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Faraday

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Faraday

HIGHER GRADE CHEMISTRY CALCULATIONS

The quantity of electrical charge flowing in a circuit is related to the current and the time.

Faraday

Where Q is the electrical charge in coulombs (C)

I is the current in amps (A)

t is the time in seconds (s)

Q = I x t

Faradays law of electrolysis states that n x F coulombs of charge are required to deposit 1 mole of a substance.

Where n is the number of electrons in the ion-electron equation and F is the Faraday (96 500C)

Faraday

Worked example 1.

Calculate the mass of copper deposited when a current of 0.5 A is passed through copper(II) sulphate solution for 1 hour.

Step 1:- Q = I x t

= 0.5 x 60 x 60

= 1800 C

Step 2:- Use the ion-electron equation to calculate the charge to give 1 mol.

Cu2+(aq) + 2e- Cu(s)

2F 1 mole

2 x 96500 C 1 mole

193000 C 1 mole

Step 3:- Calculate the mass deposited by number of coulombs in step 1.

193000 C 1 mole

193000 C 63.5 g

So 1800 C 1800/193000 x 63.5

0.59 g

Higher Grade Chemistry

- Calculations for you to try.
- Calculate the mass of hydrogen formed when a current of 0.4 A is passed through hydrochloric acid solution for 3 hours.

Q = I x t

= 0.4 x 3 x 60 x 60

= 4320 C

2H+(aq) + 2e- H2(g)

2 F 1mol

2 x 96500 C 1 mol

193000 C 1 mol

19300 C 1 mol

2 g

So 4320 C 4320/193000 x 2

0.045 g

Higher Grade Chemistry

- Calculations for you to try.
- Calculate the volume of chorine produced when a current of 2 A is passed though sodium chloride solution for 5 hours 40 minutes.
- Take the molar volume of a gas to be 23.8 l mol-1.

Q = I x t

= 2 x 340 x 60

= 408 000 C

(5 hours 40 minutes = 340 minutes)

2Cl-(aq) Cl2(g) + 2e-

1mol 2F

1 mol 2 x 96 500 C

1 mol 193 000 C

19300 C 1 mol

23.8 litres

So 408 000 C 408 000/193000 x 23.8

50.31 litres

Higher Grade Chemistry

Calculations for you to try.

3. Electrolysis of nickel(II) chloride solution produced 2.925 g of nickel per hour. What current was flowing?

58.5 g of nickel 1 mol

So 2.925 g 2.925 /58.5 mol

0.05 mol

Ni2+(aq) + 2e- Ni(s)

2 F 1mol

2 x 96500 C 1 mol

193000 C 1 mol

1 mol of nickel 193000 C

So 0.05 mole 193 000 x 0.05 C

9650 C

Q = I x t

So I = Q/t

I = 9650/60 x 60 = 2.68 A

Higher Grade Chemistry

Q = I x t

= 2 x 40 x 60

= 4800 C

Calculations for you to try.

- A chromium compound was electrolysed using a current of 2A for
- 40 minutes. The mass of chromium deposited was 0.864 g.
- Calculate the charge on the chromium ion.

52.0 g of chromium 1 mol

So 0.864 g 0.864 /52.0 mol

0.0166 mol

0.0166 mol of chromium 4800 C

So 1 mole 1 /0.0166 x 4800 C

289156 C

96500 C 1 Farady

So 289156 C 289156/96500 = 2.996 F

The charge on the ion = 3+.