Unit 16: Electrochemistry — Long Questions

Key points Oxidation Number of Atoms in Compounds and Ions
Importance of Oxidation Number

Oxidation number of atoms in compounds and ions

Oxidation Number The oxidation number is an apparent charge on an atom in a compound or an ion. It may have a positive or a negative sign.

Importance of oxidation number

It is important to assign an oxidation number to an atom in a compound because of following reasons.

(i) The assignment of oxidation number to atoms in compounds and ions helps us to identify whether the reaction involves oxidation and reduction or not.

(ii) It also helps to identify oxidizing and reducing agents in a given oxidation-reduction reaction.

(iii) Oxidation numbers are also used to write chemical formulas and names of the chemical compounds.

(iv) This concept also helps to keep track of the new distribution of electrons during an oxidation-reduction reaction.

Key point Rules for Finding Oxidation Numbers

Rules for finding oxidation numbers

The oxidation number of atoms can be found out by applying the following rules.

(i) Zero oxidation states: Atom present in a free element has an oxidation number zero.
Example: The oxidation number of sodium in sodium metal is zero. Similarly, oxidation number of oxygen in ozone (O₃) is also zero.

(ii) Monoatomic ions: For monoatomic ions, the oxidation number is equal to the charge present on the ion.
Example: The oxidation number of Ca²⁺ is +2 while that of Al³⁺ is +3.

(iii) Elements of groups 1 and 3: The elements present in group 1 (except hydrogen) in the periodic table are assigned oxidation number +1, those of group 2, +2 and those of group 3, +3.

(iv) Binary compounds: In binary compounds the more electronegative atom is always assigned a negative oxidation number.
Example: Fluorine, being the most electronegative atom, will always be given −1 oxidation number. Similarly, in NH₃ and H₂S, nitrogen and sulphur are assigned the oxidation number of −3 and −2 respectively.

(v) Elements of group 17: The elements of group 17 are normally given oxidation number −1 when they are combined with less electronegative element.
Example: Chlorine in all such compounds as HCl, PCl₅, CHCl₃ has been assigned an oxidation number −1.

(vi) Oxidation state of hydrogen: Hydrogen is assigned an oxidation number +1 when it is in combination with more electronegative atoms while it is given oxidation number −1 when it is combined with less electronegative atoms.
Example: In HCl, the oxidation number of hydrogen is +1, while in sodium hydride (NaH) it bears an oxidation number −1.

(vii) Oxidation number of oxygen: The element oxygen is assigned an oxidation number −2 in most of its compounds.

(viii) Sum of oxidation numbers in neutral compounds: The sum of oxidation numbers of all the atoms present in a neutral compound is always zero.
In polyatomic ions: In a polyatomic ion the sum of oxidation numbers is equal to the charge present on that ion.

Key point Writing Chemical Formulas of Ionic Compounds by Applying Oxidation Numbers

Writing chemical formulas of ionic compounds by applying oxidation numbers

We can write the chemical formula of a compound if we know the oxidation numbers of its constituent elements.

Prediction of principal oxidation number: The periodic table and the electronic structure of the atoms can help us in predicting the principal oxidation number of the element. The formula of the compound may then be written using the oxidation number of the atoms present in that compound.

Example of calcium chloride We shall write the formula of calcium chloride in the following way.

The element calcium is present in group II (group 2) of the periodic table and its principal oxidation number is +2.

Chlorine is present in group 17 and we know that when it is attached with less electronegative element in binary compounds its oxidation number is −1. The formula for calcium chloride will, therefore, be CaCℓ₂ because only with this formula the sum of the oxidation numbers of both the elements will add upto zero.

Ca²⁺ and Cℓ⁻ with their corresponding signs combine together.

Example of Aluminium oxide (16.4 Example)

Write the chemical formula of aluminium oxide with the help of oxidation numbers of the elements involved.

Solution

Aluminum is present in group 13 so its principal oxidation number shall be +3. Oxygen being more electronegative atom normally shows −2 oxidation state in normal oxides. The formula of aluminum oxide will then be Al₂O₃ because the some of oxidation numbers of both elements will be equal to zero with this formula.

Aℓ³⁺ and O₂⁻ with their corresponding signs cross-combine.

Key point Writing Chemical Formulas of Ionic Compounds using Ionic Charges

Writing chemical formulas of ionic compounds using ionic charges

Potassium permanganate In order to write the formula of potassium permanganate, write first the formulae of permanganate ion, MnO₄⁻ and potassium ion K⁺. Then combine the two ions together to form an electrically neutral compound.

Cation K⁺
Anion: MnO₄⁻

The formula for potassium permanganate will, therefore, be KMnO₄.

Formula of ionic compounds having more than one polyatomic ions: When writing the formulae of ionic compounds containing more than one polyatomic ions, we take the formula of the ion in brackets and indicate the number of such ions with a subscript.

The formula for ammonium carbonate will be written as follows.

NH₄⁺ and CO₃²⁻

(NH₄)₂CO₃

The formula for ammonium carbonate will, therefore, be (NH₄)₂CO₃.

16.5 Example

Write down of the formula of aluminium sulphate.

Aℓ³⁺ and (SO₄)²⁻

The formula for aluminum sulphate is Aℓ₂(SO₄)₃.

Examples of several polyatomic ions are given in table 16.1

Table 16.1 – Common Polyatomic Ions

CH₃COO⁻ | acetate | Cr₂O₇²⁻ | dichromate | NO₃⁻ | nitrite
NH₄⁺ | ammonium | HCO₃⁻ | bicarbonate | C₂O₄²⁻ | oxalate
CO₃²⁻ | carbonate | HSO₄⁻ | bisulphate | ClO₄⁻ | perchlorate
ClO₂⁻ | chlorate | HSO₃⁻ | bisulphite | MnO₄⁻ | permanganate
ClO⁻ | chlorite | OH⁻ | hydroxide | PO₄³⁻ | phosphate
CrO₄²⁻ | chromate | ClO⁻ | hypochlorite | SO₄²⁻ | sulphate
CN⁻ | cyanide | NO₃⁻ | nitrate | SO₃²⁻ | sulphite

Key point Difference between Oxidation and Reduction

Difference between oxidation and reduction

Oxidation and reduction are simultaneous but opposite processes. They always occur together. Oxidation and reduction changes are described in three principal ways.

The difference between oxidation and reduction is given:

Oxidation | Reduction

1. Oxidation is defined as a process in which oxygen is either added or hydrogen is removed. | 1. Reduction is defined as either addition of hydrogen or removal of oxygen.

Examples | Examples

(a) Oxygen combines with other elements in the formation of their oxides. | (a) In the following reactions hydrogen is heated with metal oxide when oxygen comes out of the compound and combines with hydrogen to give a molecule of water.

• Elements in these processes are said to be oxidized and oxygen is reduced. |
• Rusting of iron or burning of magnesium in air are oxidation processes. | CuO₍ₛ₎ + H₂₍ᵍ₎ --Heat--> Cu₍ₛ₎ + H₂O₍ₗ₎

4Fe₍ₛ₎ + 3O₂₍ᵍ₎ → 2Fe₂O₃₍ₛ₎ |
2Mg₍ₛ₎ + O₂₍ᵍ₎ → 2MgO₍ₛ₎ | In the above examples, Cℓ₂ and CuO are being reduced and hydrogen (in both cases) is being oxidized. Cℓ₂ and CuO are oxidizing agents while hydrogen is both the reactions will be reducing agent.

In the above reactions Fe and Mg are being oxidized while oxygen is being reduced. Oxygen is an oxidizing agent while iron and magnesium are reducing agents. |

(b) Hydrogen is removed from a compound by oxygen in the following examples. Hydrogen is removed from CH₄ and HCℓ by oxygen or a compound containing oxygen. | (b) Hydrogen is added in chlorine to give hydrogen chloride gas.

H₂₍ᵍ₎ + Cℓ₂₍ᵍ₎ → 2HCℓ₍ᵍ₎

2. Loss of electron | 2. Gain of electrons

Oxidation is a process in which electron or electrons are lost. | Reduction is a process in which electron or electrons are gained.

Example | Example

Copper may lose one or two electrons to give Cu⁺¹ or Cu²⁺ ions respectively. In both these processes, we say, copper is being oxidised. Hence, copper is called a reducing agent. | Fe³⁺ + e⁻ → Fe²⁺
Mn⁷⁺ + 5e⁻ → Mn²⁺

Cu → Cu⁺ + e | In both the above examples, ions of iron and manganese are being reduced. Hence, they are called oxidising agents.
Cu → Cu²⁺ + 2e |

3. Increase in oxidation state | 3. Decrease in oxidation state

Oxidation is process in which oxidation number or oxidation state of an element increases. | Reduction is a process in which oxidation number of an element decreases.

Cu → Cu⁺¹ + e | Fe³⁺ + e⁻ → Fe²⁺
Cu → Cu²⁺ + 2e | Mn⁷⁺ + 5e⁻ → Mn²⁺

In the above two processes, the oxidation number of copper has increased from zero to +1 [Cu (I) ] and +2 [Cu (II) ] respectively. | In the above examples oxidation number of iron is reduced from +3 [Fe(III)] to +2[Fe(II)] while oxidation number of manganese is reduced from +7[Mn(VII)] to +2 [Mn(II)].

Key point Oxidation – Reduction Reactions

Oxidation – reduction reactions

A chemical reaction which involves simultaneous oxidation and reduction is called an oxidation-reduction or redox reaction.

Explanation The reaction may involve the complete transference of electron or electrons to form ionic bonds or a partial transfer or shift of electron or electrons to form covalent bonds.

(i) Reaction of KMnO₄ with H₂C₂O₄

Potassium permanganate acts as a very powerful oxidizing agent in acidic medium. It oxidizes oxalic acid into carbon dioxide.

Disappearance of KMnO₄ During the reaction the dark purple solution of potassium permanganate changes to colourless as shown in figure 16.1. In the presence of dilute sulphuric acid the reaction proceeds as follows.

Reaction

2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5 [O]
[O] + H₂C₂O₄ → 2CO₂ + H₂O 5

2KMnO₍ₐq₎ + 3H₂SO₄₍ₐq₎ + 5H₂C₂O₄₍ₐq₎ → K₂SO₄₍ₐq₎ + 2MnSO₄₍ₐq₎ + 10CO₂₍ᵍ₎ + 8H₂O₍ₗ₎

Oxidising and reducing agent

Let us assign oxidation number to each element in these compounds to find out the oxidizing and reducing agents.

+1 +7 4(-2) 2(+1)+6 4(-2) 2(+1)+3 (+4) (-2) 2(+1)+(-2)
2KMnO₄ + 3H₂SO₄ + 5H₂C₂O₄ → K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂O

Mn⁺⁷ --+5e--> Mn⁺² (Reduction)
C⁺³ --e--> C⁺⁴ (Oxidation)

Oxidation number of manganese has decreased from +7 to +2 and that of carbon has increased from +3 to +4.

So, each manganese atom accepts five electrons to get reduced and each carbon atom loses one electron to get oxidized. KMnO₄ is therefore an oxidizing agent while oxalic acid in this reaction acts as a reducing agent.

(ii) Reaction of KMnO₄ with KI: Aqueous potassium iodide is also oxidized to iodine in acidified potassium permanganate. The formation of iodine in this reaction results in a colour change, typically from colourless to yellow-brown.

Reaction

+1 +7 4(-2) 2(+1)+1+(-2) 2(+1)+6 4(-2) 2(+1)+1 (-2) 8(+1)+1 (-2)
2KMnO₍ₐq₎ + 8H₂SO₄₍ₐq₎ + 10KI₍ₐq₎ → 6K₂SO₄₍ₐq₎ + 2 MnSO₄₍ₐq₎ + 5I₂₍ₐq₎ + 8H₂O₍ₗ₎

Mn⁺⁷ --+5e--> Mn⁺² (Reduction)
2I⁻¹ --2e--> I₂⁰ (Oxidation)

Oxidising and reducing agents In this reaction, the permanganate ion ( MnO₄⁻ ) is reduced to manganese (II) ion (Mn²⁺) and the iodide ion (I⁻) is oxidized to iodine (I₂). Potassium permanganate, therefore, acts as an oxidizing agent while potassium iodide acts as a reducing agent.

Key point Electrolysis

Electrolysis

When electricity is passed through fused ionic compounds or their aqueous solutions, they are decomposed into their constituents. Such compounds are called electrolytes and the process is called electrolysis.

During electrolysis, a non-spontaneous chemical reaction takes place with the help of electrical energy.

Electrolytic cell and electrodes The process of electrolysis can be carried out in an electrolytic bath called an electrolytic cell with the help of two electrodes through which electric current enters and leaves the cell. These electrodes may be made of graphite or any other metal.

• Anode: It is a positive electrode through which electrons enter the external circuit.
• Cathode: It is a negative electrode through which electrons leave the external circuit.

General process

• Electrolyte present in the cell is in the form of freely moving ions. During electrolysis, the electrolyte is decomposed into its components by applying an external voltage.

• Thus external voltage source (battery) forces the non-spontaneous reaction to occur by pumping electrons towards one of the electrodes, thereby, giving it a negative charge.

• The positive ions of the electrolyte are attracted towards the negative electrode and get reduced by accepting electrons from it, making it the cathode.

• The battery simultaneously withdraws electrons from the other electrode, giving it a positive charge.

• The negative ions of the electrolyte are attracted towards this and get oxidized by losing their electrons, making it the anode.

• Electrons flow in the external circuit while ions move towards their respective electrodes inside the cell.

• If the oxidation and reduction at the electrodes cease, the flow of electricity in the external circuit also stops.

THINGS TO KNOW!

S.Q What is difference between cathode and anode in battery and electrolytic cell? 10216019

Ans. Difference between cathode and anode of battery and electrolytic cell.

The terms anode and cathode always refer to the electrodes at which oxidation and reduction occur respectively. However, the charges carried by electrodes in a battery and electrolytic cell are different.

Electrode | Function | Charge
Battery | Anode | Oxidation | Negative
| Cathode | Reduction | Positive
Electrolytic Cell | Anode | Oxidation | Positive
| Cathode | Reduction | Negative

Key point Electrolysis of a Concentrated Aqueous Solution of Sodium Chloride (Brine) Using Inert Electrodes (Graphite /Pt)

Electrolysis of a concentrated aqueous solution of sodium chloride (Brine) using inert electrodes (Graphite /Pt)

The electrolysis of a concentrated solution of sodium chloride, also called brine, produces hydrogen gas at the cathode, chlorine gas at the anode, and sodium hydroxide solution as a by-product.

Electrolytic cell The electrolytic cell used in this electrolysis is shown in figure 16.3.

• Cathode: The cathode used in this cell is made of platinum.
• Anode: The anode must be made of graphite to avoid the reaction with chlorine gas.

The following ions are present in the solution.

Na⁺₍ₐq₎, Cl⁻₍ₐq₎ from sodium chloride
H⁺₍ₐq₎, OH⁻₍ₐq₎ from water

Na⁺ and H₂O, both migrate towards the cathode. Since water molecules and hydrogen ions have a greater tendency to get reduced as compared to sodium ions (Na⁺), so they pick up electrons from the cathode to form hydrogen gas.

Reaction at cathode

2H⁺₍ₐq₎ + 2e⁻ → H₂₍ᵍ₎

or

2H₂O + 2e⁻ → H₂₍ᵍ₎ + 2OH⁻

Alkaline solution around cathode

The ionic equilibrium of water is, thus, disturbed due to the discharge of H⁺ and more water molecules are ionized to restore it. The solution around the cathode becomes alkaline due to the formation of hydroxide ions (OH⁻).

Reaction at anode

2Cl⁻₍ₐq₎ → Cl₂₍ᵍ₎ + 2e⁻

The solution gradually gets alkaline due to the presence of OH⁻. The volumes of hydrogen and chlorine gases obtained are equal. The overall equation for this electrolysis is:

Overall reaction

2NaCl₍ₐq₎ + 2H₂O₍ₗ₎ → H₂₍ᵍ₎ + Cl₂₍ᵍ₎ + 2NaOH₍ₐq₎

Key point Electrolysis dilute and concentrated aqueous solutions of metal halides | Electrochemical Series

Electrolysis dilute and concentrated aqueous solutions of metal halides

• Role of water in electrolysis

When the aqueous solutions of metal halides are subjected to electrolysis, the situation becomes more complicated due to the presence of water. Water is a very weak electrolyte and upon ionization it gives very small amount of hydrogen and hydroxide ions.

H₂O₍ₗ₎ ⇌ H⁺₍ₐq₎ + OH⁻₍ₐq₎

• Choice of ions to be discharge: This will mean that the electrolyte will consist of more than one type of ions which will approach both cathode and anode. Hence there can be a choice over which ions will get discharged.

Example If an aqueous solution of sodium chloride is electrolyzed, both sodium and hydrogen cations are attracted towards the cathode and both chlorides and hydroxide ions are attracted towards the anode.

Electrochemical series

Which of ions sodium, hydrogen, chloride or hydroxide will preferentially be discharged depends upon the tendency of the elements to lose or gain electrons. The elements present at the top of electrochemical series shown in table (16.2) possess more tendency to lose electrons as compared to the elements present at the bottom.

Following are the categories of metals according to electrochemical series:

(i) Metals below hydrogen: If a metal is below hydrogen in the electrochemical series, then the metal ion is discharged. Metals like this include copper and silver. If dilute copper chloride is electrolyzed, copper metal is obtained at the cathode.

(ii) Metals higher in electrochemical series than hydrogen: If a metal is placed higher in the electrochemical series than hydrogen ion is discharged. Metals like this include lithium, sodium and magnesium. If dilute solution of sodium chloride is electrolyzed, hydrogen gas is obtained at the cathode.

(iii) Lead and nickel metals: Metals like lead and nickel present a more complicated picture.

• If aqueous solutions of halides of these metals are electrolyzed, then the products obtained at the cathode depends on the concentrations of the halide solutions.
• If the solution is concentrated, then metal is deposited at the cathode.
• If the solution is very dilute, hydrogen is evolved.
• At in-between concentrations both metal and hydrogen are obtained.

At anode The electrolysis of dilute solutions of halides will deposit oxygen because hydroxide ions are preferentially discharged due to their higher concentrations.

Reaction at cathode

2M⁺₍ₐq₎ + 2e⁻ → 2M₍ₛ₎
2H⁺₍ₐq₎ + 2e⁻ → H₂₍ᵍ₎

or

2H₂O₍ₗ₎ + 2e⁻ → H₂₍ᵍ₎ + 2OH⁻₍ₐq₎

Reaction at anode

2X⁻₍ₐq₎ → X₂₍ᵍ₎ + 2e⁻
4OH⁻₍ₐq₎ → O₂₍ᵍ₎ + 2H₂O₍ₗ₎ + 4e⁻

Ans. Key point Electrolysis of Concentrated Aqueous Solutions of Metal Halides

Electrolysis of concentrated aqueous solutions of metal halides
If the concentrated aqueous solutions of metal halides are electrolyzed the hydrogen ions are discharged at the cathode if the metal is present higher than hydrogen in electrochemical series.

Example Electrolysis of concentrated aqueous solution of sodium chloride will produce hydrogen gas at the cathode.

At cathode For less reactive metals like copper and silver the metal ions are discharged at the cathode.

At anode The halide ions are preferentially discharged at anode due to their higher concentration.

Reaction at cathode
2M+(aq) + 2e- → 2M(s)
2H+ + 2e- → H2

Reaction at anode
2X- → X2(g) + 2e-

Ans. Key point Electrolysis of molten lead (II) chloride

Electrolysis of molten lead (II) chloride

Heating of lead chloride Lead (II) chloride is solid at room temperature. To make it an electrolyte it must be heated upto 501°C to get molten lead chloride. Heat will render ions free to move and conduct electricity. The molten lead is then taken in a heated crucible where its temperature is maintained above 500°C.

Electrodes Two inert electrodes (graphite or platinum) are then immersed in it. Upon electrolysis the lead (II) ions will move towards the cathode where they are discharged forming lead in the liquid state. The chloride ions migrate towards the anode where they lose electrons to give chlorine gas. The electrolysis must be carried out at the temperature above the melting point of lead chloride.

Reaction at cathode
Pb2+(l) + 2e- → Pb(l) (Reduction)

Reaction at anode
2Cl-(l) → Cl2(g) + 2e- (Oxidation)

Overall reaction
PbCl2(l) → Pb(l) + Cl2(g)

Ans. Key point Electrolysis of dilute solution of sulphuric acid using inert electrodes (Carbon or Pt)

Electrolysis of dilute solution of sulphuric acid using inert electrodes (Carbon or Pt)

Electrolysis of dilute aqueous solution of sulphuric acid produces hydrogen and oxygen gases. In other words, this electrolysis involves the decomposition of water. Inert electrodes like platinum or graphite are used for this electrolysis because they did not react with the electrolyte.

In an aqueous solution of dilute sulphuric acid H+ and SO42- are given by sulphuric acid while the ionization of water forms H+ and OH-.

• At cathode: Hydrogen ions gain electrons to give hydrogen gas.

• At anode: Both SO42- and OH- are attracted but OH- lose electrons to give oxygen gas. Sulphate ions are not discharged at the anode because they are difficult to be oxidized compared to hydroxide ions.

• Increase in concentration of sulphuric acid: As the electrolysis moves ahead, the concentration of sulphuric acid increases because the water present is being decomposed.

• Reaction at the cathode
2H+(aq) + 2e- → H2(g) (Reduction)

• Reaction at the anode
4OH-(aq) → O2(g) + 2H2O(l) + 4e- (Oxidation)

Ans. Key point Electrolysis of Dilute Aqueous Solution of Copper Sulphate Using Inert Electrodes or Copper Electrodes

Electrolysis of dilute aqueous solution of copper sulphate using inert electrodes or copper electrodes

The electrolysis of aqueous solution of copper(II) sulphate with inert electrodes result in the decomposition of water and the copper(II) sulphate. Copper metal is deposited at the cathode (inert) and oxygen gas is produced at the anode (inert).

Reaction at cathode
Cu2+(aq) + 2e- → Cu(s) (Reduction)

Reaction at anode
4OH-(aq) → O2(g) + 2H2O(l) + 4e- (Oxidation)

Since the concentration of copper ions decreases slowly so the blue colour of the solution fades away with the passage of time.

Use of copper electrodes
If copper electrodes are used instead of inert electrodes, the copper metal atoms from the copper anode start going into the solution in the form of copper ions and the electrode mass starts decreasing.

Reaction at cathode
Cu2+(aq) + 2e- → Cu(s) (Reduction)

Reaction at anode
Cu(s) → Cu2+(aq) + 2e- (Oxidation)

Ans. Key points Fuel Cells | Construction | Working

Fuel cells
The devices that generate electricity from a chemical reaction – most commonly between hydrogen and oxygen without burning fuel are called fuel cells.

Explanation Fuel cells provide a method in which chemical energy is converted to electrical energy. These cells use gaseous fuels like hydrogen and oxygen to produce electricity.

Construction
A fuel cell is composed of two hollow tubes which act as electrodes. They are made of porous compressed carbon saturated with platinum which acts as a catalyst.

Electrolyte The electrolyte used in this cell is aqueous potassium hydroxide.

Working
At the anode, hydrogen is oxidized to water and at the cathode oxygen is reduced to hydroxide ions.

At anode
[H2(g) + 2OH-(aq) → 2H2O(l) + 2e-] 2

At cathode
O2(g) + 2H2O(l) + 4e- → 4OH-

Overall reaction
2H2(g) + O2(g) → 2H2O(l)

Products The fuel cell produces electricity and water continuously provided the reactants are continuously supplied. They are very efficient because they convert 75% of the bond energy of fuels to electricity.

Ans. Key points Advantages of Fuel Cell | Disadvantages of Fuel Cell

Advantages of fuel cell
(i) Advantages over fossil fuels: The hydrogen-oxygen fuel cells have a clear advantage over the traditional fossil fuels, petrol and diesel, used to drive the vehicles. They produce zero emission of carbon dioxide because water is the only product. Fossil fuels, on the other hand, produce carbon dioxide, nitrogen oxides and sulphur oxides which are very harmful to the environment.

(ii) More efficient: Fuel cells can convert a large proportion about 75% of chemical energy present in fuels into electrical energy as compared to the energy conversion of the fossil fuels. In other words, they are more efficient.

(iii) Less noise pollution: Fuel cells produce less noise pollution.

(iv) More sustainable: Hydrogen used in fuel cells can be obtained from renewable sources like solar or wind power. This fact renders fuel cells more sustainable.

Disadvantages of fuel cell
(i) Very expensive: Fuel cells at present, are very expensive because the production of hydrogen by electrolysis of water incurs high cost.

(ii) Handling problems of hydrogen gas: Hydrogen being highly inflammable requires very special storage and transportation facilities which further increases the cost of fuel cells.

(iii) Less durable: Fuel cells and electric motors are less durable than petrol and diesel engines so they are not so long lasting.

Ans. Key point Corrosion

Corrosion
Corrosion is a natural process due to which a metal suffers gradual decay when it interacts chemically with its environment.

Explanation During this interaction the surface of metal is coated with compounds such as oxides, sulphides and carbonates.

Rusting of iron A common example is rusting of iron, where iron reacts with oxygen and water present in the atmosphere to form hydrated iron (III) oxide. Once iron begins to rust, the process will continue as rust is porous and allows both air and moisture to come into contact with fresh metal underneath. Hence, all the metal will corrode given enough time.

Electrochemical reactions The process of corrosion of iron appears to require two important chemicals, oxygen and water, which are involved in the following electrochemical reactions.

Oxidation Iron loses electrons to form Fe (II) ions.
2Fe(s) → 2Fe2+(aq) + 4e-

Reduction Oxygen gains these electrons in the presence of water to form hydroxide ions.
O2(g) + 2H2O(aq) + 4e- → 4OH-(aq)

Conversion of Fe+2 to Fe+2: Fe2+ then react with more oxygen and water to form iron (III) hydroxide which is rust.
Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
4Fe(OH)2(s) + O2(g) + 2H2O(l) → 4Fe(OH)3(s)

Ans. Key point Prevention of Corrosion

Prevention of corrosion
(i) Coating of the metal surface: Corrosion can be avoided by coating the metal surface with some barriers that prevent the contact of metal surface with the atmosphere. For this purpose, the metal surfaces are protected by either greasing with oil or painting. Other methods of protection are galvanizing, electroplating and sacrificial protection.

(ii) Galvanization: Galvanization is a common method to protect iron or steel from rusting. It involves the dipping of an iron sheet in a bath of molten zinc. A layer of zinc metal gets deposited on the iron sheet and prevents rusting. Tin cans are actually steel cans which have been coated with a thin layer of tin.

(iii) Sacrificial protection
• In sacrificial protection a more reactive metal like zinc or magnesium is used which will corrode in place of the less reactive metal like iron.
• Magnesium is often employed to protect steel in buried fuel tanks and pipe lines and ship hulls submerged in sea water.
• The difference in the activity of the two metals (Mg and Fe) causes a current to flow between them, producing corrosion on the more active metal protecting the less active metal iron.

Ans. Key points Electroplating | Objectives of Electroplating

Electroplating
It is a process in which a metal is deposited on another metal electrolytically.

Objectives of electroplating
There are three main objectives of electroplating.
(i) Decoration: To deposit the noble metals like gold and silver on the inexpensive metals to enhance their beauty.

(ii) Protection: To protect the metals from rusting as well as from the reaction of organic acids.

(iii) Repair: To weld the broken parts of the machinery by electrodeposition of metals. Generally copper, silver, chromium, nickel and gold are electroplated.

Ans. Key point Procedure of Electroplating

Procedure of electroplating
Electroplating procedure involves the following steps.

• Washing and cleaning: The article to be electroplated is first cleaned with sand and then washed with caustic soda solution and finally with plenty of water.

• Cathode, anode: This article is then, made a cathode, while the metal to be deposited is made anode.

• Electrolyte: The electrolyte is a salt of the metal being deposited.

• Electrolytic tank: The electroplating is carried out in a tank made of cement, glass or wood. It is called an electrolytic tank.

This process of electroplating is shown in figure16.7.

Properties of electrolyte
The electrolyte chosen should have the following properties:
• Very soluble in water
• Good conductor
• Not easily oxidized or reduced or hydrolyzed
• Inexpensive

Electrodes are dipped in the electrolytic tank containing the electrolyte and electric current is passed. A thin layer of the noble metal is deposited on the article to be electroplated.

Conditions The following conditions should be fulfilled to get good deposits.
• High current density
• Low temperature
• Higher concentration of the metal in its electrolyte

Examples Silver plating, copper plating etc.

Ans. Key points Silver Plating | Copper Plating

(i) Silver plating
Electrodes are dipped in the electrolytic tank containing the electrolyte potassium argento cyanide K[Ag(CN)2], and electric current is passed. A thin layer of the silver is deposited on the article to be electroplated. The article to be electroplated is made as cathode. Anode is of pure silver.

The following ions are formed in the electrolytic solution.
K[Ag(CN)2](aq) ⇌ K+(aq) + [Ag(CN)2]-(aq)
[Ag(CN)2]-(aq) ⇌ Ag+(aq) + 2CN-(aq)

Working
When the electric current is passed, Ag+ ions migrate towards the cathode and deposit after picking up electrons. Cyanide ions move towards the anode, where they react with silver to form AgCN, which combines with KCN to reproduce the electrolyte, K[Ag(CN)2].

(ii) Copper plating:
In copper plating, 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. The article to be electroplated is made the cathode.

The following electrolytic reactions take place at the electrodes.

At anode
Cu(s) → Cu2+(aq) + 2e-

At cathode
Cu2+(aq) + 2e- → Cu(s)

Ans. Key points Galvanic Cells | Process in Galvanic cell

Galvanic cells
The cells in which chemical energy is converted into electrical energy is called galvanic cells.

Explanation
The oxidation-reduction reactions take place at the electrodes when fused ionic compounds or the aqueous solutions of ionic compounds are electrolyzed. The reverse process may also occur. In other words, oxidation-reduction reactions can be used to produce electricity. This can be understood with a very simple process.

Process in Galvanic cell
• Take a plate of zinc metal and dip it in an aqueous solution of copper sulphate.
• After some time, a pink layer of copper metal will appear on the zinc plate and the blue colour of the solution will become lighter.
• If the quantity of zinc is enough and sufficient time is allowed, the solution will become colourless. In this process, electrons are transferred from zinc to copper ions.
• Every copper ion gets two electrons from zinc atom to get deposited as a metallic copper atom.
• A zinc atom goes into the aqueous solution as Zn2+.

The reaction can be shown as follows:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

In other words, zinc is oxidized to Zn2+(aq) while Cu2+(aq) is reduced to copper metal. This reaction can be made a basis for producing electric current in a galvanic cell which is named as Daniel cell.

Ans. Key points Daniel Cell | Construction | Working

Daniel Cell
The cell which involves a spontaneous chemical reaction to generate electricity is called a galvanic cell or Daniel cell.

The cell is named after the name of English scientist F. Daniel.

Explanation Daniel cell uses oxidation-reduction reactions to produce electric current. In this process, the oxidation-reduction reactions are taking place directly on the zinc electrode, so electricity cannot be generated. However, if oxidation-reduction reactions are carried out in separate compartments, electricity can be generated. Such an arrangement is shown in figure 16.8.

Construction
• In this type of galvanic cell, a zinc plate is dipped in 1M zinc sulphate solution.
• A copper plate is dipped in 1M copper sulphate solution. The set up makes two half-cells in an electrochemical cell.

Partition or salt bridge The two solutions are kept separate by a porous partition through which ions can pass but the solutions cannot. This partition is called salt bridge.

Galvanometer The two electrodes are connected by a wire through a galvanometer which measures the electric current.

Working
• At the surface of zinc electrode, each zinc atom loses two electrons and goes into the solution as zinc ions. These electrons travel through the external wire to copper electrode where they are used to discharge copper ions into copper atoms which are deposited at this electrode.

• An oxidation reaction takes place at zinc electrode and it is called anode while reduction takes place at copper electrode and it is called cathode.

Chemical reactions
The following reactions take place at the two electrodes.

At anode Zn(s) → Zn2+(aq) + 2e- (Oxidation) Standard oxidation potential = +0.76

At cathode Cu2+(aq) + 2e- → Cu(s) (Reduction) Standard reduction potential = +0.34

Overall cell reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Cell voltage E0=1.10 volts

• Both parts of the cell are called half cells. No reaction can take place in any isolated half-cell. Both the reactions take place simultaneously at both half cells.

• In this cell, electrons travel in the external circuit while ions travel through the porous partition in the internal circuit.

• Daniel cell produces a voltage of around 1.10 volts under standard conditions.

Ans. Key point Order of Reactivity of Metals Using Voltage Data

Order of reactivity of metals using voltage data

Reactivity of metals The reactivity of metals can be found out by using voltage data obtained from galvanic cells.

• According to the available data the reactivity of metals depends upon how readily it is oxidized or lose electrons to form positive ions.

• If a galvanic cell is established between two metals, the more reactive metal will lose electrons and form positive ions.

• The voltage generated by this cell measures the comparative tendencies of the metals to lose electrons.

• A higher voltage indicates that the difference in the tendency of one metal (more reactive metal) to lose electron is far more than the other.

Example If a cell is established between magnesium and copper it produces a higher voltage than a cell established between iron and copper. This will mean that magnesium is more reactive than iron because it has a tendency to lose electrons more easily.

Electrochemical series • By measuring the voltage produced when different metal pairs are used in a voltaic cell, it is possible to determine the reactivity series of metals which is called electrochemical series.

• Metals placed higher in this series are considered more reactive and easily lose electrons than the metals placed lower in the series.