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Class 10 Chemistry Chapter 16 Electrochemistry Test 5 | Board Model Paper
Grade 10 Chemistry Chapter 16 Chapter wise Test
Prepare for your matric board examinations with the Class 10 Chemistry Chapter 16 Electrochemistry Class Test 5, covering topics 16.1 to 16.9. This practice test is designed for students preparing with the Punjab Board 2027 Model Papers and the Class 10 Chemistry syllabus. Test your understanding of Electrochemistry, revise important concepts, and improve your exam preparation through chapter-wise practice. Students can use this 10th Class Chemistry Chapter 16 Test 5 for revision, self-assessment, and preparation for Punjab Board Chemistry examinations.
Chapter Wise Class Test
T-5
Chapter-16
Electrochemistry
1.1 Choose the correct Answer.(Time: 7 Minutes)(Total Marks: 8)
1.Answer: (c)
The oxidation number of an element in its free state is:
(a) +1(b) −1(c) 0(d) +2
2.Answer: (c)
In monoatomic ions, oxidation number is equal to:
(a) Atomic mass(b) Valency(c) Charge on ion(d) Number of neutrons
3.Answer: (b)
Chlorine usually shows oxidation number −1 when combined with:
(a) More electronegative elements(b) Less electronegative elements(c) Noble gases(d) Metals only
4.
Answer: (b)
Formula of Aluminium Sulphate is
(a) Aℓ SO4(b) Aℓ2(SO4)3(c) Aℓ SO4(d) Aℓ (SO4)3
5.
Answer: (c)
Which of the following is a reducing agent?
(a) O2(b) MnO2(c) CH4(d) Fe+3
6.
Answer: (d)
If oxidation at the anode stops, the flow of electricity will:
(a) Increase(b) Decrease(c) Continue normally(d) Stop
7.
Answer: (b)
If graphite were not used as the anode, chlorine gas might:(Analytical / Conceptual)
(a) Stop forming(b) React with the electrode(c) Turn into hydrogen(d) Dissolve in water
8.
Answer: (b)
In concentrated aqueous metal halide solutions, halide ions are discharged at the:
(a) Cathode(b) Anode(c) Salt bridge(d) Battery
786Times • Chapter Wise Class Test T-5 • Chapter-16 Electrochemistry
Chapter Wise Class Test
T-5
Chapter-16
Electrochemistry
Total Marks: 17(Subjective Type)Time: 33 Minutes
2.Write short answers to any six (6) questions.[6 × 2 = 12]
- (i)Calculate the oxidation number of nitrogen in NH3.Let oxidation number of N = x. In NH3: x + 3(+1) = 0, so x + 3 = 0 and x = −3.
- (ii)Explain briefly how oxidation numbers help in writing chemical formulas.The total positive oxidation number of the cations must equal the total negative oxidation number of the anions (compound is neutral). Cross-multiplying the numbers gives the subscripts. Example: Al+3 and SO4−2 give Aℓ2(SO4)3.
- (iii)Why is the formula of ammonium carbonate written as (NH4)2CO3?NH4+ has charge +1 and CO32− has charge −2. Two NH4+ ions are needed to balance one CO32− ion. The bracket with subscript 2 shows that the whole NH4 group is taken twice.
- (iv)Explain why oxidation and reduction always occur together.Oxidation is loss of electrons and reduction is gain of electrons. Free electrons cannot exist alone, so the electrons lost by one substance must be gained by another. Hence both reactions happen at the same time (redox reaction).
- (v)Explain the colour change observed during the reaction of Mn7+ to Mn2+?Mn7+ (as MnO4−) is purple. When it gains electrons and is reduced to Mn2+, the solution becomes colourless or very pale pink, so the purple colour disappears.
- (vi)Why sodium hydroxide is formed as a by-product?In electrolysis of brine (NaCl solution), H+ ions are discharged at the cathode as H2 instead of Na+. The Na+ ions remain in solution together with OH− ions from water, so NaOH is formed as a by-product.
- (vii)Why is copper deposited at the cathode during electrolysis of dilute copper chloride?Both Cu2+ and H+ (from water) move to the cathode. Copper is lower than hydrogen in the electrochemical series, so Cu2+ ions are discharged more easily: Cu2+ + 2e− → Cu.
- (viii)Explain how electrochemical series helps in predicting products of electrolysis.The series arranges elements in order of their tendency to lose electrons. At the cathode the ion lower in the series is discharged first, so metals above hydrogen (Na, K, Ca) are not deposited from aqueous solution and H2 is formed, while metals below hydrogen (Cu, Ag) are deposited.
- (ix)Write the formula of ammonium phosphate using oxidation numbers.NH4+ (+1) and PO43− (−3). Three NH4+ balance one PO43−, so the formula is (NH4)3PO4.
Note: Attempt any one question.[5 × 1 = 5]
3.
Write the rules to assign oxidation number to the atoms. Give examples.
Rules for assigning oxidation numbers:
- Oxidation number of an atom in its free (elemental) state is zero. Examples: Na, O2, Cl2 = 0.
- For a monoatomic ion, it equals the charge on the ion. Examples: Na+ = +1, Cl− = −1, Mg2+ = +2.
- Hydrogen is usually +1 (in NaH it is −1).
- Oxygen is usually −2 (in peroxides such as H2O2 it is −1).
- Group 1 metals are +1 and group 2 metals are +2 in compounds.
- Halogens show −1 when combined with less electronegative elements. Example: Cl in NaCl = −1.
- The sum of oxidation numbers in a neutral compound is zero, and in a polyatomic ion equals its charge. H2SO4: 2(+1) + S + 4(−2) = 0 → S = +6
4.
What is electrolysis? Explain its process indetail.
Electrolysis: the chemical decomposition of an electrolyte by passing a direct electric current through it (molten or in solution).
- Apparatus: an electrolytic cell with two inert electrodes (graphite) dipped in the electrolyte, joined to a DC source. Cathode = negative, anode = positive.
- Ionisation (example: molten NaCl): NaCl(l) → Na+ + Cl−
- At cathode (reduction): positive ions gain electrons. Na+ + e− → Na
- At anode (oxidation): negative ions lose electrons. 2Cl− → Cl2 + 2e−
- Overall: 2NaCl(l) → 2Na(l) + Cl2(g)
786Times • Chapter Wise Class Test T-5 • Chapter-16 Electrochemistry
