Page 36 - ACE YR IGCSE A TOP APPROACH TO CHEM

(d) Mass of oxide = 2.19 g [1] 2 C
Number of moles of P = 0.054 mol and At the cathode, H will be reduced because it is less
+
number of moles of O = 0.137 mol [1] reactive than Na . Hydrogen is produced. At the
+
–
Ratio = P : O = 1 : 2.5 [1] anode, OH will be oxidised because there is a higher
–
Empirical formula = P O [1] concentration of OH . Oxygen is produced. Hydrogen
2 5 and oxygen are both gases.
Mass of oxide = 3.87 – 1.68 = 2.19 g
1.68 3 D
Number of moles of P = = 0.054 mol OH is oxidised at the anode. Oxygen is formed. H is
+
–
31
2.19 reduced at the cathode. Hydrogen is formed.
Number of moles of O = = 0.137 mol OH + H ➞ H O
–
+
16 2
Divide the number of moles of each by the smallest Water is being removed from the electrolyte.
number of moles. 4 C
0.054 At the cathode, hydrogen ions gain electrons to form
P = = 1 hydrogen gas. It is being reduced. At the anode, chloride
0.054
0.137 ions lose electrons to form chlorine gas. It is being
O = = 2.5 oxidised. Sodium ions and hydroxide ions are left in the
0.054
To round up the number, the ratio number × 2. solution, forming sodium hydroxide. It is alkaline.
Ratio of P : O = 2 : 5 5 B
Solid cannot be used as the electrolyte.
(e) One mole of P O = 110 g [1]
3
2
P O = 220 g [1] 6 C
4 6
10 (a) Number of moles of HCl The only product formed is water which is harmless to
the environment. Hydrogen-oxygen fuel cells are more
= 0.1 dm × 0.1 mol dm efficient than fossil fuels.
3
–3
= 0.01 mol [1]
Number of moles of CaCO 7 A
3

Electrons are supplied from the battery at the cathode.
= 0.01 mol Cations will be attracted to the cathode. Potassium ions
2 are reduced by gaining electrons.
= 0.005 mol [1]
Ratio = HCl : CaCO = 2 : 1 8 C
3
There are 4 ions present in aqueous copper(II) sulfate,

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Mass of CaCO = 0.005 mol × 100 = 0.5 g [1]
2+
2–
+
2+
3 H , Cu , SO and OH . At the cathode, Cu will be
–
4
+
(b) 0.5 × 100% = 89.3% [1] selected instead of H . This is because copper is less
0.56 reactive. Copper is deposited at the cathode. Copper
mass of pure is a brown solid. At the anode, OH will be oxidised.
–
substance Oxygen is produced. Bubbles of gas are seen. H and
+
Percentage of purity = × 100%
mass of impure SO remain in the solution forming sulfuric acid which
2–
4
substance is acidic. As the concentration of sulfuric acid increases,
(c) Molar mass of O = 16 × 3 = 48 g mol [1] the blue aqueous copper(II) sulfate turns paler.
–1
3
Percentage composition of oxygen 9 D
= 48 × 100% In the refining process of an impure copper, a pure
100 copper is placed at the cathode for copper deposition.
= 48% [1] The impure copper is placed as the anode. Since it is the
refining of copper, a copper salt solution should be used.
(d) Number of moles of CO = 0.005 mol [1] Copper(II) sulfate, copper(II) nitrate or any copper salt
2
Number of molecules which is soluble can be used as the electrolyte.
= 0.005 mol × 6.02 × 10 23 At the cathode, copper is deposited, Cu + 2e ➞ Cu. The
2+
= 0.0301 × 10 23 mass increases.
2+
= 3.01 × 10 [1] At the anode, copper is oxidised, Cu ➞ Cu + 2e. The
21
Number of molecules mass decreases.
= number of moles × Avogadro constant
10 B
There are 4 ions present in sodium bromide solution,
–
+
+
–
4 Electrochemistry Na , H , OH and Br . +
+
At the cathode, H will be selected instead of Na . This
is because hydrogen is less reactive. Hydrogen gas is
Part 1: Multiple-choice Questions produced at the cathode. Bubbles of colourless gas
–
1 C are seen. At the anode, Br will be oxidised. Bromine is
In electrolysis, positive ions gain electrons at the cathode produced. Bromine is a reddish-brown liquid.
and negative ions lose electrons at the anode.
Cambridge IGCSE TM
148 Ace Your Chemistry
Answers.indd 148 3/4/22 3:54 PM