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# Faraday - PowerPoint PPT Presentation

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).

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Presentation Transcript

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

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)

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

• 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

• 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

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

= 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