Higher Unit 3Higher Unit 3

ElectrolysisElectrolysis

After today’s lesson you After today’s lesson you should be able to:should be able to:

Use ‘Q = I x t’ to calculate Use ‘Q = I x t’ to calculate

- the value of faraday- the value of faraday

- the mass of product at an electrode- the mass of product at an electrode

- the time or current used in the experiment - the time or current used in the experiment - the volume of gas given - the volume of gas given off at an electrode off at an electrode

- the charge on an ion- the charge on an ion

ElectrolysisElectrolysis Electrolysis is a chemical reaction which Electrolysis is a chemical reaction which

occurs when an ionic compound is broken occurs when an ionic compound is broken down into its elements using electricity.down into its elements using electricity.

The ionic compound must be molten or in The ionic compound must be molten or in solution.solution.

A d.c. supply is used to separate the A d.c. supply is used to separate the products to different electrodes.products to different electrodes.

Positive ions are attracted to the negative Positive ions are attracted to the negative electrode where they undergo reduction.electrode where they undergo reduction.

Negative ions are attracted to the positive Negative ions are attracted to the positive electrode where they undergo oxidation.electrode where they undergo oxidation.

The current, symbol ‘I’, is the speed at which The current, symbol ‘I’, is the speed at which the electrical current flows round the circuit.the electrical current flows round the circuit.

Current is measured in Amperes (A).Current is measured in Amperes (A). The current is kept constant by using a The current is kept constant by using a

variable resistor in the circuit.variable resistor in the circuit. The electrical charge, symbol ‘Q’ is The electrical charge, symbol ‘Q’ is

measured in Coulombs (C).measured in Coulombs (C). The time the current has been flowing round The time the current has been flowing round

the circuit, symbol ‘t’, is measured in the circuit, symbol ‘t’, is measured in seconds (s).seconds (s).

The total charge is calculated using the The total charge is calculated using the equationequation

Q = ItQ = It

Electrons and FaradaysElectrons and Faradays The charge is carried through the wires of The charge is carried through the wires of

the circuit (the external circuit) by electrons.the circuit (the external circuit) by electrons. 1 mole of electrons = 1 Faraday = 96500C.1 mole of electrons = 1 Faraday = 96500C. The ion-electron equation for the element The ion-electron equation for the element

being calculated gives the number of moles being calculated gives the number of moles of electrons and hence, the number of of electrons and hence, the number of Faradays required to deposit 1 mole of the Faradays required to deposit 1 mole of the element.element.

e.g. Cue.g. Cu2+2+ + 2e + 2e-- → Cu.→ Cu.1 mole of copper is deposited using 2 moles 1 mole of copper is deposited using 2 moles of electrons = 2 x 96500C = 193000C.of electrons = 2 x 96500C = 193000C.

Example 1 – Calculating the Example 1 – Calculating the Faraday Faraday experimentallyexperimentally

A solution of copper(II) nitrate was A solution of copper(II) nitrate was electrolysed for 20 mins using a current of electrolysed for 20 mins using a current of 0.2A. A mass of 0.08g of copper was 0.2A. A mass of 0.08g of copper was deposited on the negative electrode.deposited on the negative electrode.

Calculate the value of the Faraday.Calculate the value of the Faraday.

Step 1 – calculate the total charge using the Step 1 – calculate the total charge using the equation ‘Q = It’ equation ‘Q = It’

Q = ?Q = ?

I = 0.2AI = 0.2A

t = 20 mins = (20 x 60)s = 1200st = 20 mins = (20 x 60)s = 1200s

Q = 0.2 x 1200 = 240CQ = 0.2 x 1200 = 240C

Step 2 – calculate the number of moles of Step 2 – calculate the number of moles of the element deposited using the the element deposited using the

equation n = m equation n = m ÷ gfm÷ gfm..

n = ?n = ?

m = 0.08gm = 0.08g

gfm Cu = 63.5ggfm Cu = 63.5g

n = 0.08 n = 0.08 ÷÷ 63.5 = 0.00126mol 63.5 = 0.00126mol

Step 3 – calculate the value of FaradayStep 3 – calculate the value of Faraday

0.00126 mol Cu is deposited by 240C0.00126 mol Cu is deposited by 240C

∴ ∴ 1 mol Cu 1 mol Cu → → 240240 = = 190476C190476C

0.001260.00126

From the equation From the equation CuCu2+2+ + 2e + 2e-- → Cu→ Cu

1 mole of Cu is deposited using 2 moles of 1 mole of Cu is deposited using 2 moles of

electronselectrons

but 1 mole of electrons = 1 Faradaybut 1 mole of electrons = 1 Faraday

∴ ∴ Faraday = Faraday = 190476190476 = = 95238C95238C

22

ExerciseExercise

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

Example 2 – calculating the Example 2 – calculating the mass deposited mass deposited

What mass of nickel is deposited in the What mass of nickel is deposited in the

electrolysis of nickel(II) sulphate solution if a electrolysis of nickel(II) sulphate solution if a

current of 0.4A is passed for 120mins.current of 0.4A is passed for 120mins.

Step 1 – calculate the total charge using the Step 1 – calculate the total charge using the equation ‘Q = It’ equation ‘Q = It’

Q = ?Q = ?

I = 0.4AI = 0.4A

t = 120 mins = (120 x 60)s = t = 120 mins = (120 x 60)s = 7200s7200s

Q = 0.4 x 7200 = 2880CQ = 0.4 x 7200 = 2880C

Step 2 – calculate the number of moles of Step 2 – calculate the number of moles of electrons that this charge electrons that this charge

representsrepresents

96500 C96500 C →→ 1 mole1 mole

∴∴ 1C1C →→ 11

9650096500

∴∴ 2880C2880C →→ 11 x 2880 x 2880

9650096500

= 0.03 mol= 0.03 mol

Step 3 – calculate the mass of the element depositedStep 3 – calculate the mass of the element deposited

From the equation From the equation NiNi2+2+ + 2e + 2e-- → Ni→ Ni1 mole of Ni is deposited using 2 moles of electrons i.e. 1 mole of Ni is deposited using 2 moles of electrons i.e.

2 moles e2 moles e-- → 1mole Ni→ 1mole Ni2 mol e2 mol e-- → 58.7g Ni → 58.7g Ni1mol e1mol e-- → → 58.758.7

22

0.03mol e0.03mol e-- → → 58.7 58.7 x 0.03 = x 0.03 = 0.8805g0.8805g 22

ExerciseExercise

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

Example 3Example 3How long must a current of 0.25A flow in the How long must a current of 0.25A flow in the

electrolysis of molten aluminium oxide to electrolysis of molten aluminium oxide to

cause the deposition of 1.08g of aluminium at cause the deposition of 1.08g of aluminium at

the negative electrode?the negative electrode?

Step 1 – calculate the number of moles of Step 1 – calculate the number of moles of element deposited element deposited

n = ?n = ?

m = 1.08gm = 1.08g

gfm = 27.0ggfm = 27.0g

n = 1.08 n = 1.08 ÷÷ 27.0 = 0.04 mol 27.0 = 0.04 mol

Step 2 – calculate the number of moles of Step 2 – calculate the number of moles of electrons involved. electrons involved.

From the equation From the equation AlAl3+3+ + 3e + 3e-- → Al→ Al1 mole of Al is deposited using 3 moles of 1 mole of Al is deposited using 3 moles of

electrons electrons i.e. i.e.

1mole Al → 1mole Al → 3 moles e3 moles e-- 0.04mol Al → 0.04 x 3 = 0.12mol0.04mol Al → 0.04 x 3 = 0.12mol

Step 3 – calculate the charge this representsStep 3 – calculate the charge this represents

1 moles e1 moles e-- → 96500C→ 96500C0.12 mol e0.12 mol e-- → 96500 x 0.12 → 96500 x 0.12

= 11580C= 11580C

Step 4 – calculate t.Step 4 – calculate t.

t = t = QQ II

= = 1158011580 = 46320s OR 772mins = 46320s OR 772mins 0.250.25

ExerciseExercise

P130 - 132 of ‘Test your Higher Chemistry P130 - 132 of ‘Test your Higher Chemistry

Calculations’ Calculations’

Q15.46, 15.48- 15.51, 15.53 – 15.60Q15.46, 15.48- 15.51, 15.53 – 15.60

Example 4Example 4What volume of hydrogen would be given off What volume of hydrogen would be given off

at the negative electrode in the electrolysis of at the negative electrode in the electrolysis of

an aqueous solution if a current of 0.25A an aqueous solution if a current of 0.25A

flowed for 2hrs. flowed for 2hrs.

(molar volume of hydrogen = 24 l mol(molar volume of hydrogen = 24 l mol-1-1))

Step 1 – calculate the total charge using the Step 1 – calculate the total charge using the equation ‘Q = It’ equation ‘Q = It’

Q = ?Q = ?

I = 0.25AI = 0.25A

t = 2hrs = (2 x 60 x 60)s = 7200st = 2hrs = (2 x 60 x 60)s = 7200s

Q = 0.25 x 7200 = 1800CQ = 0.25 x 7200 = 1800C

Step 2 – calculate the number of moles of Step 2 – calculate the number of moles of the electrons used the electrons used

96500 C96500 C →→ 1 mole1 mole

∴∴ 1C1C →→ 11

9650096500

∴∴ 1800C1800C →→ 11 x 1800 x 1800

9650096500

= 0.019 mol= 0.019 mol

Step 3 – calculate the number of moles of Step 3 – calculate the number of moles of hydrogen given off hydrogen given off

From the equation From the equation 2H2H++ + 2e + 2e-- → H→ H22

1 mole of H1 mole of H22 is deposited using 2 moles of is deposited using 2 moles of

electrons i.e. electrons i.e.

2 moles e2 moles e-- → 1mole H→ 1mole H22

1 mole e1 mole e-- → → ½ ½ mole Hmole H22

0.019 moles e0.019 moles e-- → 0.019 X → 0.019 X ½ ½

= 0.0095 mole = 0.0095 mole HH22

Step 4 – calculate the volume of Step 4 – calculate the volume of hydrogen hydrogen given off given off

1 moles 1 moles HH2 2 → 24L→ 24L

0.0095 mole 0.0095 mole HH2 2 → 0.0095 x 24→ 0.0095 x 24

0.228L OR 228cm0.228L OR 228cm33

ExerciseExercise

P128 - 129 of ‘Test your Higher Chemistry P128 - 129 of ‘Test your Higher Chemistry

Calculations’ Calculations’

Q15.36 – 15.45Q15.36 – 15.45

Example 5Example 5A molten iron compound is electrolysed using A molten iron compound is electrolysed using

a current of 4.73A for 30mins during which a current of 4.73A for 30mins during which

1.64g of iron is deposited. Calculate the 1.64g of iron is deposited. Calculate the

number of positive charges on each iron ion.number of positive charges on each iron ion.

Step 1 – calculate the total charge using the Step 1 – calculate the total charge using the equation ‘Q = It’ equation ‘Q = It’

Q = ?Q = ?

I = 4.73AI = 4.73A

t = 30 = (30 x 60)s = 1800st = 30 = (30 x 60)s = 1800s

Q = 4.73 x 1800 = 8514CQ = 4.73 x 1800 = 8514C

Step 2 – calculate the number of moles of Step 2 – calculate the number of moles of the electrons used the electrons used

96500 C96500 C →→ 1 mole1 mole

∴∴ 1C1C →→ 11

9650096500

∴∴ 8514C8514C →→ 11 x 8514 x 8514

9650096500

= 0.088 mol= 0.088 mol

Step 3 – calculate the number of moles of Step 3 – calculate the number of moles of iron deposited iron deposited

n = ?n = ?

m = 1.64gm = 1.64g

gfm Fe = 55.8ggfm Fe = 55.8g

n = 1.64 n = 1.64 ÷÷ 55.8 = 0.03mol 55.8 = 0.03mol

Step 3 – calculate the charge on the ionStep 3 – calculate the charge on the ion

0.03 mol Fe is deposited by 0.088mol e0.03 mol Fe is deposited by 0.088mol e--

∴ ∴ 1 mol Fe 1 mol Fe → → 0.0880.088

0.030.03

= 2.93 ≃ 3 mol= 2.93 ≃ 3 mol

The ion-electron equation must be:The ion-electron equation must be:

FeFe3+3+ + 3e + 3e-- → Fe→ Fe

ExerciseExercise

P133 of ‘Test your Higher Chemistry P133 of ‘Test your Higher Chemistry

Calculations’ Calculations’

Q15.61 – 15.65Q15.61 – 15.65