Electrolysis Problems and their Solutions

Electrolysis is the process in which electrical energy changes into chemical energy. It passes the electric current that produces a chemical change.

Ex. (i) Breakdown of water into hydrogen.

      (ii) Breakdown of water into oxygen.

Electrolysis

Components that are mainly required for producing electrolysis are Electrolyte, Electrodes and external power source.Process of electrolysis is usually done in the electrolytic cell which contains two electrodes cathode and anode. These two electrodes are connected to a direct electric current and electrolyte. The electrodes are made up of metals and semiconductors.

Electrolysis Problems and their Solutions

Question 1. An iron pipe with a 14 cm in diameter and length of an 1 metre is galvanised to a thickness of 0.01 cm using zinc nitrate solutions and a current of 25 A. Find the loss of mass from zinc ,anode and what will be the time required for the electrolysis? 

Density of zinc is 7.14g/cm3. The equivalent weight of zinc is 32.8.

Ans. Mass of the zinc to be coated = V * D( where ‘V’ is volume of zinc) Volume of the zinc = 2rxl * 0.01 cm3(where ‘r’ is the radius and ‘l’ is the length).

= 2*22/7*14/2*1000*0.01

= 440 g

Now, Mass of the zinc lost from the anode = 440 g

By Faraday first  law of electrolysis,

Formula of faraday first law of electrolysis.

M = EQ/96485 

= EIt/96485(here ‘m’ is the mass, ‘E’ is the equivalent mass, ‘I’ is the current, ‘t’ is the time)

So,

t = 96485 * m/E * I

= 96485 * 440/32.8 * 25

= 51772 sec. Or 14.4 hrs

Question 2. What volume of O2(g) and H2 are produced at STP when 30 A of current is passed through a dilute aqueous solution of K2SO4 for 193 min ?

Ans. Current = 30 A 

         Time = 193 min

The quantity of current passed = 193 * 60 * 30 = 347400 columbus.

                                                         = 347400/96500

                                                         = 3.6 Faraday.

1 mole of hydrogen and 0.5 mole of oxygen( by faraday law)

When 3.6 Faraday of electricity in 3.6 moles of hydrogen and (3.6 * 0.5) = 1.8 moles of oxygen.

So, the volume of Hydrogen at STP = 3.6 * 22.4 = 80.64 litres.

Volume of Oxygen at STP = 1.8 * 22.4 = 40.32 litres

Question 3. What mass of nickel metal can be plated on the cathode from a Ni(NO3)2 solution using a current of 12 A for 20 min.

Ans. Given,

          I = 12 A

         Time = 20*60 = 1200 s

Charge  = 12 * 1200

              = 14400 C

According to the reaction.

Ni2+(aq.) + 2e → Ni(s)

Nickel deposit by (2 * 96487) C = 58.7 g

Nickel deposit by 14400 C = 58.7 * 14400/2 * 96487

                                         = 40,779 g

Question 4. During the electrolysis of molten NaCl, the time required to produce 0.10mol of chlorine gas using  2 A current.

Ans. I= 3 A

          2Cl → Cl2 +2e 

         W = E/96500 * it

        0.1 * 71 = 35.5/96500 * 3 * t

                t = 76.570(approx)sec.

Question 5.  If 10g of chlorine is evolved in 6 hour from the NACl solution by the current.

Ans. 2Cl   → Cl2 +2e  

        2F → 1mol of Cl2

          10/35.5 → 10/35.5*2F 

  10 * 2/35.5 = change/96500 = 6 * 60 * 60 * I/96500

I = 4 A.(approx)

Question 6. With a 20 A current, how long will it take to plate out 50.00 g Al from a solution of AlCl3?

Ans. First,  Convert gram into coulomb.     

     50.00 g Al * 1 mol/26.98 g * 3 / 1 Al * 96500/1 mol 

               = 563,500 C

Now, calculate the amount of time required 

     I = q/t

        20,00 A = 563,500 C/t

                       = 281.75 sec.

Question 7. The same amount of electricity is passed through two cells containing molten Al2O3 and molten NACl. If 1.6 g of Al were liberated in one cell, the amount of Na liberated in another cell.

Ans. When, 27 g of Al → 3F

           1.8 g →  1.6/27 * 3F

Now, the number of mole Na produced.

   Electricity = 1.6/27 * 3 * 23

             Mass = 1.164 g

Question 8. Density of copper is 8.94 g/cm3, number of Faradays required to plate an area is 20 * 10 cm2 of thickness of 10-2 cm using CuSO4 solution.

Ans. Volume of copper  is = 20 * 10 * 10-2 = 2 cm3

  Density = Mass/Volume

            Mass of the copper is = 8.94 g

        Cu2+ + 2e → 2 mol F

           63.5 g →  2 mol F

            8.94 g →  2/63.5 * 8.94 = 0.28 F

Question 9. Calculate the mass of magnesium  which was deposited by electrolysis of molten MgCl2by passing 10A of current in 600 minutes.

Ans. Mass of Mg = (24 * 193 * 600 * 60)/(2 * 96500)

                            = 4.168g

Question 10. A current of 3.50 A is passed through a wire of 4.00 min.Find quantity of electric charge that passes through a wire?

Ans. Current that passes through an wire= 3.50 A

           I = q/t

        3,50 A = q/180 sec.

       q = 51.42 C   

Question 11. If the cathode is an Hg electrode Find the maximum mass of amalgam from a solution?

Ans. Na + 1e + Hg → NaHg

      2.0 moles of NACl are present in the solution of 2 moles of Na.

    The mass of 2 moles of NaHg = 2(23 + 200)

                                                     = 446 g

Conclusion

Electrodes and external power source.Process of electrolysis is usually done in the electrolytic cell which contains two electrodes cathode and anode. These two electrodes are connected to a direct electric current and electrolyte. The electrodes are made up of metals and semiconductors.