Electrolysis Dalal Simplified ICSE Chemistry Class-10

Electrolysis Dalal Simplified ICSE Chemistry Class-10 Solutions Chapter 5 . Solutions of Dr Dalal Simplified ICSE Chemistry by Dr Viraf and J Dalal for Class 10. Step by step Solutions of Electrolysis ICSE Chemistry by Dr Viraf and J Dalal Simplified Dalal Chemistry.

 Electrolysis Dalal Simplified ICSE Chemistry Class-10

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Additional Chemistry Electrolysis Dalal Simplified ICSE

Question 1.

Define:

1. Electrolysis
2. Electrodes
3. Ions
4. Electrolytic dissociation.

Answer:
(1) Electrolysis: “The process of decomposition of a chemical compound in aqueous solution or in molten state accompanied by chemical change.”
(2) Electrodes: Electrodes allow the electric current to enter or leave the electrolyte solution.
(3) Ions: They are atoms which carry a positive or negative charge and become free and mobile when an electric current is passed through an aqueous solution of a chemical compound.
(4) Electrolytic dissociation: The process due to which an ionic compound in the fused or in aqueous solution dissociates into ions by passage of electric current through it is called electrolytic dissociation.

Question 2.

Differentiate between

1. Electrolytes and Non-electrolytes
2. Strong and Weak electrolytes
3. Anode and Cathode
4. Electrolytic dissociation and Ionisation with suitable examples.

Answer:

Question 2(i).

Electrolytes and Non-electrolytes

Electrolytes:

Chemical compound – which conduct electricity in the fused or in aq. solution state and -undergo chemical decomposition due to the flow of current through it.
Electrolytes – are ionic compounds
Particles in Electrolytes – ions only or Ions and molecules only
Examples:
Acids – dil. HCl, HNO₂ H₂SO₄ Alkalis – KOH, NaOH solutions Ionic salts – PbBr₂ [molten],
CuSO₄ [aq.]

Non-electrolytes:

Chemical compound – which do not conduct electricity in the fused or aq. soln. state and – do not undergo chemical decom­position due to the flow of current through it.
Non-electrolytes – are covalent compounds
Particles in non-electrolytes – Molecules only
Example:
Pure or distilled water, Alcohol, Kerosene, Carbon disulphide, carbon tetrachloride, sucrose, glucose, sugar solution.

Question 2(ii).

Strong and Weak electrolytes:

Strong electrolytes:

1. The compounds which in their aqueous solution or in fused state are almost completely ionised are called strong electrolytes.
2. They allow a large amount of electricity to flow through them and hence are good conductors of electricity.
3. In aqueous solution or molten state, only ions are present.
   Examples:
   Strong acids: HCl, H₂SO₄,HNO₃
   Strong bases: NaOH, KOH
   Salts: NaCl, NaNO₃, K₂S0₄

Weak electrolytes:

1. The compound which in their aqueous solution or in fused state are partially ionised are called weak electrolytes.
   
2. They allow small amount of electricity to flow through them and hence are poor conductors of electricity.The compound which in their aqueous solution or in fused state are partially ionised are called weak electrolytes.
   In aqueous solution or molten state ions as well as unionised molecules are present.
   Examples:
   Weak acids: CH₃COOH, H₂CO₃
   Weak bases: NH₄OH, Ca(OH)₂
   Salts: CH₃COONH₄

Question 2(iii).

Anode and Cathode:

Anode:

1. It is the electrode connected to the positive terminal of the battery
2. Anions migrate to anode.
3. The anions donate excess electrons to the anode and they are oxidised to neutral atoms.

Cathode:

1. It is the electrode connected to the negative terminal of the battery.
2. Cations migrate to cathode.
3. The cations gain excess electrons from the cathode and they are reduced to neutral atoms.

Question 2(iv).

Electrolytic dissociation and ionisation with suitable examples.

Dissociation:

1. Separation of ions which are already present in an ionic compounds.
2. Electrovalent compounds show dissociation e.g. potassium chloride, lead bromide.
   KCl → K⁺ + Cl^–

Ionisation

1. Formation of positively or negatively charged ions from molecules which are not intially in the ionic state.
   
2. Polar covalent compounds show ionisation.
   e.g. HCl, H₄CO₃ , NH₄OH

Question 3.

Compare the flow of electricity through a nickel wire and nickel sulphate solution.
Answer:
Flow of electricity through nickel wire

1. It is due to the flow of electrons.
2. It is a physical change.
3. It can take place in solid state.
4. Electrical conductivity is more.

low of electricity through nickel sulphate solution

1. It is due to the flow of ions.
2. It is a chemical change.
3. It cannot take place in solid state.
4. Electrical conductivity is less.

Question 4.

Name three organic compounds and one neutral liquid which are non-electrolytes.
Answer:
Organic compounds which are non-electrolyte:
Sugar, glucose, naphthalene
Neutral liquid which are non-electrolytes:
Carbon tetrachloride, carbon disulphide

Question 5.

State which of the following solutions are weak electrolytes – dil. HCI ; carbonic acid ; NH₄OH ; dil. H₂SO₄ ; AgNO₃ ; Na₂CO₃ ; PbBr₂ ; KOH ; HI ; oxalic acid, NaHCO₃ ; sodium acetate ; Na₂SO₄ ; NaOH.
Answer:
Weak Electrolytes: Carbonic acid, NH₄OH, Na₂CO₃ oxalic acid.

Question 6.

State which of the following solutions contain (1) molecules only (2) ions only (3) both molecules and ions – CS₂ ; CH₃COOH ; NH₄OH ; NaOH ; dil. HNO₃ ; Na₂CO₃ ; CuCl₂ ; oxalic acid ; pure H₂O,
kerosene ; HI.
Answer:

 

Question 7.

State giving reasons, in which state or medium does
(1) NaCl, (2) HCI gas (3) NH₃ gas conduct electricity.
Answer:

1. NaCl will conduct electricity only molten state or when dissolved in water. This is because the Na⁺ and Cl^– ions present in solid NaCl are too big to move under the influence of applied electric field.
2. HCI gas is a polar covalent compound when dissolved in water, it will ionise to give H⁺ and CP ions. Under the influence of applied electric field these ions can easily move in an aqueous solution and thus conduct electricity.
   Thus HCI (g) when dissolved in water conducts electricity.
3. NH₃(g) will dissolved in water to give NH₄OH.
   AND NH₃(g) + H₂O (l) → NH₄OH (aq)

1. NH₄OH will ionise to give NH₄⁺ and OH^– ions. Under the influence of applied electric field these ions can migrate in an aqueous solution and hence conduct electricity. Thus NH₃ (g) when dissolved in water conduct electricity.

Question 8.

State on what basis are acids, bases and salts classified as strong and weak electrolytes.
Answer:
Strong electrolytes – Compound which in the fused or in the aqueous solution state are almost completely dissociated and are good conductors of electricity are called.

Weak Electrolytes – Compound which in the fused or in the aqueous solution state are feebly or partially dissociated and are poor conductors of electricity are called – weak electrolytes.

Question 9.

Explain the terms 

1. metal activity or electro chemical series
2. selective discharge of ions.

Answer:

1. Depending on the ease with the metals lose their electrons and form ions – they are arranged in a series known as – metal activity series or electro chemical series.
   The arrangement is so done that the elements that – ionize most readily [discharged with great difficulty]- are placed at the top of the series and other elements in the descending order.
2. Selective discharge of ions: The preferential discharge of ions present in an electrolyte at the respective electrodes is known as selective discharge of ions.

It depends upon the following factors:

• Relative position of the ion in the electo chemical series: Lower the position of the ion in the series, easier to discharge.
• Concentration of the ion: More the concentration of the ion, easier to discharge.
• Nature of the electrode: Inert electrodes (graphite, platinum) do not take part in the electrolytic reaction.
  Active electrodes (Cu electrodes for electrolysis of aq. CuSO₄, Ag electrodes for electrolysis of aq. AgNO₃ etc.) take part in electrolytic reactions.

Question 10.

From the ions –

1. SO₄^2- and OH^1-
2. Cu²⁺ and H¹⁺
   
3. Ag¹⁺ and H¹⁺ state giving reasons which ion is discharged at the respective electrode in each case.

Answer:

1. Out of SO₄^2- and OH^– ions, OH^– ions will be discharged at anode forming O₂ gas.
2. In Out of Cu²⁺ and H⁺, Cu²⁺ ions will be discharged because Cu is lower in the electro chemical series than H.
3. Out of Ag⁺ and H⁺, Ag⁺ will be discharged because Ag is lower in the electro chemical
   series than H.

Question 11.

With reference to nature of electrodes — name three inert and three active electrodes.
Answer:
Inert electrodes

• Platinum
• Iron
• Graphite

Active electrodes

• Copper
• Nickel
• Silver

Question 12.

State the reason for difference in product formed at the anode during electrolysis of aq. CuSO₄ using

1. active electrode – copper anode
2. inert electrode – platinum anode.

Answer:

1. Electrolysis of aq. CuS0₄ using copper anode:
   As copper can easily lose electron, copper from anode will dissolve as Cu²⁺ ions.
   Cu (s) – 2e^–→ Cu²⁺ (aq)
2. Electrolysis of aq. CuS0₄ using inert platinum anode: Due to very low tendency of platinum to lose electron platinum anode does not take part in electrolytic reaction. Further tendency of SO₄^2- to lose electron is much less than that of OH^–(from feebly ionised water). Thus OH^– ions get oxidised in preference to SO₄^2- ions to  give O₂
   4OH – 4e → 4OH
   40H→ 2H₂O + O_2↑

Question 13.

Give the electrode reactions for formation of

1. Lead metal and bromine vapours from molten PbBr₂ using inert electrodes
2. H₂ and O₂ gas (2:1) from acidified water using inert Pt electrodes.

Answer:

1. Electrolysis of molten PbBr₂ using inert electrodes.

1. At cathode: Pb²⁺ (l) + 2e^– →Pb (s)
   At anode: 2Br^– (l) – 2e^–→ 2Br
   Br + Br→ Br₂ (g)
2. Electrolysis of acidified water using Pt electrodes.
   At cathode:

1. 4H⁺ (aq) + 4e^– → 4H
   2H + 2H^–→ 2H₂
   At anode:
   4OH⁺ (aq) – 4e^– → 4OH
   4OH→ 2H₂O (l) + O₂ (g)

Question 14.

Starting from aq. copper (II) sulphate solution, give equations for the reactions at the cathode and anode during electrolysis of aq. CuSO₄ using active copper electrodes.
Answer:

Electrode reaction:

Reaction at Cathode: Cu²⁺ + 2e^– → Cu

Cu²⁺ and H¹⁺ ions migrate to the cathode.
Reaction at anode: Cu – 2e^– → Cu²⁺
S0₄^2- and OH^1- ions migrate to anode but not discharged.

Question 15.

Give reasons for the following changes –

(1) pure water a non-electrolyte – becomes an electrolyte on addition of dil. H₂SO₄
(2) Blue colour of aq. CuSO₄ – turns almost colourless on its electrolysis using Pt electrodes.
Answer:

1. Pure water is a non electrolyte. It consists of entirely of molecules. It can be electrically decomposed by Addition of traces of dil. Sulphuric acid. Which dissociates into H¹⁺ and sulphate (S0₄^2-) ions.
2. If platinum anode is used the blue colour of CuS0₄ solution fades since the blue Cu²⁺ ions which are discharged at the cathode are not replaced or added at the anode.

Question 16.

‘Iron is electroplated with silver’ –

1. define the term in italics
2. state two reasons for electroplating
3. state why the iron is not placed at the anode and silver at the cathode during electroplating.

Answer:

1. Electroplating: The electrolytic process of deposition of a superior metal on the surface of a baser metal or article is called electroplating.
2. Reasons for electroplating:
   
   • Prevents corrosion or rusting.
   • Makes the article attractive and gives it an expensive appearance.
3. The article to be electroplated is always placed at the cathode because during electrolytic reaction the metal is always deposited at the cathode by gain of electrons.

Question 17.

Draw a diagram for –
(1) electroplating an article with silver;
(2) electrorefining or purification of copper.

Answer:
(1)

 

Question 18.

State the

1. electrolyte
2. cathode used
3. anode used
4. electrode reaction at cathode
5. electrode reaction at anode
6. product at cathode and anode – during
   (a) electroplating an article with nickel
   (b) electroplating a spoon with silver
   (c) purification of impure copper.

Answer:

Products at cathode
(a) Ni (ions)
(b) Silver
(c) Copper
Products at anode
(a) SO₄²– and OH^1--(ions)
(b) CN^– and OH^1- (ions)
(c) SO₄^2- and OH^1- (ions)

Question 19.

Give a reason why metals –
copper, silver and lead are electrorefined but K, Na and Ca are not.
Answer:
Extraction of K, Na and Ca are done by electrolysis and are extracted in their fused state. Their oxides are highly stable and the metal has a strong affinity for oxygen. They do not decompose on thermal decomposition. .

Question 20.

Explain the term ‘electrometallurgy’. At what electrode is the extracted metal always deposited ?
Answer:
Electrometallurgy – is the process of extraction of metals by electrolysis. Metals comparatively higher in the electrochemical series are extracted by electrolysis. During this process, the extracted metal is always deposited on cathode.

Question 21.

State how activity series of metals plays a role in extraction of metals from oxides.
Answer:
In activity series, metals are placed in decreasing order of reactivity with most reactive metal at the top and least reactive metal at the bottom. Depending upon the reactivity of metals, different methods are used for their extraction from respective ores.
For example:

1. Extraction by electrolysis: Metals at the top of the activity series are extracted from their ores by electrolysis. Being highly reactive, they cannot be reduced by common reducing agent (C, CO, H₂) For example: K, Na, Ca, Mg,Al.
2. Extraction by common reducing agents: Metals at the middle of the activity series, being less reactive, can be extracted from their ores by reduction with common reducing agents like C, CO, H₂
   For example: Zn, Fe, Pb, Cu.
3. Extraction by thermal decomposition: Metals near the bottom of the activity series, due to their very low reactivity, can be extracted from their ores, by heating only.
   For example: Hg, Ag.
4. Metal at the bottom of the activity series exist in native state (Au, Pt).
5. State the electrode reaction at the respective electrodes during extraction of Al from Al₂ O₃.

Question 22.

State the electrode reaction at the respective electrodes during extraction of Al from Al₂ O₃.
Answer:
At cathod:   2Al³⁺ + 6e^–  → 2Al
At anode:    3O^2- – 6e^– → 3|0|
3|0|+3|0|→3O₂

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