# Higher Unit 3 Electrolysis. 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.

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Higher Unit 3 Electrolysis

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

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.

 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

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.

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.

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

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

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

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

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

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.

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

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

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

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

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?

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

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

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

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

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 )

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

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

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

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

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

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.

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

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

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

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

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

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