Frank Brothers Solutions for ICSE Class 10 Chemistry Chapter 6 Electrolysis

Chapter 6. Electrolysis

Solution 1:

(a) Electrolysis: It is the process of decomposition of the electrolyte in the molten or aqueous state by discharge of ions at the electrodes on passage of an electric current.

(b) Electrolyte: It is a compound which either in aqueous solution or in the molten state allows an electric current to pass through it and is accompanied by discharge of ions and finally into neutral atoms at the two electrodes. For example: Hydrochloric acid.

(c) Non-electrolyte: They are substances which do not conduct electricity in the fused or aqueous state. They contain only molecules and do not ionize. For example: Petrol.

(d) Cation and anion:
Cation: Positively charged ions are called cations. For example: \( Na^+ \)
Anion: Negatively charged ions are called anions. For example: \( Cl^- \)

📝 Teacher's Note: Use simple analogies - electrolysis is like splitting a compound using electricity, just like using a hammer to break something. Draw diagrams showing positive ions moving to negative electrode and vice versa.

🎯 Exam Tip: Always give examples for each definition and mention both "molten" and "aqueous" states when defining electrolytes to get full marks.

Solution 2:

(a) Substances which will behave as strong electrolytes: \( NaCl \), \( NaOH \), \( H_2O \)(pure), dil. \( H_2SO_4 \)

(b) Substances which will behave as weak electrolytes: \( NH_4OH \), acetic acid, \( H_2CO_3 \)

(c) Substances which are non-electrolytes: urea, glucose

📝 Teacher's Note: Explain that strong electrolytes completely break into ions while weak ones only partially break. Use the analogy of a completely broken vs cracked glass to show the difference.

🎯 Exam Tip: Remember that all acids, bases, and salts are electrolytes, but their strength depends on how much they ionize. Organic compounds like glucose are usually non-electrolytes.

Solution 3:

(a) A strong electrolyte and a weak electrolyte

(b) Electrolytic dissociation and ionization

📝 Teacher's Note: Use the bulb test demonstration to show conductivity differences. Explain that ionization creates new ions while dissociation separates existing ions - like making vs separating LEGO blocks.

🎯 Exam Tip: Focus on the key difference - ionization forms new ions from neutral molecules, dissociation separates already existing ions. Always mention examples for full marks.

Solution 4:

A steel rod is connected to the negative terminal and a graphite rod to positive terminal of a battery. A silica crucible is filled to about two thirds with solid lead bromide and the rods are dipped into it and is then melted by heating it over a Bunsen burner.

Lead bromide contains: \( PbBr_2 \rightleftharpoons Pb^{2+} + 2Br^- \)

Reaction at cathode: Lead ions are positively charged they get discharged at cathode. It receives two electrons from cathode and changes to atom of lead and get deposited at the cathode.
\( Pb^{2+} + 2e^- \rightarrow Pb \)

Reaction at the anode: Bromide ions, being negatively charged discharge at anode. It loses its only electron and becomes an atom of bromine. Thus the atoms of bromine combine in pairs to form molecules of bromine which escape as red vapours, at the anode.
\( Br^- + e^- \rightarrow Br \)
\( Br + Br \rightarrow Br_2 \)

📝 Teacher's Note: Emphasize that molten lead bromide must be used because solid doesn't conduct. Show students how electrons flow from cathode to positively charged ions and from negatively charged ions to anode.

🎯 Exam Tip: Always write both half-reactions separately and balance electrons. Remember: cathode attracts cations (positive ions), anode attracts anions (negative ions).

Solution 5:

Three applications of electrolysis are:

1. Electro plating with metals
2. Electrorefining of metals
3. Extraction of metals

📝 Teacher's Note: Give real-life examples for each application - jewelry plating, purifying copper for electrical wires, and extracting aluminum from bauxite. This makes the concepts more relatable.

🎯 Exam Tip: These three applications are the most commonly asked. Always elaborate with one example of each if time permits - like copper plating on iron, purification of crude copper, and extraction of sodium.

Solution 6:

(a) When solid copper sulphate is electrolysed between platinum electrodes the electrolytic reaction will not take place as no electrolytes are formed in solid state.

(b) When the electrolysis of copper sulphate(aqueous) is electrolysed between platinum electrodes the following reaction follows:
\( CuSO_4 \rightleftharpoons Cu^{2+} + SO_4^{2-} \)
\( H_2O \rightleftharpoons H^+ + OH^- \)

Reaction at cathode: Copper ions and hydrogen ions migrate towards the cathode. \( Cu^{2+} \) ions being lower in the electrochemical series are preferentially discharged to hydrogen ions to form neutral copper atoms at platinum cathode.
\( Cu^{2+} + 2e^- \rightarrow Cu \)

Reaction at anode: Sulphate ions and hydroxyl ions migrate towards the anode. Hydroxyl ions are preferentially discharged to sulphate ions at anode, to form neutral particles of \( OH^- \). The electrically neutral hydroxyl reacts among themselves to give water and oxygen.
\( OH^- - e^- \rightarrow OH \)
\( 4(OH) \rightarrow 2H_2O + O_2 \)

(c) When Aqueous copper sulphate is electrolysed between copper electrodes then the ions formed are
\( CuSO_4 \rightleftharpoons Cu^{2+} + SO_4^{2-} \)
\( H_2O \rightleftharpoons H^+ + OH^- \)

Reaction at the cathode: Copper and hydrogen ions both being positively charged migrate towards the cathode. Copper ions are discharged in preference to hydrogen ions. Copper gains two electrons from the cathode and changes into an atom of copper. The atoms of copper are deposited at the cathode. The atoms of copper are deposited at the cathode ad form a layer of pink copper metal which gradually turns reddish brown.
\( Cu^{2+} + 2e^- \rightarrow Cu \)

Reaction at the anode: Anode receives electrons from the ions and supplies them to the cathode. The atoms of copper from the anode changes into ions of copper which go into the solution and the electrons liberated in this change are taken up by the anode.
\( Cu - 2e^- \rightarrow Cu^{2+} \)

Thus for every copper ion discharged at the cathode, an ion of copper is formed at the anode which goes into the solution. Thus the atoms of copper are deposited at the cathode, the cathode becomes thicker and the atoms of copper from the anode change into ions of copper, the anode becomes thinner.

Therefore there is transference of copper atoms from anode to cathode.

📝 Teacher's Note: Emphasize the difference between inert (platinum) and active (copper) electrodes. With copper electrodes, the anode participates in the reaction, making it perfect for electroplating and purification.

🎯 Exam Tip: For copper sulphate electrolysis, remember the key difference: with platinum electrodes, oxygen is produced at anode; with copper electrodes, copper dissolves from anode. This is crucial for scoring full marks.

Solution 8:

The following are the factors that influence the preferential discharge of ions at the electrode:

i) Position of the metallic ion in electrochemical series: If all the factors remain the same, an ion placed lower in the electrochemical series gets preferentially discharged at the respective electrode in comparison to all those ions, which are placed above it in the series.

ii) Concentration of ions in the electrolyte: Higher the concentration of negative ion in the electrolytic solution, greater is its probability of being discharged at the anode.

iii) Nature of the electrode: If the electrode used is inert i.e made of less reactive material such as graphite, platinum etc, the electrode does not play any role in deciding the preferential discharge of an ion at it.

If the electrode used is active i.e made of active material such as Cu, Ag, Ni etc it takes part in the electrode reaction and plays an important role in deciding the ions which will preferentially be discharged. In such a case, anions migrate to the anode but do not get discharged. Instead the active anode itself loses electrons and form ions.

📝 Teacher's Note: Use the electrochemical series chart frequently in class. Explain that higher concentration means more ions available for discharge, like having more people in a queue. Active vs inert electrodes is like having a participating vs non-participating referee.

🎯 Exam Tip: Remember the three factors in order: electrochemical series position, concentration, and electrode nature. Always mention that lower position in series means easier discharge for better marks.

Solution 9:
Answer:
a) Distilled water is a non electrolyte because it does not contain any ions.
b) The electrolytic cell used in electrolysis of acidulated water is called Hoffmann's Apparatus.
c) (i) At cathode: Hydrogen ions are \( H^+ \) ions. They migrate to the cathode and discharge there. The ions gain electrons from the cathode to form atoms of Hydrogen which combine in pairs to form molecules of Hydrogen.
\( H^+ + e^- \rightarrow H \)
\( H + H \rightarrow H_2 \)
or \( 4H^+ + 4e^- \rightarrow 2H_2 \)
(ii) At anode: Sulphate ions and hydroxyl ions are the anions present in the solution. Both migrate to the anode. \( OH^- \) ions are discharged in preference to sulphate ions. The \( OH^- \) ions loose their electrons and become electrically neutral particles of OH, which react among themselves to give water and oxygen.
\( 2OH^- - 2e^- \rightarrow H_2O + O \)
or \( 4OH^- - 4e^- \rightarrow 4OH \)
\( 4OH \rightarrow 2H_2O + O_2 \)
d) Electrolytic Cell: Hoffmann Voltameter
Electrolyte: Acidified water
Electrode: Cathode - Platinum Foil, Anode - Platinum Foil
Dissociation of acidified water:
\( H_2SO_4 \rightleftharpoons 2H^+ + SO_4^{2-} \)
\( H_2O \rightleftharpoons H^+ + OH^- \)
Reaction at cathode:
\( H^+ + e^- \rightarrow H \)
\( H + H \rightarrow H_2 \)
or \( 4H^+ + 4e^- \rightarrow 2H_2 \)
Reaction at anode:
\( 2OH^- - 2e^- \rightarrow H_2O + O \)
or \( 4OH^- - 4e^- \rightarrow 4OH \)
\( 4OH \rightarrow 2H_2O + O_2 \)
e) Electrolysis of acidulated water is considered as an example of catalysis because the reaction is catalyzed by acids.
In simple words: This solution explains how water with acid added can be broken down using electricity to produce hydrogen and oxygen gases at different electrodes.

📝 Teacher's Note: Use the Hoffmann apparatus diagram to show students how hydrogen collects in one tube and oxygen in the other. The 2:1 volume ratio visually demonstrates the chemical formula of water.

🎯 Exam Tip: Always mention that pure water doesn't conduct electricity and explain why acid is added - examiners look for understanding of ionization and conductivity.

Solution 10:
Answer: In order to get a spoon plated with silver a solution of sodium silver cyanide can be taken as electrolyte. If silver nitrate was chosen as electrolyte then the deposition of silver will be very fast and hence not very uniform and smooth.
In simple words: To coat a spoon with silver properly, we use a special silver solution that deposits the silver coating slowly and evenly, making it smooth and shiny.

📝 Teacher's Note: Show students actual electroplated items and explain why slow deposition gives better finish. Compare it to painting - rushing gives uneven results.

🎯 Exam Tip: Mention both the correct electrolyte choice and explain why other options won't work - this shows deeper understanding of the process.

Solution 11:
Answer:
(i) Electro refining: It is the process of refining the impure metals through the use of electric current through the use of electric current or electrolysis.
(ii) Electro metallurgy: It is the process of extraction of metal from its ore through the use of electric current.
(iii) Anode mud: During electro refining some impurities which are insoluble fall down near the anode and is known as anode mud.
In simple words: These are three different ways electricity is used to get pure metals - either cleaning dirty metals, extracting metals from rocks, or dealing with leftover waste.

📝 Teacher's Note: Use copper purification as a concrete example for electro refining. Show how impure copper becomes pure through this electrical process.

🎯 Exam Tip: Define each term clearly and give one practical example for each - electro metallurgy (aluminum), electro refining (copper), anode mud (impurities).

Solution 12:
Answer:
a) Anode, lack
b) current, Cathode, excess
c) gain, Cations
d) loose, anions
e) Electropositive, loose
f) Electronegative, gain
In simple words: This fill-in-the-blanks shows how electricity moves through solutions and how different charged particles behave at positive and negative electrodes.

📝 Teacher's Note: Use the memory trick: "Cats go to negative" (Cations go to Cathode which is negative) and "Ants go to positive" (Anions go to Anode which is positive).

🎯 Exam Tip: Remember that electrons flow from anode to cathode, making cathode negative and anode positive in electrolytic cells - opposite to galvanic cells.

Solution 13:
Answer:
a) The order of discharge of ions is \( Cu^{2+}, Zn^{2+}, Al^{3+}, Na^+ \) because as the concentration is same, an ion placed lower in electrochemical series get preferentially discharged.
b) \( OH^- \) is likely to discharge first in comparison to \( Br^- \) ions.
In simple words: Some metals and ions are more eager to gain or lose electrons than others, just like some people are more eager to get in line first - there's a definite order.

📝 Teacher's Note: Teach the electrochemical series as a "preference list" - ions higher up get discharged first when concentration is equal. Use practical examples like copper plating.

🎯 Exam Tip: Learn the electrochemical series order and always mention that this applies when concentrations are equal - concentration can override the series.

Solution 14:
Answer: Aluminium is extracted by the electrolysis of fused pure alumina dissolved in fused mixture of cryolite and fluorspar. This mixtures lowers the melting point of alumina and increases the electrical conductivity and the electrolysis is carried out in an iron tank lined with gas carbon. This lining of a gas carbon serves as cathode. The anode consists of a number of carbon rods which dip in fused electrolyte. On passing electric current following reaction takes place.
At cathode: \( Al^{3+} + 3e^- \rightarrow Al \)
At anode: \( O^{2-} - 2e^- \rightarrow O \)
\( O + O \rightarrow O_2 \)
Certain metal oxides are highly stable and so cannot be reduced by conventional reducing agents like coke, carbon monoxide or hydrogen. They are extracted from their oxides or salts by the electrolysis in fused state. For e.g., metals like aluminium, sodium, potassium, magnesium, titanium, calcium etc. are extracted by this method.
In simple words: Some metals like aluminum are so tightly stuck to oxygen that only electricity can separate them - heat and chemicals alone won't work.

📝 Teacher's Note: Explain why aluminum cannot be extracted by simple heating unlike iron. The aluminum-oxygen bond is much stronger, requiring electrical energy to break.

🎯 Exam Tip: Mention the role of cryolite (lowers melting point) and write both cathode and anode reactions clearly with proper electron transfer.

Solution 15:
Answer: During the electrolysis of copper sulphate using copper electrode, the number of copper ions remains same as that for every copper ions discharged at the cathode, an ion of copper is formed at the anode which goes in the solution. Thus the colour intensity does not change.
In simple words: When using copper electrodes with copper sulphate solution, the blue color stays the same because for every copper ion removed, a new one is added back.

📝 Teacher's Note: Demonstrate this with actual copper sulphate solution and show how the color remains constant unlike when using inert electrodes where the color fades.

🎯 Exam Tip: Contrast this with inert electrodes where the color would fade - explain the difference clearly to show understanding.

Solution 16:
Answer: Electrolytic refining is used for the purification of copper. In this case
Cathode: Pure copper strip
Anode: Impure copper
Electrolyte: A solution of copper sulphate and dilute sulphuric acid
When the current is passed through the electrolyte, the copper ions of the copper sulphate solution are attracted to the cathode where they gain electrons and are deposited on pure copper strips. The impure copper loses electrons and passes into solution as soluble copper ions.
At cathode: \( Cu^{2+} + 2e^- \rightarrow Cu \)
At anode: \( Cu + 2e^- \rightarrow Cu^{2+} \)
In simple words: This is like moving all the pure copper from dirty copper to clean copper using electricity - the impurities get left behind.

📝 Teacher's Note: Explain that impurities either remain in solution or fall as anode mud. Pure copper deposits only at the cathode, leaving impurities behind.

🎯 Exam Tip: Clearly state what happens to impurities (some dissolve, some form anode mud) and explain why only pure copper deposits at the cathode.

Solution 17:
Answer: The main applications of electrolysis are:
1. Electro plating with metals
2. Electrofining of metals
3. Extraction of metals
In simple words: Electrolysis is used to coat objects with metals, clean up dirty metals, and extract metals from rocks.

📝 Teacher's Note: Give practical examples for each application - jewelry plating, copper purification, and aluminum production from bauxite ore.

🎯 Exam Tip: Always give one specific example for each application to demonstrate understanding - electroplating (silverware), refining (copper), extraction (aluminum).

Solution 18:
Answer: Electrolytic Dissociation is the dissociation of an electrovalent compound into ions in the fused state or in aqueous solution state. Reversible breakdown of a chemical compound into simpler substances by heating it. The splitting of ammonium chloride into ammonia and hydrogen chloride is an example. On cooling, they recombine to form the salt.
In simple words: Some compounds can break apart into charged pieces when dissolved in water or melted, and some can be split by heat but stick back together when cooled.

📝 Teacher's Note: Distinguish between electrolytic dissociation (ionic compounds in solution) and thermal decomposition (reversible breakdown by heat).

🎯 Exam Tip: Give the ammonium chloride example for thermal decomposition and explain how it differs from ionic dissociation in water.

Solution 19:
Answer: A solution of NaOH dissociates by electrolytic dissociation. And fused NaOH dissociates by thermal dissociation. The similarity between both of these is that they both liberate same ions.
\( NaOH \rightleftharpoons Na^+ + OH^- \)
(aq. or fused)
In simple words: Whether you dissolve sodium hydroxide in water or melt it with heat, it breaks into the same charged pieces - sodium and hydroxide ions.

📝 Teacher's Note: Emphasize that the end result (Na⁺ and OH⁻ ions) is the same regardless of whether you use water or heat to break the compound apart.

🎯 Exam Tip: Write the dissociation equation and mention both methods produce the same ions - this shows understanding of the similarity.

Solution 20:
Answer:
1. \( Na_2CO_3 \)
2. \( NH_3 \)
3. Graphite, Cu electrode
4. \( NH_4^+ \)
5. Graphite
In simple words: These are answers to fill-in-the-blank questions about various chemical formulas and electrode materials used in electrolysis.

📝 Teacher's Note: Review each chemical formula and electrode type, ensuring students understand why each specific material is chosen for particular reactions.

🎯 Exam Tip: Learn common chemical formulas and electrode materials - graphite is often used because it's inert and conducts electricity well.

Solution 1994-1:
Answer:
1. The article to be plated must be made Cathode.
2. The ions of the metal which is to be electroplated must be present in the electrolyte.
3. The metal to be plated on the article must be made anode. It needs to be periodically replaced.
In simple words: For electroplating, the object goes on the negative side, the coating metal goes on the positive side, and the solution must contain ions of the coating metal.

📝 Teacher's Note: Use a simple setup like plating a key with copper to demonstrate these three essential requirements in practice.

🎯 Exam Tip: Remember all three conditions must be met - many students forget that the anode must be made of the plating metal.

Solution 1994-2:
Answer: The passage of electricity through an electrolyte occurs through ions furnished by the electrolyte where as the passage of electricity through a copper wire occurs through electrons.
In simple words: Electricity moves through liquids using charged particles (ions) but moves through metal wires using electrons.

📝 Teacher's Note: Draw diagrams showing electron flow in wires versus ion movement in solutions to visualize this fundamental difference.

🎯 Exam Tip: Always mention both - ions in electrolytes and electrons in metals - to show complete understanding of electrical conduction.

Solution 1995-1:
Answer:
(a) The ions of the silver must be present in the electrolyte.
(b) Anode should be made of pure clean silver.
(c) The cathode is made of copper wire.
(d) The equation for the reaction which takes place at the cathode is
\( Ag^+ + e^- \rightarrow Ag \)
In simple words: To plate something with silver, you need silver ions in the solution, pure silver as the positive electrode, and the reaction shows silver ions gaining electrons to become solid silver.

📝 Teacher's Note: Emphasize the electron transfer equation - this is fundamental to understanding all electroplating reactions.

🎯 Exam Tip: Always write the electrode reaction with proper electron notation (e⁻) and balanced equations for full marks.

Solution 1995-2:
Answer: It is the process of decomposition of an electrolyte in the molten or aqueous state by discharge of ions at the electrodes on the passage of an electric current.
In simple words: Electrolysis is using electricity to break apart compounds that are either melted or dissolved in water.

📝 Teacher's Note: Emphasize both conditions - molten OR aqueous state - and that ions must be free to move for electrolysis to occur.

🎯 Exam Tip: Include all key terms: decomposition, electrolyte, discharge of ions, and electric current in your definition for complete marks.

Solution 1995-3:
Answer: Pure water does not conduct electricity because the degree of ionization is low. Thus to make it a good conductor of electricity acid is added to it which will increase the degree of ionization.
In simple words: Pure water has very few charged particles, so electricity can't flow through it easily - adding acid creates more charged particles.

📝 Teacher's Note: Demonstrate with a conductivity tester showing pure water versus acidulated water to prove the difference in electrical conduction.

🎯 Exam Tip: Mention that pure water has very few ions due to low ionization, and acid increases the number of ions available for conduction.

Solution 1996-1:
Answer: Substance which contain
1. Ions only:- HCl
2. Molecules only:- Petrol
3. Both ions and molecules:- CH₃COOH
In simple words: Some substances have only charged particles (like HCl), some have only neutral molecules (like petrol), and some have both (like vinegar).

📝 Teacher's Note: Explain why HCl completely ionizes, petrol doesn't ionize at all, and acetic acid partially ionizes in water.

🎯 Exam Tip: Learn examples for each category - strong acids/bases (complete ionization), organic compounds (no ionization), weak acids/bases (partial ionization).

Solution 1996-2:
Answer:
1. Electrolyte is a compound which either in aqueous solution or in molten state allows an electric current to pass through it and is accompanied by discharge of ions and finally into neutral atoms at the two electrodes.
2. Non-electrolyte are substances which do not conduct electricity in fused or aqueous state. They contain only molecules and do not ionize. For example: petrol, alcohol.
3. If the electrolyte is described as 'strong electrolyte' it means it completely dissociates into its constituting ions in aqueous solution.
In simple words: Electrolytes let electricity pass through and break into ions, non-electrolytes don't conduct at all, and strong electrolytes break apart completely.

📝 Teacher's Note: Use practical examples like salt water (electrolyte), sugar water (non-electrolyte), and explain the difference in terms of ion formation.

🎯 Exam Tip: Give specific examples for each type and explain the key difference - presence or absence of free-moving ions.

Solution 1996-3:
Answer:
1. As for every copper ion discharged at the cathode, an ion of copper is formed at the anode which goes into the solution. Since atoms of copper are deposited at the cathode, the cathode becomes thicker and as the atoms of copper from the anode change into ions of copper, the anode becomes thinner.
2. When platinum rods are used as electrodes, then the blue colour of copper sulphate solution fades and sulphuric acid is formed. This is because oxygen is liberated at anode and copper metal is deposited at cathode.
3. Practical application of electrolysis of copper sulphate solution: This is the basis for purification of copper. Other metals like Zinc, Nickel, Silver, Lead can also be purified.
In simple words: With copper electrodes, the copper moves from one electrode to the other keeping the solution the same, but with other electrodes, the solution changes as it breaks down into its parts.

📝 Teacher's Note: Compare the two scenarios side by side - copper electrodes (no color change) versus inert electrodes (color fades) to highlight the difference.

🎯 Exam Tip: Always specify which type of electrode is being used - the behavior is completely different between copper and inert electrodes.

Solution 1997-1:
Answer: Lead Bromide should be in the molten state if it has to conduct electricity.
In simple words: Lead bromide can only conduct electricity when it's melted because that's when its ions can move freely.

📝 Teacher's Note: Explain that in solid state, ions are locked in crystal structure and cannot move, but melting frees them to carry current.

🎯 Exam Tip: Always mention that ionic compounds conduct only when ions are free to move - either molten or in aqueous solution.

Solution 1997-2:
Answer: Lead Bromide contains only positively charged lead ions and negatively charged bromide ions.
\( PbBr_2 \leftrightarrow Pb^{2+} + 2Br^- \)
In simple words: When lead bromide breaks apart, it forms positive lead ions and negative bromine ions in a 1:2 ratio.

📝 Teacher's Note: Show the ionic formula and emphasize the 2+ charge on lead and 1- charge on bromide, explaining why you need two bromide ions per lead ion.

🎯 Exam Tip: Write the balanced dissociation equation showing correct charges and stoichiometry for full marks.

Solution 1997-3:
Answer: Equations for the reactions, which takes place at the electrodes:-
At Cathode:
\( Pb^{2+} + 2e^- \rightarrow Pb \)
At anode:
\( Br^- + e^- \rightarrow Br \)
\( Br + Br \rightarrow Br_2 \)
In simple words: At the negative electrode, lead ions gain electrons to become solid lead, while at the positive electrode, bromine ions lose electrons to form bromine gas.

📝 Teacher's Note: Emphasize electron gain at cathode (reduction) and electron loss at anode (oxidation) - use the mnemonic "An Ox, Red Cat".

🎯 Exam Tip: Always show electron transfer clearly and balance the equations properly - include the formation of Br₂ molecules from Br atoms.

Solution 1998-1:
Answer:
1. Electrolyte
2. Nickel
3. Cathode
4. Anode
5. Cations
In simple words: These are fill-in-the-blank answers related to electroplating and electrolysis terminology.

📝 Teacher's Note: Review each term in context - ensure students understand the role of each component in electrochemical processes.

🎯 Exam Tip: Learn the basic vocabulary of electrolysis - these terms appear frequently in exam questions across different contexts.

Solution 1999-1:
Answer: The electrolysis of lead bromide liberates lead at cathode and bromine at anode.
In simple words: When electricity passes through molten lead bromide, solid lead forms at one electrode and bromine gas forms at the other.

📝 Teacher's Note: This is a classic example of electrolysis of a simple ionic compound - use it to teach the general principles.

🎯 Exam Tip: Remember that metals always form at the cathode and non-metals at the anode in simple ionic compound electrolysis.

Solution 1999-2:
Answer: When a fused metallic chloride is electrolyzed, the metal is obtained at cathode.
In simple words: In any molten metal chloride electrolysis, the pure metal will always appear at the negative electrode.

📝 Teacher's Note: This is a general rule for all metallic chlorides - metals always deposit at cathode regardless of which specific metal it is.

🎯 Exam Tip: This applies to all metallic halides - the metal always goes to the cathode where it gains electrons to become neutral atoms.

Solution 2000-1:

1. Strong electrolytes – dilute hydrochloric acid, dilute sulphuric acid, Ammonium chloride
2. Weak electrolyte – Acetic acid, Ammonium hydroxide
3. Non-electrolytes – Carbon tetrachloride

Solution 2002-1:

1. molecules.
2. will not

Solution 2002-2:

1. When sulphuric acid is added to water it becomes good conductor as addition of sulphuric acid causes dissociation of water molecules into H+ and OH- ions which are then responsible for conduction of electricity by pure water. The water thus obtained is called acidified water.
2. Cathode, Anode

Solution 2002-3:

Solution 2003-1:

electricity, chemical

Solution 2004-2:

1. Molecules are found in a liquid compound which is a non-electrolyte.
2. Non ionized molecules; H+ and X- particles will be present in dilute solution.
3. Loss, Gain
4. The ions of the metal which is to be electroplated on the article must be present in a solution.
5. Redox reaction is one in which oxidation and reduction occurs simultaneously.
6. Similarly in case of electrolysis:
At cathode: The cations gain electron and become neutral. As the electrons are gained the ion is said to be reduced.
At anode: The anions lose electron to form neutral atoms. As the electrons are lost the ion is said to be oxidized.
Hence in electrolysis also the oxidation and reduction occurs hence it is an example of Redox reaction.

Solution 2004-1:

(a) X \( \rightarrow \) X²⁺ + 2e⁻
Y + 3e⁻ \( \rightarrow \) Y³⁻

(b) 3X + Y₂ \( \rightarrow \) X₃Y₂

(c) (i). Electroplating of metals.
(ii). Electrorefining of metals.

(d) If the compound formed between X and Y is melted and an electric current passed through the molten compound, the element X will be obtained at the cathode and Y at the anode of the electrolytic cell.

Solution 2005-1:

1. Copper metal is solid and has no mobile ions whereas an electrolyte should dissociate into oppositely charged ions to conduct the electric current.
2. Hydrogen is released at the cathode when acidulated water is electrolyzed.
3. In sodium chloride, Na+ and Cl- ions are not free to carry the electric current.
4. (a) Reduced
(b) Higher

Solution 2006-1:

(a) (i) Electrode A – Anode
Electrode B – Cathode
(ii) Electrode A.

(b) AgNO₃ will turn blue.

(c) Reaction at cathode:
H⁺ + e⁻ → H
H + H → H₂
or 4H⁺ + 4e⁻ → 2H₂

Reaction at anode:
2OH⁻ – 2e⁻ → H₂O + O
or 4OH⁻ – 4e⁻ → 4OH
4OH → 2H₂O + O₂

Solution 2006-2:

(a) (i) Electrode A – Anode
Electrode B – Cathode
(ii) Electrode A.

(b) AgNO₃ will turn blue.

(c) Reaction at cathode:
H⁺ + e⁻ → H
H + H → H₂
or 4H⁺ + 4e⁻ → 2H₂

Reaction at anode:
2OH⁻ – 2e⁻ → H₂O + O
or 4OH⁻ – 4e⁻ → 4OH
4OH → 2H₂O + O₂

Solution 2007-1:

1. Molten ionic compound – Strong electrolyte
2. Carbon tetrachloride– Non-electrolyte
3. An aluminium wire– Metallic conductor
4. A solution containing solvent molecules, solute molecules and ions formed by the dissociation of solute molecules– weak electrolyte
5. A sugar solution with sugar molecules and water molecules– Non-electrolyte

Solution 2007-2:

(a) Bauxite is reacted with sodium hydroxide to obtain pure aluminium oxide.
(b) 2Al(OH)₃ → Al₂O₃ + 3H₂O
(c) Carbon serves both as anode and cathode.
(d) Reaction at cathode:
Al³⁺ + 3e⁻ → Al

(e) Reaction at cathode:
Al³⁺ + 3e⁻ → Al

(f) Reaction at anode:
O²⁻ – 2e⁻ → O
O + O → O₂

Solution 2008-1:

(d) Lead is deposited at the cathode

Solution 2008-2:

(a) The reaction takes place at anode. Yes, this is an example of oxidation.

(b) Cu²⁺ will be discharged first.
Cu²⁺ + 2e⁻ → Cu

(c) Carbon tetrachloride is a non-electrolyte as it is a covalent compound and contains only molecules.

Solution 2009-2:

Mg(OH)2 as it is basic while rest are amphoteric.

Solution 2009-3:

Molten Lead bromide conducts electricity.

Solution 2009-4:

1. Nickel ions move towards cathode.
2. Nickel ions.