Class 10 Chemistry Chapter 16 Electrochemistry Test 6 | Board Model Paper
Grade 10 Chemistry chapter 16 Topic 16.10 to 16.17

Prepare for your matric board examinations with the Class 10 Chemistry Chapter 16 Electrochemistry Class Test 6, based on Topics 16.10 to 16.17 and the complete chapter exercise. This practice test is part of the Punjab Board 2027 Model Papers series and is designed to help students revise important concepts, practise textbook exercise questions, and strengthen their preparation for the Class 10 Chemistry examination. Students can use this Electrochemistry Test 6 for chapter-wise revision, self-assessment, and board exam practice according to the Punjab Board syllabus.
If the temperature falls below the melting point during electrolysis, the process will:
Sulphate ions are not discharged at the anode because they:
If copper electrodes are used instead of inert electrodes in CuSO4 solution, the mass of the anode:
Platinum in the electrode acts as:
Which two substances are necessary for rusting of iron?
The electrolyte used should be:
A higher cell voltage indicates:
The voltage of a galvanic cell measures:
- (i)Explain how electrochemical series helps in predicting cathode products.In the series, elements are arranged by their tendency to lose electrons. At the cathode, the ion that is lower in the series (less reactive) is discharged first. Metals below hydrogen (Cu, Ag) are deposited, while for metals above hydrogen (Na, K, Ca) in aqueous solution H2 gas is produced instead.
- (ii)Why must electrolysis of molten lead chloride be carried out above 500°C?Solid PbCl2 is ionic, but its ions are held in fixed positions and cannot conduct. Its melting point is about 501°C, so it must be kept above 500°C to remain molten and keep Pb2+ and Cl− ions free to move.
- (iii)Why are inert electrodes used in this electrolysis?Inert electrodes (graphite or platinum) only carry current to and from the electrolyte. They do not react with the products, such as hot chlorine gas, and do not dissolve, so only the electrolyte is decomposed.
- (iv)Why copper electrode is used instead of inert electrode in electrolysis of copper sulphate?With a copper anode, Cu atoms dissolve as Cu2+ ions (Cu → Cu2+ + 2e−) while Cu is deposited at the cathode. The solution stays blue and constant in concentration, and no O2 is produced. This is used for purification of copper and electroplating.
- (v)Why fuel cells are considered efficient energy devices?A fuel cell changes the chemical energy of fuel (H2 and O2) directly into electrical energy, without a heat stage. So less energy is wasted than in engines. They work continuously while fuel is supplied and give only water as product.
- (vi)Why does rusting continue once it starts?Rust is porous and flaky, so it does not protect the metal beneath. Moisture and oxygen keep reaching fresh iron, and the electrochemical rusting process goes on until the iron is consumed.
- (vii)How repairing take place by electroplating?The worn or damaged article is made the cathode, the repair metal is the anode, and a solution of its salt is the electrolyte. On passing current, metal ions are reduced and deposited on the damaged part, filling it with a fresh layer of metal.
- (viii)Explain why a Mg–Cu cell produces higher voltage than an Fe–Cu cell.Cell voltage depends on the difference in tendency of the two metals to lose electrons. Mg is much higher in the electrochemical series than Fe, so the difference between Mg and Cu is greater than between Fe and Cu. Hence the Mg–Cu cell gives higher voltage.
- (ix)Why is zinc called the anode in a Daniel cell?Zinc is more reactive than copper and loses electrons: Zn → Zn2+ + 2e−. The electrode at which oxidation occurs is called the anode, so zinc is the anode (negative electrode).
- Ionisation: PbCl2(l) → Pb2+ + 2Cl−
- At cathode (reduction): Pb2+ + 2e− → Pb(l) Molten grey lead collects at the bottom.
- At anode (oxidation): 2Cl− → Cl2(g) + 2e− Greenish-yellow chlorine gas is released.
- Overall: PbCl2(l) → Pb(l) + Cl2(g)
- Ions present: H2SO4 → 2H+ + SO42− H2O → H+ + OH−
- At cathode (reduction): H+ ions are discharged. 2H+ + 2e− → H2(g)
- At anode (oxidation): SO42− ions are difficult to oxidize, so OH− ions are discharged. 4OH− → 2H2O + O2(g) + 4e−
- Overall: 2H2O(l) → 2H2(g) + O2(g) Hydrogen and oxygen are collected in volume ratio 2 : 1. Only water is used up, so the acid becomes more concentrated.

