Answers:

i) Cii) Biii) Biv) Av) B
vi) Cvii) Aviii) Dix) Bx) B

 

  1. Give short answers

(i)         What is oxidation state?

Ans.     Oxidation state or oxidation number is defined as the number of apparent charges that an atom will have in the molecule.

The elements that show an increase in oxidation number are oxidized. The elements that show a decrease in oxidation number are reduced.

(ii)        What is the oxidation number of Cr in chromic acid (H2CrO4)?

The sum of oxidation state of all the atoms is zero.

2×1 + x + 4 (-2)           = 0

2 + x -8 = 0

x – 6 = 0

x = 6

(iii)       Identify reducing agent in the following reaction.

CuO + H2            Cu + H2O

Ans.     First assign oxidation number to each atom.

+2 -2       0              0         2(+1) -2

CuO + H2     Cu + H2O

The oxidation number of Cu decreases, Cu is reduced so CuO3 is an oxidizing agent. Similarly the oxidation number of H increases, H is oxidized. Therefore H2 is reducing agent.

(iv)       Write chemical reactions that occur in Nelson’s cell.

Following reactions occur in the Nelson’s cell:

At anode:

2Cl(aq)               →        Cl2(g) + 2ē

At cathode:

2H2O (l) + 2ē    →        H2(g) + 2OH (aq)

Overall reaction:

2Cl(aq) + 2H2O (l) → Cl2(g) + H2(g) + 2OH (aq)

 

(v)        Why tin plated steel is used to make food cans?

Ans.     Tin plated steel is used to make food cans. This is because the components of food beverages and the preservatives contain organic acids or their salts. They may form toxic substances by reacting with iron. These acids and salts are corrosive. Tin plating is non-poisonous and prevents corrosion.

(vi)       Explain one example from daily life which involves oxidation-reduction?

Ans.     Examples from daily life which involves oxidation-reduction is rusting of iron. Oxygen and water are necessary for iron to rust. A region of metal surface that has relatively less moisture, act as anode.

Fe (s)                 →        Fe+2 (aq) + 2ē

Another region on the surface of metal that has relatively more moisture act as cathode. The electrons released in the oxidation process reduce atmospheric oxygen to hydroxyl ions.

O2 + 2H2O +4ē                         →        4OH

The Fe+2 ions formed at the anodic regions flow to the cathodic regions through the moisture on the surface. Fe+2 ions further react with oxygen to form rust, Fe2O3. xH2O

Q.3      Define oxidation and reduction in terms of loss or gain of oxygen or hydrogen.

Ans.     Oxidation:

Oxidation is defined as the gain of oxygen atoms or loss of hydrogen atoms by an element.

Reduction:

Reduction is defined as the loss of oxygen atoms or gain of hydrogen atoms by an element.

 

Q.4      Define oxidation and reduction in terms of loss or gain of electrons.

Ans.     Oxidation:

The process that involves the loss of electrons by an element is called oxidation.

Reduction:

A process that involves the gain of electrons by a substance is called reduction.

 

Q.5      List the possible uses of electrolytic cell.

Ans.     Possible uses of electrolytic cells are as follows:

  • Down’s cell is used for the commercial preparation of sodium metal. It produces chlorine gas as by product.
  • Nelson’s Cell is used for the commercial preparation of sodium hydroxide. It also produces chlorine and hydrogen gas as by product.
  • Electrolytic cells are used for the commercial preparation of calcium and magnesium metals.
  • It is used to produce aluminum metal commercially.
  • It is used for the purification of copper.
  • Electrolytic cells are used to electroplate metals such as tin, silver, nickel etc on steel.
  • Electrolytic cells are used to prepare anodized aluminum.

Q.6      Sketch a Daniel Cell, labeling the cathode , anode and the direction of flow of electrons

 

Q.7      Describe how a battery produces electrical energy?

Ans.     Battery produces electric current by a redox reaction. When connected in a circuit, its anode oxidizes by releasing electrons. These electrons through the external circuit begin to flow towards the cathode. At cathode these electrons reduce oxidizing agent present in the electrolyte. Examples of batteries are dry cell, storage cell, mercury battery etc.

 

Q.8      Describe the method of recovering metal from is ores.

Ans.     The process of producing a metal from its ores, always involves oxidation-reduction reaction. Most of the metals are found in nature as oxides or sulphide ores. After mining the ore, desired mineral is of the metals are found in nature as oxides or sulphide ores. After mining the ore, desired mineral is separated from the other materials. Purified metal oxides are reduced to free metals by using a reducing agent.

Reducing Agents:

Aluminum, coke, carbon monoxide gas and hydrogen as are used as reducing agents.

Examples:

Extraction of iron:

Extraction of iron involves the chemical reduction of Hematite (Fe2O3) ore by coke (carbon)

4C + 3O2          →        2CO + 2CO2

These reactions are highly exothermic. As hot CO rises, it reacts with iron oxide and reduces it to iron. Molten iron collects at the bottom of the furnace. Lime stone removes impurities from iron as slag.

Fe2O3 + 3CO    →        2Fe + 3CO2

Extraction of lead and zinc:-

Lead and Zinc metals occur naturally as sulphide ores. These ores first converted into corresponding oxide by heating in oxygen.

2ZnS2 + O2       →        2ZnO + 4S

2PbS2 + O2       →        2PbO + 4S

In this process sulphur is oxidized and oxygen is reduced. These oxides are then

ZnO + C           →        Zn+ CO

PbO + CO        →        Pb + CO2

Q.9.     Explain electrolytic refining of copper.

Ans.     The copper metal obtained from its ores is usually impure. It contains impurities such as zinc, iron, silver and gold. These impurities are removed by the process of electrolysis. In this process impure copper bars act as anode and pure copper bars at cathode. CuSo4 solution containing little sulphuric acid is used as the electrolyte.

 

On passing electricity, copper anode dissolves forming Cu+2 ions. Cations move towards the cathode at which only Cu+2 ions are reduced. Thus pure copper deposits at cathode. The less electropositive metals, silver and gold fall to the bottom of the cell. Copper obtained in this process is 99.5%. Following reaction occur in this process.

At anode:

Cu(s)                 →        Cu+2 (aq) +2ē

At cathode:

Cu+2 (aq) + 2 ē   →        Cu(S)

 

Q.10    Compare the effects of Al2O3 and Fe2O3 formation on their parent metals and cite examples from daily life.

Ans.                 Aluminum has a great tendency to corrosion. However, aluminum corrosion is aluminum oxide (Al2O3) a very hard material that actually protects the aluminum from further corrosion. Aluminum oxide corrosion also looks a lot more like aluminum, so it is not as easy to notice as rusted iron.

When iron corrodes the color changes and it actually expands. This expanding and color change can produce large red flakes that we all know as rust. Unlike aluminum oxide the expanding and flaking of rust in iron exposes new metal to further rusting. That is why it is so important to provide a barrier to stop rust.

 

Q.11    Explain how food and beverage industries deal with corrosion.

Ans.     Tin plated steel is used to make cans. Food and beverages industries use tinplated steel cans. This is because the components of food beverages and the preservatives contain organic acids or their salts. They may form toxic substances by reacting with iron. These acids and salts are corrosive. Tin plating is non-poisonous and prevents corrosion.

 

Q.12    Explain how chemistry interacts with photography.

Ans.     A photographic film is basically an emulsion of silver bromide, (AgBr) in gelatin. When the film is exposed to light, Silver bromide granules become activated. This activation depends on the intensity of the light falling upon them. When exposed film is placed in the developer solution that is actually a reducing agent. Hydroquinone which is a mild reducing agent is used as developer. In hydroquinone the activated granules of silver bromide are reduced to black metallic silver. Reduced silver atoms form image.

Ag+ +1e → Ag (s)

Inactivated silver bromide is removed from the film by using a solvent called a fixer. Sodium thiosulphate is used for this purpose. The areas of the film exposed to the light appear darkest because they have the highest concentration of metallic Silver. Thus photography involves oxidation-reduction reaction.

Q.13    Electrolysis has major role in electrochemical industries.

(a)        Sketch an electrolytic cell, label the anode and cathode and indicate the direction of electron transfer.

Ans.     Sketch of an electrolytic cell.

 

(b)        Describe the nature of electrochemical process.

Ans.     Electrochemical processes are oxidation-reduction reactions in which chemical energy released by a spontaneous reaction is converted to electricity or in which electrical energy is used to drive a non-spontaneous reaction. Whether an electrochemical process releases or requires energy, it always involves the transfer of electrons from one substance to another. This means that this process always involves an oxidation-reduction or a redox reaction.

 

(c)        Distinguish between electrolytic and voltaic cell.

Ans.     Difference between electrolytic and voltaic cell.

 Electrolytic CellGalvanic Cell
1.It consists of one complete cell connected to a batteryIt consist of two half cells connected through a salt bridge.
2.Anode has positive charge while cathode has negative chargeAnode has negative charge while cathode has positive charge.
3.Electrical energy is converted into chemical energy.Chemical energy is converted into electrical energy.
4.Current is used for a non-spontaneous chemical reaction.Redox reaction takes place spontaneously and produces electric current.
5.Example: Down’s cell, Nelson’s cellExample: Daniel’s cell

 

Q.14    State the substances which are oxidized or reduced. Give reason for your answer.

(a)        N2 + 3H2          →        NH3

0        0                                  -3 +3

Ans.                 N2 + 3H2          →        NH3

There is a decrease in oxidation state of N. Therefore N atoms undergo reduction.

There is a increase in oxidation state of H. Therefore H-atoms undergo oxidation.

 

(b)                    CO2 + 2Mg     →        2MgO + C

Ans.                +4 2(-2)   0                                 +2   -2       0

CO2 + 2Mg       →        2MgO + C

 

There is a decrease in oxidation state of O. Therefore O-atoms undergo reduction.

 

There is a increase in oxidation state of Mg. Therefore Mg-atom undergo oxidation.

 

(c)        Mg + H2O       →        MgO + H2

Ans.     + →        +

 

There is a decrease in oxidation state of H. Therefore H-atoms undergo reduction.

There is an increase in oxidation state of Mg. Therefore Mg-atoms undergo oxidation.

(d)        H2S + Cl2         →        2HCl + S

Ans.     + →         2 +

 

There is a decrease in oxidation state of Cl. Therefore Cl-atoms undergo reduction.

There is an increase in oxidation state of S. Therefore S-atoms undergo oxidation.

 

(e)        2NH3 + 3CuO →       3Cu + N2 + 3H2O

Ans.       2 + 3 → 3+ +

There is a decrease in oxidation state of Cu. Therefore Cu-atoms undergo reduction.

There is an increase in oxidation state of N. Therefore N-atoms undergo oxidation.

 

Q.15(a)            Define oxidation number or oxidation state.

Ans.     Oxidation number or oxidation state is defined as the number of apparent charges that an atom will have in a molecule.

(b)        Find the oxidation state of Nitrogen in the following compounds.

(i)         NO2

The sum of oxidation state of all the atoms is zero

x + 2 (-2) = 0

x – 4 = 0

x -4            = 0

x    = 4

(ii)        N2O

The sum of oxidation state of all the atoms is zero

2x + (-2) = 0

2x – 2 = 0

2x  = 2

2x  = 4

2       2

X   = 1

(iii)       N2O3

The sum of oxidation state of all the atoms is zero

2x + 3 (-2) = 0

2x –6 = 0

2x  = 6

2x  = 6

2         2

X   = 3

(iv)       HNO3

The sum of oxidation state of all the atoms is zero

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

1 + x – 6 = 0

x -5            = 0

x    = 5

 

Q.16    Find the oxidation state of S in the following compound

(a)        H2S

The sum of oxidation state of all the atoms is zero

2(+1) x   = 0

2 + x = 0

x    = -2

Oxidation number for N in H2S is -2.

(b)        H2SO3

The sum of oxidation state of all the atoms is zero.

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

2 + x – 6 = 0

X – 4 = 4

Oxidation number for S in H2SO3 is +4.

(c)        Na2S2O3

The sum of oxidation state of all the atoms is zero

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

2 + 2x – 6 = 0

2x -4   = 0

2x       = 4

2x      = 4

2             2

X   = 2

Oxidation number for S in Na2S2O3 is +2

 

Q.17    Define oxidizing and reducing agents.

(a)        Oxidizing agents:

An oxidizing agent is the reactant containing the element that is reduced (gains electrons) in a chemical reaction.

Reducing agents:

A reducing agent is the reactant containing the element that is oxidized (loses electrons) in the chemical reaction.

 

(b)        Identify the oxidizing agents and reducing agents in the following reactions.

(i)         H2S + Cl2   →  2HCl + S

Ans.     First assign oxidation number to each atom.

2(+1) -2     0                        +1 -1       0

H2S + Cl2   →        2HCl + S

The oxidation number of Cl decreases, Cl is reduced so Cl2 is an oxidizing agent.

Similarly the oxidation number of S increases, S is oxidized, therefore S is reducing agent.

 

  1. ii) 2 FeCl2 + Cl2 →        2FeCl3

Ans.     First assign oxidation number to each atom

+2 2(-1)       0                       +3 3(-1)

2 FeCl2 + Cl2    →        2FeCl3

The oxidation number of Cl decreases, Cl is reduced so Cl2 is an oxidizing agent. Similarly the oxidation number of Fe increases, Fe is therefore oxidized. FeCl2 is reducing agent.

 

(iii)       2KI + Cl2        →        2KCI + I2

Ans.     First assign oxidation number to each atom.

2 + → 2 +

The oxidation number of Cl decreases, Cl is reduced. So Cl2 is an oxidizing agent. Similarly the oxidation number of I increases, I is oxidized. Therefore KI is reducing agent.

 

(iv)       Mg + 2HCl       →        MgCl2 + H2

First assign oxidation number to each atom.

+ 2     →        +

The oxidation number of Cl decreases, Cl is reduced. So Cl2 is oxidizing agent.

Similarly the oxidation number of Mg increases, Mg is oxidized. Therefore Mg is reducing agent.

 

Q.8      Hydrogen peroxide reacts with silver oxide and lead (II) sulphide according to the following equations.

(i)         H2O2 + Ag2O  →        2Ag + H2O + O2

(ii)        4H2O2 + PbS   →        PbSO4 + 4H2O

Is hydrogen peroxide an oxidizing or reducing agent in these reactions? Give your reasons.

Ans.     2(+1) 2(-1) 2(+1) -2                                                0         +2 -2       0

  1. i)               H2O2 + Ag2O → 2Ag + H2O + O2

Hydrogen peroxide is reducing agent. Because the oxidation number of Ag decreases, Ag is reduced. So Ag2O is oxidizing agent.

Similarly the H is oxidized. Therefore H2O2 is reducing agent.

  1. ii) 2(+2) 2(-2)   +2 -2 +2 +6 4(-2)     +2   -2

4H2O2 + PbS   →        PbSO4 + 4H2O

Hydrogen per oxide is an oxidizing agent. Because oxidation number of S increases, S is oxidized. So PbS is reducing agent.

Similarly oxidation number of H decreases, H is reduced. Therefore H2O2 is oxidizing agent.

 

 

ThinkTank

  1. What materials do you need to electro plate copper onto an iron nail? Make a diagram showing how these materials should be arranged.

Ans.         The anode is made of copper metal and the electrolyte is CuSO4 to which a few drops of sulphuric acid are added to prevent its hydrolysis.

Iron nail to be electroplated is made the cathode.

Electrolytic Reactions:

At Anode: Cu(s)                         Cu2+(aq)+2e-1

At Cathode: Cu2+(aq)+2e-1                         Cu(s)

 

  1. Describe the process that is occurring in the following illustration. Shoe has steel strips.

Ans.       In the give figure copper plating is being used since electrolyte is copper sulphate.

Anode is made of copper metal, shoe is made cathode and is to be copper plated, electrolyte used is copper sulphate solution. Following electrolytic reactions take place.

At Anode: Cu(s)                         Cu2+(aq)+2e-1

At Cathode: Cu2+(aq)+2e-1                         Cu(s)

  1. Following redox reaction occurs in the voltaic cell illustrated below:

Ni+2(aq) + Fe(s)                                 Fe+2(aq)+Ni(s)

Identify the anode, cathode and indicate the direction of flow of electrons.

Ans.       Anode is Fe electrode, oxidation will take place at anode.

Cathode is Ni electrode, reduction takes place at cathode.

Following electrolytic reactions will take place

At Anode: Fe(s)                         Fe2+(aq)+2e-1

At Cathode: Ni2+(aq)+2e-1                          Ni(s)

Electrons will flow from iron anode to Ni Cathode.

  1. Design an experiment to demonstrate cathodic protection from corrosion.

Ans.       Cathodic protection can be demonstrated by connecting more active metals such as Mg with metal to be saved from rust or corrosion like Fe. Since Mg is more active than iron, so it acts as anode and iron as cathode. Mg will oxidize itself and save iron from corrosion. Cathodic protection is used to protect pipelines, underground steel structures etc and Mg is connected to pipelines at regular intervals. Oxidized Mg is also replaced after every few weeks.