1. Faraday

2. 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)

3. 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. Cu 2+ (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

4. 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 -  H 2 (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

5. 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)  Cl 2 (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

6. 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? Ni 2+ (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

7. 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 C 96500 C  1 Farady So 289156 C  289156 / 96500 = 2.996 F The charge on the ion = 3+.