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Chemistry [3 Mark Question Answer]

Electrolysis · ICSE STD 10 (ICSE - English Medium). 8 questions.

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[3 Mark Question Answer]

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Question 13 Marks
Choose A, B, C or D to match the descriptions (i) to (v) below. Some letters may be repeated.
A. non-electrolyte
B. strong electrolyte
C. weak electrolyte
D. metallic conductor
i. Molten ionic compound
ii. Carbon tetrachloride
iii. An aluminum wire
iv. A solution containing solvent molecules, solute molecules, and ions formed by the dissociation of solute molecules.
v. A sugar solution with sugar molecules and water molecules.
Answer
i. Molten ionic compound - strong electrolyte
ii. Carbon tetrachloride - non-electrolyte
iii. Aluminum wire - metallic conductor
iv. A solution containing solvent molecules, solute molecules, and ions formed by the dissociation of solute molecules. - weak electrolyte
v. A sugar solution with sugar molecules and water molecules. - non-electrolyte
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Question 23 Marks
Classify the following substance under three headings:
(a) strong electrolytes (b) Weak electrolytes (c) Non-electrolytes.
Acetic acid, ammonium chloride, ammonium hydroxide, carbon tetrachloride, dilute hydrochloric acid, sodium acetate, dilute sulphuric acid.
Answer
(a) Strong electrolyte: Dilute hydrochloric acid, dilute sulphuric acid, ammonium chloride, sodium acetate
(b) Weak electrolyte: Acetic acid, ammonium hydroxide
(c) Non-electrolyte: Carbon tetrachloride
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Question 33 Marks
A solution of caustic soda (NaOH) in water or when fused, conducts an electric current. What is the similarity in these two cases?
Answer
In the aqueous state, the slightly negatively charged oxygen atoms of the polar water molecule exerts a pull on the positively charged sodium ions. A similar pull is exerted by the slightly charged hydrogen atoms of the water on the negatively charged chloride ions. Thus the ions become free in solution. These free ions conduct electricity.
In the molten state, the high temperatures required to melt the solid weakens the bond between the particles and the ions are set free.
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Question 43 Marks
Electrolysis is a redox process. Explain.
Answer
Electrolysis is a redox process. The reaction at the cathode involves reduction of cations as they gain of electrons while the reaction at anode involves oxidation of anions as they loss of electrons to become neutral.
Example: Dissociation of sodium chloride during electrolysis.
$NaCl ⇌ Na^+ + Cl^-$​​​​​​​
Cathode : $Na^+ + e^- ⟶ Na (reduction)$
$Cl^- − e^- ⟶ Cl(oxidation)$
$Cl + Cl ⟶ Cl_2​​​​​​​$​​​​​​​
Overall reaction : $2NaCl ⟶ 2Na + Cl_2​​​​​​​$
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Question 53 Marks
What is the difference between:
A cation and an anion
Answer
A cation and anion:
CationAnion
1. Are positively charged ions.Are negatively charged ions.
2. Migrate to cathode during electrolysis.Migrate to anode during electrolysis.
3. Gain electron from the cathode and get reduced to become a neutral atom.Lose electrons to the anode and get oxidized to become a neutral atom.
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Question 63 Marks
Describe what happens to the iron object and the copper rod.
Answer
The copper anode continuously dissolves as ions in solution and is replaced periodically. The electrolyte dissociates into $Cu ^{+2}$ ions which migrate towards the iron object taken as the cathode and are deposited as neutral copper atoms on the cathode.
Electrolyte: Aqueous solution of nickel sulphate
Dissociation : $CuSO _4 \rightleftharpoons Cu ^{2+}+ SO _4{ }^{2-}$
$
H _2 O \rightleftharpoons H ^{+}+ OH ^{-}
$
Electrodes:
Cathode: Article to be electroplated
Anode: Block of pure copper
Electrode reactions:
Reaction at cathode: $Cu ^{2+}+2 e ^{-} \rightarrow Cu$ (deposited)
Reaction at anode: $Cu -2 e ^{-} \rightarrow Cu ^{2+}$
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Question 83 Marks
Give reason for the following:
The ratio of hydrogen and oxygen formed at the cathode and anode is 2: 1 by volume.
Answer
As per electrolytic reactions, $4 H ^{+1}$ are needed at cathode and $4 OH ^{-}$at the anode and two molecules of water are produced at the anode. Hence for every two molecules of water, two molecules of hydrogen and one molecule of oxygen are liberated at the cathode and anode respectively.
$
2 H _2 \xrightarrow[\text { Current }]{ H _2 SO _4} 2 H _2[\text { cathode }]+ O _2[\text { anode }]
$
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[3 Mark Question Answer] - Chemistry STD 10 Questions - Vidyadip