Download presentation

Presentation is loading. Please wait.

Published byYadiel Stinson Modified over 2 years ago

1
Higher Unit 3 Electrolysis

2
After today’s lesson you should be able to: Use ‘Q = I x t’ to calculate - the value of faraday - the mass of product at an electrode - the time or current used in the experiment - the volume of gas given off at an electrode - the charge on an ion

3
Electrolysis Electrolysis is a chemical reaction which occurs when an ionic compound is broken down into its elements using electricity. The ionic compound must be molten or in solution. A d.c. supply is used to separate the products to different electrodes. Positive ions are attracted to the negative electrode where they undergo reduction. Negative ions are attracted to the positive electrode where they undergo oxidation.

4
The current, symbol ‘I’, is the speed at which the electrical current flows round the circuit. Current is measured in Amperes (A). The current is kept constant by using a variable resistor in the circuit. The electrical charge, symbol ‘Q’ is measured in Coulombs (C). The time the current has been flowing round the circuit, symbol ‘t’, is measured in seconds (s). The total charge is calculated using the equation Q = It

5
Electrons and Faradays The charge is carried through the wires of the circuit (the external circuit) by electrons. 1 mole of electrons = 1 Faraday = 96500C. The ion-electron equation for the element being calculated gives the number of moles of electrons and hence, the number of Faradays required to deposit 1 mole of the element. e.g. Cu 2+ + 2e - → Cu. 1 mole of copper is deposited using 2 moles of electrons = 2 x 96500C = 193000C.

6
Example 1 – Calculating the Faraday experimentally A solution of copper(II) nitrate was electrolysed for 20 mins using a current of 0.2A. A mass of 0.08g of copper was deposited on the negative electrode. Calculate the value of the Faraday.

7
Step 1 – calculate the total charge using the equation ‘Q = It’ Q = ? I = 0.2A t = 20 mins = (20 x 60)s = 1200s Q = 0.2 x 1200 = 240C

8
Step 2 – calculate the number of moles of the element deposited using the equation n = m ÷ gfm. n = ? m = 0.08g gfm Cu = 63.5g n = 0.08 ÷ 63.5 = 0.00126mol

9
Step 3 – calculate the value of Faraday 0.00126 mol Cu is deposited by 240C ∴ 1 mol Cu → 240 = 190476C 0.00126 0.00126

10
From the equation Cu 2+ + 2e - → Cu 1 mole of Cu is deposited using 2 moles of electrons but 1 mole of electrons = 1 Faraday ∴ Faraday = 190476 = 95238C 2

11
Exercise P118 of ‘Test your Higher Chemistry Calculations’ Q15.1 – 15.5

12
Example 2 – calculating the mass deposited What mass of nickel is deposited in the electrolysis of nickel(II) sulphate solution if a current of 0.4A is passed for 120mins.

13
Step 1 – calculate the total charge using the equation ‘Q = It’ Q = ? I = 0.4A t = 120 mins = (120 x 60)s = 7200s Q = 0.4 x 7200 = 2880C

14
Step 2 – calculate the number of moles of electrons that this charge represents 96500 C →1 mole ∴ 1C → 1 96500 ∴ 2880C→ 1 x 2880 96500 = 0.03 mol

15
Step 3 – calculate the mass of the element deposited From the equation Ni 2+ + 2e - → Ni 1 mole of Ni is deposited using 2 moles of electrons i.e. 2 moles e - → 1mole Ni 2 mol e - → 58.7g Ni 1mol e - → 58.7 2 0.03mol e - → 58.7 x 0.03 = 0.8805g 2

16
Exercise P124 - 125 of ‘Test your Higher Chemistry Calculations’ Q15.21 – 15.35

17
Example 3 How long must a current of 0.25A flow in the electrolysis of molten aluminium oxide to cause the deposition of 1.08g of aluminium at the negative electrode?

18
Step 1 – calculate the number of moles of element deposited n = ? m = 1.08g gfm = 27.0g n = 1.08 ÷ 27.0 = 0.04 mol

19
Step 2 – calculate the number of moles of electrons involved. From the equation Al 3+ + 3e - → Al 1 mole of Al is deposited using 3 moles of electrons i.e. 1mole Al → 3 moles e - 1mole Al → 3 moles e - 0.04mol Al → 0.04 x 3 = 0.12mol

20
Step 3 – calculate the charge this represents 1 moles e - → 96500C 0.12 mol e - → 96500 x 0.12 = 11580C = 11580C Step 4 – calculate t. t = Q I = 11580 = 46320s OR 772mins = 11580 = 46320s OR 772mins 0.25 0.25

21
Exercise P130 - 132 of ‘Test your Higher Chemistry Calculations’ Q15.46, 15.48- 15.51, 15.53 – 15.60

22
Example 4 What volume of hydrogen would be given off at the negative electrode in the electrolysis of an aqueous solution if a current of 0.25A flowed for 2hrs. (molar volume of hydrogen = 24 l mol -1 )

23
Step 1 – calculate the total charge using the equation ‘Q = It’ Q = ? I = 0.25A t = 2hrs = (2 x 60 x 60)s = 7200s Q = 0.25 x 7200 = 1800C

24
Step 2 – calculate the number of moles of the electrons used 96500 C →1 mole ∴ 1C → 1 96500 ∴ 1800C→ 1 x 1800 96500 = 0.019 mol

25
Step 3 – calculate the number of moles of hydrogen given off From the equation 2H + + 2e - → H 2 1 mole of H 2 is deposited using 2 moles of electrons i.e. 2 moles e - → 1mole H 2 2 moles e - → 1mole H 2 1 mole e - → ½ mole H 2 1 mole e - → ½ mole H 2 0.019 moles e - → 0.019 X ½ = 0.0095 mole H 2 = 0.0095 mole H 2

26
Step 4 – calculate the volume of hydrogen given off 1 moles H 2 → 24L 1 moles H 2 → 24L 0.0095 mole H 2 → 0.0095 x 24 0.0095 mole H 2 → 0.0095 x 24 0.228L OR 228cm 3 0.228L OR 228cm 3

27
Exercise P128 - 129 of ‘Test your Higher Chemistry Calculations’ Q15.36 – 15.45

28
Example 5 A molten iron compound is electrolysed using a current of 4.73A for 30mins during which 1.64g of iron is deposited. Calculate the number of positive charges on each iron ion.

29
Step 1 – calculate the total charge using the equation ‘Q = It’ Q = ? I = 4.73A t = 30 = (30 x 60)s = 1800s Q = 4.73 x 1800 = 8514C

30
Step 2 – calculate the number of moles of the electrons used 96500 C →1 mole ∴ 1C → 1 96500 ∴ 8514C→ 1 x 8514 96500 = 0.088 mol

31
Step 3 – calculate the number of moles of iron deposited n = ? n = ? m = 1.64g gfm Fe = 55.8g n = 1.64 ÷ 55.8 = 0.03mol

32
Step 3 – calculate the charge on the ion 0.03 mol Fe is deposited by 0.088mol e - ∴ 1 mol Fe → 0.088 0.03 0.03 = 2.93 ≃ 3 mol = 2.93 ≃ 3 mol The ion-electron equation must be: Fe 3+ + 3e - → Fe

33
Exercise P133 of ‘Test your Higher Chemistry Calculations’ Q15.61 – 15.65

Similar presentations

OK

Chapter 19: Electrochemistry: Voltaic Cells Generate Electricity which can do electrical work. Voltaic or galvanic cells are devices in which electron.

Chapter 19: Electrochemistry: Voltaic Cells Generate Electricity which can do electrical work. Voltaic or galvanic cells are devices in which electron.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on mauryan india Ppt on water activity vs moisture Ppt on our changing earth Ppt on kingdom monera powerpoint Ppt on social contract theory declaration Ppt on personal computer museum Ppt on energy cogeneration plant Free ppt on forest society and colonialism in india Ppt on education schemes in india Ppt on cross docking facility