![After the complete deposited of copper, the electrolysis will discharge
hydrogen at cathode
and oxygen at anode. The amount of charge passed = 1.2 × 7 × 60 = 504
coulomb
So, Oxygen liberated = 1/96500 × 504 = 0.00523 g equivalent
= 8/32 × 0.00523 = 0.001307 g mole
Hydrogen liberated = 0.00523 g equivalent
= 1/2 × 0.00523 = 0.00261 g mole
Total gases evolved = (0.00315 + 0.001307 + 0.00261) g mole
= 0.007067 g mole
Volume of gases evolved at NTP = 22400 × 0.007067 mL = 158.3 mL
Example
An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until
all the copper is deposited. The electrolysis is continued for seven more
minutes with volume of solution kept at 100 mL and the current at 1.2 amp.
Calculate the gases evolved at NTP during the entire electrolysis
Solution:
Amount of charge passed = 1.70 × 230 coulomb Amount of actual charge
passed = 90/100 × 1.70 × 230 = 351.9 coulomb
No. of moles of Zn deposited by passing 351.9 coulomb of charge =
1/(2x96500) × 351.9 = 0.000182
Molarity of Zn2+ ions after deposition of zinc = [0.160-
(0.000182×1000)/300]M = 0.154 M
Example
Calculate the electricity that would be required to reduce 12.3 g of
nitrobenzene to aniline, if the current efficiency for the process is 50 per cent.
If the potential drop across the cell is 3.0 volt, how much energy will be
consumed?
Solution:
The reduction reaction is
C6H5NO2 + 3H2 C6H5NH2 + 2H2O
Hydrogen required for reduction of 12.3/123 or 0.1 mole of nitrobenzene = 0.1
× 3 = 0.3 mole
Amount of charge required for liberation of 0.3 mole of hydrogen = 2 × 96500
× 0.3 = 57900 coulomb
Actual amount of charge required as efficiency is 50% = 2 × 57900 = 115800
coulomb](https://image.slidesharecdn.com/lect18examplesonelectrolysis-241010204213-fe95d401/85/lect-ures-no-18_Examples-on-Electrolysis-pdf-10-320.jpg)
lect ures no. 18_Examples on Electrolysis.pdf
- 1. lect 18;Examples on Electrolysis Example Calculate the current intensity passing through diluted HCl solution within 1 hour by using Pt electrodes which liberates 0.336L of O2 and H2 gases mixture at electrodes under standards conditions. Solution 1F causes librates 1eq(8g of O2 at anode) and 1eq( 1g of H2 at cathode) 96500 C causes librates of 1/4mol of O2 and 1/2mol of H2 Total moles 1/4+1/2=3/4mole 1mole of gas occupy 22.4L Volume of liberated gases 3/4molex22.4 Lmol-1 =16.8L 0.336Lx965ooC)/16.8L=1930 C ( Q= I=Q/t=1930C/3600sec=0.536 amp Λ= κ/c =(0.009751Ohm-1 cm-1 )/(0.1 eq/1000cm3 ) 97.51eq-1 Ohm-1 cm2 = Example A metallic object to be plated with copper is placed in a solution of CuSO4. 1- To which electrode of a direct current power supply should the object be connected? 2- What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours? Solution: 1- Since Cu2+ ions are being reduced, the object acts as a cathode and must be connected to the negative terminal (where the electrons come from!) 2- The amount of charge passing through the cell is Ixt=Q
- 2. (0.22 amp) × (5400 sec) = 1200 C or (1200 c) ÷ (96500 C F–1 ) = 0.012 F Since the reduction of one mole of Cu2+ ion requires the addition of two moles of electrons, the mass of Cu deposited will be Cu 2+ +2e → Cu 1F 0.5mol of Cu 0.012F ? Moles of deposited Cu = 0.006mol 1mol of Cu 63.54 g 0.006mol ? W=0.381g Example How much electric power is required to produce 1 metric ton (1000 kg) of chlorine from brine, assuming the cells operate at 2.0 volts and assuming 100 % efficiency? Solution: moles of Cl2 produced: (106 g) ÷ 70 g mol–1 = 14300 mol Cl2 faradays of charge: (2 F/mol) × (14300 mol) = 28600 F charge in coulombs: (96500 C/F) × (28600 F) = 2.76 × 109 c duration of electrolysis: (3600 s/h) x (24 h) = 86400 s current (rate of charge delivery): (2.76 × 109 amp/sec) ÷ (86400 sec) = 32300 amps power= (volt xamps): (2.0 v) × (32300 amp) = 64.6 kw energy in kW/h: (64.6 kw) × (24 h) = 1550 kw/h energy in joules: (1550 kw/h) × (3.6Mj/kw/h) = 5580 Mj (megajoules) (In the last step, recall that 1 w = 1 j/s, so 1 kw/h = 3.6 Mj) Example
- 3. Find the charge in coulomb on 1 g-ion of N3- . Solution: Charge on one ion of N3- = 3 × 1.6 × 10-19 coulomb Thus, charge on one g-ion of N3- = 3 × 1.6 10-19 × 6.02 × 1023 = 2.89 × 105 coulomb Example How much charge is required to reduce (a) 1 mole of Al3+ to Al and (b)1 mole of Mn4- to Mn2+ ? Al3+ + 3e- → Al Mn4- + 8H+ 5e- → Mn2 + + 4H2O Solution: (a) The reduction reaction is Al3+ + 3e- → Al Thus, 3 mole of electrons are needed to reduce 1 mole of Al3+ Q = 3 × F = 3 × 96500 = 289500 coulomb (b) The reduction is Mn4- + 8H+ 5e- → Mn2 + + 4H2O 1 mole 5 mole Q = 5 × F = 5 × 96500 = 482500 coulomb Example How much electric charge is required to oxidise (a) 1 mole of H2O to O2 and (b)1 mole of FeO to Fe2O3?
- 4. H2O → 1/2 O2 + 2H+ + 2e- FeO + 1/2 H2O → 1/2 Fe2O3 + H+ +e- Solution: (a) The oxidation reaction is H2O → 1/2 O2 + 2H+ + 2e- Q = 2 × F = 2 × 96500=193000 coulomb (b) The oxidation reaction is FeO + 1/2 H2O → 1/2 Fe2O3 + H+ +e- Q = F = 96500 coulomb Example Exactly 0.4 faraday electric charge is passed through three electrolytic cells in series, first containing AgNO3, second CuSO4 and third FeCl3 solution. How many gram of each metal will be deposited assuming only cathodic reaction in each cell? Solution: The cathodic reactions in the cells are respectively. Ag+ + e- → Ag Cu2+ + 2e- → >Cu and Fe3+ + 3e- --> Fe Hence, Ag deposited = 108 × 0.4 = 43.2 g Cu deposited = 63.5/2×0.4=12.7 g and Fe deposited = 56/3×0.4=7.47 g Example
- 5. An electric current of 100 ampere is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP. Solution: The reaction taking place at anode is 2Cl- +2e→ Cl2 - Q = I × t = 100 × 5 × 60 ×60 coulomb =1800000 coulomb 1F 96500 coulomb 1mole of e 2F 193000 coulomb 1mole of Cl2 1800000coulomb ? ?=9.3264 mole Volume of Cl2 liberated at NTP = 9.3264 × 22.4 = 208.91 L Example A 100 watt, 110 volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hours? The cathodic reaction is Cd2+ + 2e- → Cd Solution: power = I× V 100 watt = I × 110 v I = 100/110 =0.90 amp Q = I × t = 0.90amp×10×60×60 sec= 3240 amp.sec-1 =3240 coulomb 1F 96500 coulomb 1mole of e 2F 193000 coulomb 1mole of Cd 3240 coulomb ? ?= 0.1678 mole of Cd 1mole of Cd= 112.4g 0.1678 mole of Cd=18.86g
- 6. Example In an electrolysis experiment, a current was passed for 5 hours through two cells connected in series. The first cell contains a solution gold salt and the second cell contains copper sulphate solution. 9.85 g of gold was deposited in the first cell. If the oxidation number of gold is +3, find the amount of copper deposited on the cathode in the second cell. Also calculate the magnitude of the current in ampere. Solution: (Mass of Au deposited)/(Mass of Cu deposited)=(Eq.mass of Au)/(Eq.Mass of Cu) Eq. mass of Au = 197/3 Eq. mass of Cu 63.5/2 (9.85g)/(Mass of Cu deposited) = 63.5/2 x 3/197 g Mass of Cu deposited= 4.7625 g Let Z be the electrochemical equivalent of Cu. E = Z × 96500 or Z =E/96500=63.5/(2×96500) Applying W = Z × I × t T = 5 hour = 5 × 3600 second 4.7625 = 63.5/(2×96500) × I × 5 × 3600 or I = (4.7625 × 2 × 96500)/(63.5 × 5 × 3600)=0.0804 ampere Example How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of 80 cm2 with 0.005 cm thick layer? Density of silver is 10.5 g/cm3 . Solution: Mass of silver to be deposited = Volume × density = Area ×thickness × density Given: Area = 80 cm2 thickness = 0.0005 cm and density = 10.5 g/cm3 Mass of silver to be deposited = 80 × 0.0005 × 10.5 = 0.42 g 1F 96500 coulomb 1mole of e 1F 96500 coulomb 1mole of Ag(108g)
- 7. ? (0.42g) ?=357.40coulomb Q=I ×t t=Q/I t=375.4/3=125.1sec Example What current strength in ampere will be required to liberate 10 g of chlorine from sodium chloride solution in one hour? 2Cl- → Cl2 + 2e- Solution: 2F 193000 coulomb liberates 35.5g of Cl2 ? 10g ?=54366.1 coulomb Q=I×t I=Qlt=54366 coulomb/3600sec I=15amp Example 0.2964 g of copper was deposited on passage of a current of 0.5 ampere for 30 minutes through a solution of copper sulphate. Calculate the atomic mass of copper. (1 faraday = 96500 coulomb) Solution: Quantity of charge passed 0.5 × 30 × 60 = 900 coulomb 900 coulomb deposit copper = 0.2964 g 96500 coulomb deposit copper = 0.2964/900×96500=31.78 g Thus, 31.78 is the equivalent mass of copper. At. mass = Eq. mass × Valency = 31.78 × 2 = 63.56
- 8. Example 19 g of molten SnCI2 is electrolysed for some time using inert electrodes until 0.119 g of Sn is deposited at the cathode. No substance is lost during electrolysis. Find the ratio of the masses of SnCI2 : SnCI4 after electrolysis. Solution: The chemical reaction occurring during electrolysis is 2SnCl2 → SnCl4 + Sn 2×190 g 261 g 119 g 119 g of Sn is deposited by the decomposition of 380 g of SnCl2 So, 0.119 g of SnCl2 of Sn is deposited by the decomposition of 380/119×0.119=0.380 g of SnCl2 Remaining amount of SnCl2 = (19-0.380) = 18.62 g 380 g of SnCl2 produce = 261 g of SnCl4 So 0.380 g of SnCl2 produce = 261/380×0.380=0.261 g of SnCl Thus, the ratio SnCl2 : SnCl4 =18.2/0.261 , i.e., 71.34 : 1 Example A current of 2.68 ampere is passed for one hour through an aqueous solution of copper sulphate using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At. mass of copper = 63.5). Solution: The electrode reactions are: Cu2+ + 2e- → Cu (Cathode) 1 mole 2 × 96500 C Cu → Cu2+ + 2e- (Anode) Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves. Charge passed through cell = 2.68 × 60 × 60 coulomb Copper deposited or dissolved = 63,5/(2×96500)×2.68×60×60 =3.174 g Increase in mass of cathode = Decrease in mass of anode = 3.174 g Example An ammeter and a copper voltameter are connected in series through which a constant current flows. The ammeter shows 0.52 ampere. If 0.635 g of copper is deposited in one hour, what is the percentage error of the ammeter? (At. mass of copper = 63.5) Solution : The electrode reaction is: Cu2+ + 2e → Cu
- 9. 1 mole 2 × 96500 C 63.5 g of copper deposited by passing charge = 2 × 96500 Coulomb 0.635 g of copper deposited by passing charge =(2×96500)/63.5×0.653 coulomb = 2 × 965 coulomb = 1930 coulomb We know that Q = l × t 1930 = I × 60 × 60 I= 1930/3600=0.536 ampere Percentage error = ((0.536-0.52))/0.536×100=2.985 Example A current of 3.7 ampere is passed for 6 hours between platinum electrodes in 0.5 litre of a 2 M solution of Ni(NO3)2. What will be the molarity of the solution at the end of electrolysis? What will be the molarity of solution if nickel electrodes are used? (1 F = 96500 coulomb; Ni = 58.7) Solution: The electrode reaction is Ni2+ + 2e- → Ni 1 mole 2 × 96500 C Quantity of electric charge passed = 3.7 × 6 × 60 × 60 coulomb = 79920 coulomb Number of moles of Ni(NO3)2 decomposed or nickel deposited = (1.0 - 0.4140) = 0.586 Since 0.586 moles are present in 0.5 litre, Molarity of the solution = 2 × 0.586 = 1.72 M When nickel electrodes are used, anodic nickel will dissolve and get deposited at the cathode. The molarity of the solution will, thus, remain unaffected Example An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with volume of solution kept at 100 mL and the current at 1.2 amp. Calculate the gases evolved at NTP during the entire electrolysis. Solution: 0.4 g of Cu2+ = 0.4/31.75 = 0.0126 g equivalent At the same time, the oxygen deposited at anode = 0.0126 g equivalent = 8/32 × 0.0126 = 0.00315 g mol
- 10. After the complete deposited of copper, the electrolysis will discharge hydrogen at cathode and oxygen at anode. The amount of charge passed = 1.2 × 7 × 60 = 504 coulomb So, Oxygen liberated = 1/96500 × 504 = 0.00523 g equivalent = 8/32 × 0.00523 = 0.001307 g mole Hydrogen liberated = 0.00523 g equivalent = 1/2 × 0.00523 = 0.00261 g mole Total gases evolved = (0.00315 + 0.001307 + 0.00261) g mole = 0.007067 g mole Volume of gases evolved at NTP = 22400 × 0.007067 mL = 158.3 mL Example An acidic solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with volume of solution kept at 100 mL and the current at 1.2 amp. Calculate the gases evolved at NTP during the entire electrolysis Solution: Amount of charge passed = 1.70 × 230 coulomb Amount of actual charge passed = 90/100 × 1.70 × 230 = 351.9 coulomb No. of moles of Zn deposited by passing 351.9 coulomb of charge = 1/(2x96500) × 351.9 = 0.000182 Molarity of Zn2+ ions after deposition of zinc = [0.160- (0.000182×1000)/300]M = 0.154 M Example Calculate the electricity that would be required to reduce 12.3 g of nitrobenzene to aniline, if the current efficiency for the process is 50 per cent. If the potential drop across the cell is 3.0 volt, how much energy will be consumed? Solution: The reduction reaction is C6H5NO2 + 3H2 C6H5NH2 + 2H2O Hydrogen required for reduction of 12.3/123 or 0.1 mole of nitrobenzene = 0.1 × 3 = 0.3 mole Amount of charge required for liberation of 0.3 mole of hydrogen = 2 × 96500 × 0.3 = 57900 coulomb Actual amount of charge required as efficiency is 50% = 2 × 57900 = 115800 coulomb
- 11. Energy consumed = 115800 × 3.0 = 347400 J = 347.4 kJ Example After electrolysis of a sodium chloride solution with inert electrodes for a certain period of time, 600 mL of the solution was left which was found to be 1 N in NaOH. During the same period 31.75 g of copper was deposited in the copper voltameter in series with the electrolytic cell. Calculate the percentage theoretical yield of NaOH obtained. Solution: Equivalent mass of NaOH = 40/1000 × 600 = 24 g Amount of NaOH formed = 40/1000 × 600 = 24 g 31.75 g of Cu = 1 g equivalent of Cu. During the same period, 1 g equivalent of NaOH should have been formed. 1 g equivalent of NaOH = 40 g % yield = 24/40 × 100 = 60 Example After electrolysis of a sodium chloride solution with inert electrodes for a certain period of time, 600 mL of the solution was left which was found to be 1 N in NaOH. During the same period 31.75 g of copper was deposited in the copper voltameter in series with the electrolytic cell. Calculate the percentage theoretical yield of NaOH obtained. Solution: Equivalent mass of NaOH = 40/1000 × 600 = 24 g Amount of NaOH formed = 40/1000 × 600 = 24 g 31.75 g of Cu = 1 g equivalent of Cu. During the same period, 1 g equivalent of NaOH should have been formed. 1 g equivalent of NaOH = 40 g % yield = 24/40 × 100 = 60 Example Cadmium amalgam is prepared by electrolysis of a solution of CdCl2 using a mercury cathode. Find how long a current of 5 ampere should be passed in order to prepare 12% Cd-Hg amalgam on a cathode of 2 g mercury. At mass of Cd = 112.40. Solution: 2 g Hg require Cd to prepare 12% amalgam = 12/88 × 2 = 0.273 g Cd2+ + 2e- → Cd 1 mole 2 × 96500C 112.40g Charge required to deposit 0.273 g of Cd = 2*96500/112.40 × 0.273 coulomb
- 12. Charge = ampere × second Second = 2*96500*0.273/112.40*5 = 93.75 Example :Electrolysis of water Pure water is an insulator and cannot undergo significant electrolysis without adding an electrolyte. If the object is to produce hydrogen and oxygen, the electrolyte must be energetically more difficult to oxidize or reduce than water itself. Electrolysis of a solution of sulfuric acid or of a salt such as NaNO3 results in the decomposition of water at both electrodes: cathode: 2 H2O + 2 e– → H2(g) + 2 OH– E° = –0.83 v anode: H2O → ½ O2(g) + 2 H+ + 2 e– E° = –1.23 v net: 3 H2O(l) → H2(g) + ½ O2(g) E = –2.06 v