CIE Chemistry Paper-1: Specimen Questions 6 - 7 of 99

Question number: 6

MCQ▾

Question

  1. The table shows the results of adding three metals, P, Q and R, to dilute hydrochloric acid and to water.

results of adding three metals, P, Q and R, to dilute hydrochloric acid

Finding the result if HCl is added to 3 different metals

metal

dilute hydrochloric acid

Water

P

hydrogen produced

hydrogen produced

Q

no reaction

no reaction

R

hydrogen produced

no reaction

What is the order of reactivity of the metals?

Choices

Choice (4) Response
a.

most reactive is P next is R and least reactive is Q

b.

most reactive is R next is P and least reactive is Q

c.

most reactive is P next is Q and least reactive is R

d.

most reactive is R next is Q and least reactive is P

Answer

a.

Explanation

most metals are usually electropositive in nature and lose electrons in a chemical reaction they do not react with the same vigour or speed. Metals display different reactions towards different substances. The greater the ease with which an element loses its electrons and acquires a positive charge, the greater is its reactivity. In the above reactions we can see P reacted with diluted HCL and water rigorously to produce hydrogen.

Question number: 7

MCQ▾

Question

Three electrolysis cells are set up. Each cell has inert electrodes. The electrolytes are listed below.

  • cell 1 aqueous sodium chloride
  • cell 2 dilute sulfuric acid
  • cell 3 molten lead (II) bromide

    In which of these cells is a gas formed at both electrodes?

Choices

Choice (4) Response
a.

3 only

b.

2 only

c.

1 and 3

d.

1 and 2

Answer

d.

Explanation

Electrolysis of aqueous sodium chloride yields hydrogen and chlorine, with aqueous sodium hydroxide remaining in solution. The reaction at the cathode is:

H2O (l) +2e−→H2 (g) +2OH−

The reaction at the anode is:

Cl−→12Cl2 (g) +1e−

The overall reaction is as follows:

NaCl (aq) +H2O (l) →Na+ (aq) +OH− (aq) +H2 (g) +12Cl2 (g) * Electrolysis of dilute sulphuric acid yields oxygen:

The positive anode electrode reaction:

2H2O (l) – 4e == > 4H+ (aq) + O2 (g) (oxygen gas)

The negative cathode electrode reaction:

2H+ (aq) + 2e == > H2 (g) (hydrogen gas)

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