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Class 10 Chemistry Chapter 16 Electrochemistry Test 7 – full unit Test
Grade 10 Chemistry chapter 16 Full Unit Test
Prepare for your matric board examinations with the Class 10 Chemistry Chapter 16 Electrochemistry Whole Unit Test 7, based on the Punjab Board 2027 Model Papers. This complete unit test covers the Electrochemistry chapter and helps students revise important concepts, practise exam-style questions, and assess their understanding of Class 10 Chemistry. Students preparing for the Punjab Board Class 10 Chemistry Examination 2027 can use this Electrochemistry whole chapter test for revision, self-assessment, and effective exam preparation according to their syllabus and paper pattern.
Full Chapter Test
T-7
CHAPTER # 16
Time:15 mint(Objective Type)Total Marks: 12
1
2
3
4
5
6
7
8
9
10
11
12
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.
1.Answer: (a) — S: 2(+1) + x = 0, so x = −2
The oxidation number of sulphur in H2S is:
(a) −2(b) +2(c) +6(d) 0
2.Answer: (c) — A neutral compound has total charge zero
The sum of oxidation numbers in a neutral compound is always:
(a) +1(b) −1(c) 0(d) +2
3.Answer: (b) — Metals below H are discharged more easily at the cathode
If a metal is placed below hydrogen in the electrochemical series, it will:
(a) Form hydrogen gas(b) Be discharged at cathode(c) Move to anode(d) Remain in solution
4.
Answer: (b) — Oxidation (OH− loses electrons) occurs at the anode
In this electrolysis, oxygen gas is produced at the:
(a) Cathode(b) Anode(c) Copper plate(d) Battery
5.
Answer: (c) — Chemical energy changes directly to electrical energy
Fuel cells are highly efficient because they convert about:
(a) 25% of bond energy(b) 50% of bond energy(c) 75% of bond energy(d) 100% of bond energy
6.
Answer: (c) — Zinc coating protects iron from rusting
Galvanization involves coating iron with:
(a) Tin(b) Copper(c) Zinc(d) Aluminium
7.
Answer: (c) — 0.34 V + 0.76 V = 1.10 V
The standard cell voltage of a Daniel cell is approximately:
(a) 0.34 V(b) 0.76 V(c) 1.10 V(d) 2.00 V
8.
Answer: (c) — 2(+1) + x + 4(−2) = 0, so x = +6
The oxidation number of Mn in K2MnO4 :
(a) +7(b) +5(c) +6(d) -6
9.
Answer: (d) — Cu anode dissolves: Cu → Cu2+ + 2e−
Which product you will get at the anode when an aqueous solution of CuSO4 is electrolyzed using copper electrodes?
(a) Cu(b) O2(c) H2(d) No product, instead copper metal will go into the solution as Cu2+.
10.Answer: (a) — Cl− ions are discharged: 2Cl− → Cl2 + 2e−
Which product will be obtained at the anode when a concentrated solution of NaCl will be electrolyzed?
(a) Cl2(b) O2(c) More O2 and less Cl2(d) More Cl2 and less O2
11.Answer: (d) — O2 + 2H2O + 4e− → 4OH−
What is produced at the cathode in hydrogen-oxygen fuel cell?
(a) H+(b) O2−(c) H2O(d) OH1−
12.Answer: (c) — Mg is the most reactive of the four
Which of the following metals is highest in the electrochemical series?
(a) Tin(b) Iron(c) Magnesium(d) Zinc
786Times • Full Chapter Test T-7 • Chapter # 16
Here is Subjective Part of Model paper.
Marks: 48
CHEMISTRY (SUBJECTIVE TYPE)
Time: 1:45 minute
Part – I
2.Write short answers to any Five (5) questions:[5×2=10]
- (i)Find out the oxidation number of manganese in KMnO4.Let Mn = x. In KMnO4: (+1) + x + 4(−2) = 0, so x − 7 = 0 and x = +7.
- (ii)Write the formula of aluminum sulfate and explain the steps.Formula: Al2(SO4)3.
- Write ions with charges: Al3+ and SO42−.
- Criss-cross the numbers: 3 goes to SO4 and 2 goes to Al.
- Put the polyatomic ion in brackets: Al2(SO4)3.
- Check: 2(+3) + 3(−2) = 0.
- (iii)Define reduction in three ways.Reduction is:
- Gain of electrons.
- Decrease in oxidation number.
- Gain of hydrogen or loss of oxygen.
- (iv)Differentiate between anode and cathode in an electrolytic cell.Anode: connected to the positive terminal of the battery; oxidation occurs here; negative ions (anions) move towards it.
Cathode: connected to the negative terminal; reduction occurs here; positive ions (cations) move towards it. - (v)Why must lead (II) chloride be heated before electrolysis?Solid PbCl2 is ionic but its ions are fixed in the lattice and cannot conduct. Heating above its melting point (about 501°C) makes it molten, so Pb2+ and Cl− ions become free to move and carry current.
- (vi)Why sulphuric acid is added to water before electrolysis?Pure water is a very poor conductor because it has very few ions. Sulphuric acid ionises to give H+ and SO42− ions, which increase the conductivity so that electrolysis can take place.
- (vii)Write chemical reactions involved in electrolysis of copper sulphate.CuSO4 → Cu2+ + SO42− and H2O → H+ + OH−Cathode: Cu2+ + 2e− → CuAnode: 4OH− → 2H2O + O2 + 4e−(with inert electrodes).
- (viii)Why is hydrogen production costly?Hydrogen does not occur free in nature. It is made by electrolysis of water, which uses a large amount of electricity, or from fossil fuels using much energy. Its storage and transport (high pressure or very low temperature) are also costly.
3.Write short answers to any Five (5) questions:[5×2=10]
- (i)Write chemical reaction in which iron reacts with oxygen and water to form rust.4Fe + 3O2 + 2xH2O → 2Fe2O3·xH2O (rust)
- (ii)What ions move towards the cathode and anode during silver plating?Positive silver ions (Ag+) move towards the cathode (the object to be plated). Negative ions (such as NO3−) move towards the anode (the silver bar).
- (iii)On which metals chromium plating is usually carried out?Chromium plating is usually done on iron or steel articles that are first plated with copper or nickel (also on brass), because chromium does not stick directly to iron. Example: car parts, taps.
- (iv)Why must oxidation and reduction take place in separate compartments?If both reactions occur in the same place, electrons pass directly between the substances and the energy is lost as heat. In separate compartments, electrons must flow through the external wire, so an electric current is obtained.
- (v)Why does the more reactive metal lose electrons in a galvanic cell?A more reactive metal is higher in the electrochemical series and has a greater tendency to lose electrons. So it is oxidised at the anode and sends electrons through the wire to the less reactive metal.
- (vi)Assign oxidation number to the marked atom of the following compound: (NH4)2 SO4Ammonium ion NH4+ = +1. Let S = x: 2(+1) + x + 4(−2) = 0, so x = +6.
- (vii)Write the formulae of following ionic compound with the help of charges present on their ions: Sodium SulfiteNa+ and SO32− → criss-cross → Na2SO3.
- (viii)Identify the oxidising agent and the reducing agent in the following reaction.2CuSO4 + 4KI 2CuI + I2 + 2K2SO4Cu changes from +2 (in CuSO4) to +1 (in CuI): it is reduced, so CuSO4 is the oxidising agent. I− changes from −1 (in KI) to 0 (in I2): it is oxidised, so KI is the reducing agent.
4.Write short answers to any Five (5) questions:[5×2=10]
- (i)Which other electrolyte can be used in hydrogen-oxygen fuel cell?Besides aqueous KOH, an acidic electrolyte such as dilute sulphuric acid or phosphoric acid can be used.
- (ii)If a sample of iron and a sample of zinc come into contact, which will corrode? Explain your answer.Zinc will corrode. Zinc is higher in the electrochemical series, so it loses electrons more easily and acts as the anode, while iron is protected. This is why zinc is used for galvanization.
- (iii)How electroplating is used to repair worn-out areas of metal objects?The worn-out object is made the cathode and the repair metal is the anode in a solution of its salt. On passing current, metal is deposited on the worn area and fills it with a new layer.
- (iv)Mention one difference between an electrolytic cell and a galvanic cell.An electrolytic cell changes electrical energy into chemical energy (non-spontaneous reaction), while a galvanic cell changes chemical energy into electrical energy (spontaneous reaction).
- (v)Why are fuel cells regarded as environment friendly?The only product of H2–O2 fuel cells is water. They give no smoke, CO2 or other pollutants.
- (vi)How can H2O2 behave as an oxidizing agent?In H2O2 oxygen has oxidation number −1 and can decrease to −2 by gaining electrons: H2O2 + 2H+ + 2e− → 2H2O. It oxidises other substances, e.g. Fe2+ to Fe3+.
- (vii)Why does corrosion become fast during rainy season?In the rainy season moisture and dissolved gases and salts are plentiful. Water acts as an electrolyte, so the electrochemical reactions of corrosion become faster.
- (viii)What is the function of a porous partition in a galvanic cell?It keeps the two solutions from mixing, but lets ions pass through. This completes the circuit and keeps both half-cells electrically neutral.
Part – II
NOTE:- Attempt any two questions.[9×2=18]
5.
(a)
Define oxidation number. Write its significance.
Oxidation number is the apparent charge on an atom of an element in a compound or ion, assuming all shared electrons are completely transferred to the more electronegative atom.
Significance:
Significance:
- Identifies oxidation (increase) and reduction (decrease) in a reaction.
- Helps to write formulae and names of compounds.
- Helps to identify oxidising and reducing agents and to balance redox equations.
(b)
Explain the electrolysis of sodium chloride (Brine).
Brine is concentrated NaCl solution. Ions present: Na+, Cl−, H+ and OH−. The electrodes are graphite (anode) and iron or steel (cathode).NaCl → Na+ + Cl− H2O → H+ + OH−Cathode: H+ is discharged in preference to Na+.2H+ + 2e− → H2(g)Anode: Cl− is discharged (concentrated solution).2Cl− → Cl2(g) + 2e−Na+ and OH− remain in solution, giving NaOH.2NaCl + 2H2O → 2NaOH + H2 + Cl2
6.
(a)
Define electrochemical series write its applications.
The electrochemical series is the arrangement of elements (metals) in order of their decreasing tendency to lose electrons (or by their standard electrode potentials).
Applications:
Applications:
- Compare the reactivity of metals.
- Predict which metal displaces another from its salt solution.
- Predict the anode, cathode and voltage of a galvanic cell.
- Predict the products at the cathode in electrolysis.
- Tell which metals react with acids to give hydrogen (those above H).
(b)
Describe the working of a fuel cell.
A hydrogen-oxygen fuel cell has two porous electrodes (with platinum catalyst) dipped in KOH solution. H2 is supplied to the anode and O2 to the cathode.Anode: 2H2 + 4OH− → 4H2O + 4e−Cathode: O2 + 2H2O + 4e− → 4OH−Overall: 2H2 + O2 → 2H2OElectrons flow from the anode to the cathode through the external circuit, giving electric current. The cell works continuously while fuel is supplied, and water is the only product.
7.
(a)
What is corrosion? Explain the reactions involved in corrosion of iron.
Corrosion is the slow destruction of a metal by chemical or electrochemical reaction with its surroundings (e.g. rusting of iron). Iron rusts in the presence of oxygen and water:Anode: Fe → Fe2+ + 2e−Cathode: O2 + 2H2O + 4e− → 4OH−Fe2+ + 2OH− → Fe(OH)24Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3 → Fe2O3·xH2O (rust)
(b)
Write procedure of electroplating with the help of diagram and chemical reactions.
Example: silver plating of an iron object.
- Clean the object well and make it the cathode (negative terminal).
- Use a pure silver bar as the anode (positive terminal).
- Use silver nitrate (AgNO3) solution as the electrolyte and pass direct current.
786Times • Chemistry (Subjective Type) • Marks: 48
