faraday. higher grade chemistry calculations faraday the quantity of electrical charge flowing in a...
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Faraday
HIGHER GRADE CHEMISTRY CALCULATIONS
Faraday
The quantity of electrical charge flowing in a circuit is related to the current and the time.
Q = I x t Where Q is the electrical charge in coulombs (C)
I is the current in amps (A)
t is the time in seconds (s)
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)
Higher Grade Chemistry
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
Faraday
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.
1. 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.
2. 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?
Ni2+(aq) + 2e- Ni(s)
2 F 1mol
2 x 96500 C 1 mol
193000 C 1 mol
58.5 g of nickel 1 mol
So 2.925 g 2.925 /58.5 mol
0.05 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
Calculations for you to try.
4. 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.Q = I x t
= 2 x 40 x 60
= 4800 C
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 C96500 C 1 Farady
So 289156 C 289156/96500 = 2.996 F
The charge on the ion = 3+.