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Class 10 Chemistry Chapter 16 Electrochemistry Full Unit Test 8 – 2027
Chemistry 10th Class chapter 16 Full Unit Test
Prepare for your matric board examinations with the Class 10 Chemistry Chapter 16 Electrochemistry Full Unit Test 8, designed for students preparing for the Punjab Board 2027 Model Papers. This complete unit test covers the important concepts of Electrochemistry from the Class 10 Chemistry syllabus and helps students practise examination-style questions, revise key topics, and assess their subject knowledge. Use this Class 10 Chemistry Full Unit Test for comprehensive revision, self-assessment, and effective preparation for the Punjab Board Class 10 Chemistry examination under the 2026–27 syllabus.
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PAPER NO.
08
CHEMISTRY
CHAPTER
WISE
WISE
Roll No.(in Figures):
(in Words):
ABCD
1ABCD
2ABCD
3ABCD
4ABCD
ABCD
5ABCD
6ABCD
7ABCD
8ABCD
ABCD
9ABCD
10ABCD
11ABCD
12ABCD
NOTE:Four possible answers A, B, C and D to each question are given. The choice which you think is correct, fill that circle in front of that question with Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in that question.
Q1.12
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1.
The oxidation number of Mn in K2MnO4:(A) +7(B) +5(C) +6(D) −6Answer: (C) +6 | 2(+1) + Mn + 4(−2) = 0 → Mn = +6
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2.
For monoatomic ions, the oxidation number is equal to:(A) The atomic mass of the ion(B) The charge present on the ion (C) The number of electrons(D) The number of protonsAnswer: (B) The charge present on the ion
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3.
Which element in the following reaction is being reduced?4HBr(aq)+MnO2 (s) MnBr2 (aq)+Br2 (g)+2H2O(l)(A) Br(B) Mn(C) H(D) Both Mn and HAnswer: (B) Mn | Mn goes from +4 (in MnO2) to +2 (in MnBr2), so it gains electrons.
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4.
The formula of ammonium carbonate is:(A) NH4CO3(B) (NH4)2CO3(C) (NH4)3CO3(D) NH4(CO3)2Answer: (B) (NH4)2CO3 | 2 NH4+ balance 1 CO32−
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5.
During electrolysis of dilute solution of H2SO4 using Pt electrodes, which process takes place at the cathode?(A) Reduction(B) Neither oxidation nor reduction (C) Oxidation(D) First oxidation and then reductionAnswer: (A) Reduction | 2H+ + 2e− → H2 at the cathode.
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6.
Reduction is a process in which electrons are:(A) Lost(B) Shared(C) Gained(D) None of theseAnswer: (C) Gained
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7.
Which of the following statements is NOT correct about Zn-Cu electrochemical cell?(A) Anode is negatively charged(B) Reduction occurs at anode (C) Cathode is positively charged(D) Reduction occurs at cathodeAnswer: (B) | Oxidation (not reduction) occurs at the anode.
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8.
Which ions are preferentially discharged at cathode when a metal is below hydrogen in the electrochemical series?(A) Hydrogen ions(B) Metal ions(C) Hydroxide ions(D) Halide ionsAnswer: (B) Metal ions | Metals below hydrogen are reduced more easily than H+.
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9.
Which product will be obtained at the anode when a concentrated solution of NaCl will be electrolysed?(A) Cl2(B) O2 (C) More O2 and less Cl2(D) More Cl2 and less O2Answer: (A) Cl2 | 2Cl− → Cl2 + 2e− (chlorine gas at the anode).
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10.
During electrolysis of CuSO4, as the concentration of copper ions decreases, the blue colour of the solution:(A) Becomes darker(B) Remains the same(C) Fades away(D) Turns redAnswer: (C) Fades away | Blue colour is due to Cu2+ ions.
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11.
Which of the following metals is highest in the electrochemical series?(A) Tin(B) Iron(C) Magnesium(D) ZincAnswer: (C) Magnesium | Order: Mg > Zn > Fe > Sn
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12.
The reactivity series of metals determined by voltage data is called:(A) Activity series(B) Periodic table (C) Electrochemical series(D) Reactivity tableAnswer: (C) Electrochemical series
Answer Key
1. C2. B3. B4. B5. A6. C
7. B8. B9. A10. C11. C12. C
Here is Subjective Part of Paper.
Roll No.(in Figures):
(in Words):
Chapter 16:
Electrochemistry
Electrochemistry
SUBJECTIVE TYPE
(PART – I)
Maximum Marks: 48
Time Allowed :1.45 Hours
Q2. Write short answers to any FIVE (5) questions:(5×2=10)
(i)What happens when electricity is passed through an aqueous solution of NaCl?
Ans. NaCl solution is split into its products (electrolysis). Hydrogen gas is liberated at the cathode, chlorine gas at the anode (in concentrated solution) and NaOH is left in the solution.
Cathode: 2H+ + 2e− → H2↑
Anode: 2Cl− → Cl2↑ + 2e−
(ii)What is the oxidation number of an atom present in a free element?
Ans. It is zero (e.g. Na, H2, O2, S8, Cl2), because the atoms are not bonded to atoms of a different element and there is no transfer of electrons.
(iii)Mention one difference between an electrolytic cell and a galvanic cell.
Ans. In an electrolytic cell electrical energy is converted into chemical energy (non-spontaneous reaction), whereas in a galvanic cell chemical energy is converted into electrical energy (spontaneous reaction).
(iv)Write the formula of aluminum oxide using oxidation numbers.
Ans. Al = +3, O = −2. Criss-cross the numbers:
Al+3 O−2 → Al2O3
(v)Through which electrode, electrons leave the electrolytic cell?
Ans. Electrons leave the electrolytic cell through the anode (where oxidation takes place) and flow to the positive terminal of the battery. They enter the cell through the cathode.
(vi)Define oxidation and reduction in terms of electrons.
Ans. Oxidation: loss of electrons by an atom, ion or molecule, e.g. Na → Na+ + e−.
Reduction: gain of electrons by an atom, ion or molecule, e.g. Cl2 + 2e− → 2Cl−.
Reduction: gain of electrons by an atom, ion or molecule, e.g. Cl2 + 2e− → 2Cl−.
(vii)How does an electrolyte conduct electricity?
Ans. An electrolyte conducts electricity through the movement of ions. In molten state or aqueous solution its ions are free: cations move towards the cathode and anions towards the anode, so charge is carried through the solution.
(viii)Differentiate between anode and cathode.
Ans.
| Anode | Cathode |
|---|---|
| Oxidation occurs here | Reduction occurs here |
| Anions move towards it; electrons leave from it | Cations move towards it; electrons enter at it |
| Positive in electrolytic cell (negative in galvanic cell) | Negative in electrolytic cell (positive in galvanic cell) |
Q3. Write short answers to any FIVE (5) questions:(5×2=10)
(i)What will happen if a strip of Cu metal is dipped in FeSO4 solution?
Ans. Nothing will happen. Copper is less reactive than iron (it lies below Fe in the electrochemical series), so it cannot displace Fe2+ ions from the solution.
Cu + FeSO4 → No reaction
(ii)Which determines what ions will be preferentially discharged at the electrodes?
Ans. The position of the ions in the electrochemical series (ease of discharge), together with the concentration of the ions in the solution and the nature of the electrode.
(iii)Why fuel cells are regarded as environment friendly?
Ans. In a H2–O2 fuel cell the only product is water (2H2 + O2 → 2H2O). No smoke, CO2 or other pollutants are released, so they do not pollute the environment.
(iv)At what temperature must the electrolysis of molten PbCl2 be carried out and why?
Ans. Above its melting point, i.e. about 501°C (above 500°C). Solid PbCl2 does not conduct because its ions cannot move; only in the molten state the ions become free to move and carry current.
(v)How can H2O2 behave as an oxidizing agent?
Ans. Oxygen in H2O2 has oxidation number −1 and can decrease to −2 by gaining electrons, so H2O2 is itself reduced and oxidizes other substances.
H2O2 + 2H+ + 2e− → 2H2O
Example: H2O2 + 2KI → 2KOH + I2
(vi)What happens at the anode during electrolysis of dilute H2SO4?
Ans. OH− ions (from water) are discharged in preference to SO42− ions, and oxygen gas is liberated.
4OH− → 2H2O + O2↑ + 4e−
(vii)Why different products are formed at the electrodes when a dilute and concentrated aqueous solutions of NaCl are electrolysed?
Ans. Because of the concentration of ions. In dilute NaCl, OH− ions of water are discharged at the anode and O2 is formed; in concentrated NaCl the Cl− ions are present in large number and are discharged, giving Cl2. At the cathode H2 is formed in both cases.
(viii)Describe the composition of the electrodes in a fuel cell.
Ans. The electrodes are porous carbon (graphite) or nickel plates containing a catalyst such as platinum or palladium. H2 is supplied at the anode and O2 at the cathode, with an electrolyte such as KOH between them.
Q4. Write short answers to any FIVE (5) questions:(5×2=10)
(i)Why corrosion becomes fast during rainy season?
Ans. Rain water and humid air supply moisture, which acts as an electrolyte. It also dissolves CO2 and SO2 from the air, becoming acidic and more conducting, so the electrochemical process of rusting speeds up.
(ii)Why does rust continue to spread once it starts?
Ans. Rust (Fe2O3·xH2O) is porous and flaky; it does not form a protective layer and peels off. Fresh iron beneath stays exposed to air and moisture, so rusting goes on.
(iii)If a galvanic cell is established between Cu and Ag, what reactions do you expect at the electrodes?
Ans. Cu is more reactive than Ag, so Cu is the anode and Ag the cathode.
Anode (oxidation): Cu → Cu2+ + 2e−
Cathode (reduction): 2Ag+ + 2e− → 2Ag
Overall: Cu + 2Ag+ → Cu2+ + 2Ag
(iv)Describe the process of silver plating.
Ans. The object to be plated is made the cathode, a pure silver rod is the anode, and the electrolyte is AgNO3 solution. On passing current, Ag dissolves from the anode and is deposited on the object.
Anode: Ag → Ag+ + e−
Cathode: Ag+ + e− → Ag
(v)What is galvanic cell?
Ans. A galvanic (voltaic) cell is a device in which a spontaneous redox reaction produces electrical energy, e.g. the Daniel cell.
(vi)What is the function of the porous partition in a Daniel cell?
Ans. It separates the two solutions and prevents their mixing, while allowing ions to pass through so the circuit is completed and the solutions stay electrically neutral.
(vii)How can the chemical reactivity of metals be compared by using electrochemical series?
Ans. Metals higher in the series lose electrons more easily, so they are more reactive and stronger reducing agents. A metal placed higher in the series displaces a lower metal from its salt solution (e.g. Zn displaces Cu from CuSO4).
(viii)Define electrochemical series.
Ans. The arrangement of elements (metals) in order of their tendency to lose electrons, i.e. according to their standard electrode potentials, is called the electrochemical series.
(PART – II)
Note: Attempt any TWO questions.(2 × 9 = 18)
Q5.(a) Explain with examples how the formula of Ionic compound can be written using oxidation numbers.5
Ans. The formula of an ionic compound is written so that the total positive and negative oxidation numbers are equal. Steps:
- Write the symbol of the positive part (metal / cation) first, then the negative part, with their oxidation numbers.
- Criss-cross the numbers (without sign) and write them as subscripts.
- Reduce to the simplest ratio and put polyatomic ions in brackets if their subscript is more than 1.
Examples:
1. Na+1 Cl−1 → NaCl 2. Al+3 O−2 → Al2O3 3. Ca+2 (OH)−1 → Ca(OH)2 4. Mg+2 (PO4)−3 → Mg3(PO4)2 5. Ca+2 O−2 → Ca2O2 → CaO (reduced) (b) Find the oxidation numbers of the indicated elements in the following compounds.4
Al (NO3)3, H2S2O7, Na2 S2 O3
Ans.
(1) N in Al(NO3)3: Let N = x. (+3) + 3(x) + 9(−2) = 0 ⇒ 3x = 15 ⇒ x = +5
(2) S in H2S2O7: Let S = x. 2(+1) + 2(x) + 7(−2) = 0 ⇒ 2x = 12 ⇒ x = +6
(3) S in Na2S2O3: Let S = x. 2(+1) + 2(x) + 3(−2) = 0 ⇒ 2x = 4 ⇒ x = +2
Q6.(a) Write down the formulae of the following compounds using Ionic charges.5
Sodium hypochlorite, calcium oxalate, ammonium phosphate and calcium nitrite
Ans.
| Compound | Ions | Formula |
|---|---|---|
| Sodium hypochlorite | Na+, ClO− | NaClO |
| Calcium oxalate | Ca2+, C2O42− | CaC2O4 |
| Ammonium phosphate | NH4+, PO43− | (NH4)3PO4 |
| Calcium nitrite | Ca2+, NO2− | Ca(NO2)2 |
(b) Discuss the electrolysis of concentrated solutions of metal halides.4
Ans. In a concentrated metal halide solution (e.g. brine, NaCl) the ions present are Na+, Cl−, H+ and OH−.
At cathode: H+ is lower than Na+ in the electrochemical series, so H+ is discharged and hydrogen is evolved.
2H+ + 2e− → H2↑At anode: because of its high concentration Cl− is discharged in preference to OH−, giving chlorine gas.
2Cl− → Cl2↑ + 2e−Na+ and OH− remain in solution, so NaOH is obtained. Overall:
2NaCl + 2H2O → 2NaOH + H2 + Cl2If the metal is less active than hydrogen (e.g. CuCl2), the metal itself is deposited at the cathode (Cu2+ + 2e− → Cu) while the halogen is liberated at the anode.
Q7.(a) Explain electroplating in details.5
Ans. Electroplating is the process of depositing a thin layer of a metal on the surface of another object (usually a cheaper metal) by electrolysis. It is done to prevent corrosion, improve appearance and increase hardness.
Requirements:
- Object to be plated → cathode (connected to negative terminal)
- Pure plating metal → anode (connected to positive terminal)
- Electrolyte → solution of a salt of the plating metal
Example – silver plating of a spoon: the electrolyte is AgNO3 (or K[Ag(CN)2]) solution.
Anode: Ag → Ag+ + e− Cathode: Ag+ + e− → Ag (deposited on the spoon)The concentration of Ag+ in solution remains constant because the anode dissolves as fast as silver is deposited. A smooth, even coating needs a low current and a clean object.
Fig: Silver plating of an object
(b) What is corrosion? Explain the reactions involved in corrosion of iron.4
Ans. Corrosion: the gradual destruction (eating away) of a metal by the chemical or electrochemical action of its surroundings (air, moisture, acids) is called corrosion. The corrosion of iron is called rusting.
Reactions: A drop of water on iron forms a tiny galvanic cell.
Anode (oxidation): 2Fe → 2Fe2+ + 4e− Cathode (reduction): O2 + 2H2O + 4e− → 4OH− Fe2+ + 2OH− → Fe(OH)2 4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3 2Fe(OH)3 → Fe2O3·xH2O (rust) + 3H2OHydrated ferric oxide, Fe2O3·xH2O, is the reddish-brown rust.
Fig: Daniel cell (a galvanic cell) – for Q4(vi)
