Faraday's Laws of Electrolysis

During electrolysis, substances like oxygen gas, chlorine gas, bromine etc. are liberated at the anode depending on the electrolytes used, while substances like hydrogen gas, copper, silver etc. are liberated or deposited at the cathode. The volume of gases liberated or mass of metals deposited depend on the amount of electricity that is passed, be it carried out on one or more electrolytes. These relationships were summarized by Michael Faraday into what are now known as the Laws of Electrolysis.
Faraday's 1st Law of Electrolysis
This law states that the mass of a substance deposited or liberated at the electrodes during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. Mathematically, this can be expressed as:
                          m α Q ..................................(i)
where, m = mass in grams (g); and Q = quantity of electricity in Coulombs (C)
but,
                         Q = I x t ................................(ii)
where, I = current in Amperes (A); and t = time in seconds (s)
From equation (ii), it implies that 1 Coulomb = 1 Ampere x 1 sec
                                                                             = 1 Ampere sec (As)
Substituting equation (ii) into (i) will give us:
                          m α It ...................................(iii)
Removing the sign of proportionality, by introducing the equal sign and a constant of proportionality, equation (iii) changes to:
                         m = EIt ...........................................(iv)
where, E = electrochemical equivalence, and most times has a value of 1. Its unit is grams per Coulomb (g C^-1)
Equation (i) can also be written as:
                          m = EQ ............................................(v)
From equation (v), a plot of mass (m) against quantity of electricity (Q) gives a straight line graph passing through the origin, with a slope equal to the electrochemical equivalence of the element involved.
(Please refer to your textbooks for the verification of Faraday's 1st Law of Electrolysis)
The amount of electricity required to discharge one electron during electrolysis is known as the Faraday and is equivalent to 96500Coulombs, i.e., 1F = 96500C
The quantity of electricity required to deposit 1 mole of various classes of ions is summarized below:
For a univalent monoatomic ion, (e.g., Na+, Ag+, Au+) = 1mol of e- = 1F is required
      a divalent monoatomic ion, (e.g., Zn2+, Cu2+, S2-) = 2mol of e- = 2F are required
      a trivalent monoatomic ion, (e.g., Al3+, P3-) = 3mol of e- = 3F are required
      a univalent diatomic ion, (e.g., Cl-, F-, Br-) = 2mol of e- = 2F are required
      a divalent diatomic ion, (e.g., O2-) = 4mol of e- = 4F are required
     a trivalent diatomic ion, (e.g., N3-) = 6mol of e- = 6F are required
Examples
Question 1
What is the mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s? [Ag = 108, 1F = 96500 C]
Answer
Quantity of electricity passed = Current x Time
                                                   = 10 A x 4830s
                                                   = 48300 C
Silver ions undergo reduction by gaining one electron at the cathode, according to the equation:
                          Ag+(aq) + e- ----> Ag(s)
                        108g          1mol       108g
From the above equation, 1 mole of electron is required to deposit 1 mole of Ag,
i.e., 1F = 1 mole of Ag
but, 1F = 96500C and 1mole = 108 g of Ag
Therefore,
If, 96500 C of electricity are required to deposit 108 g of Ag, then
    48300 C will be required to deposit xg of Ag
                            x = 48300 x 108/96500
                               = 54.05 g of Ag
Question 2
If a given quantity of electricity deposits 0.65g of Zn2+, what amount of Hg2+ would be deposited by the same quantity of current? [Zn = 65, Hg = 201]
Answer
                      Zn2+(aq) + 2e- ----> Zn(s)
                     1mol            2F             1mol
                      65g             2F              65g
From the above equation, 2F of electricity are needed to deposit 1 mole of zinc metal, and 1 mole of zinc = 65g.
If, 65g of zinc are deposited by 2F of electricity, then
0.65g of zinc will be deposited by x F of electricity
                                x = 0.65 x 2/65
                                   = 0.02F
For mercury, Hg;
                      Hg2+(aq) + 2e- ----> Hg(l)
                      1mol            2F            1mol
                      201g            2F            201g
If, 2F of electricity deposit 201g of Hg
0.02F of electricity will deposit xg of Hg
                             x = 0.02 x 201/2
                                = 2.01g of Hg
Question 3
What current will deposit 3.25g of zinc in 2hrs? [Zn = 65, F = 96500 C mol^-1]
Answer
From the equation of the reduction of zinc ion to zinc metal in question 2 above, we can see that:
65g of zinc = 2F = 2 x 96500C (recall 1F = 96500C)
If, 65g of Zn are deposited by 2 x 96500C of electricity
3.25g of Zn will be deposited by xC of electricity
                              x = 3.25 x 193000/65
                                 = 9650C
but,
Quantity of electricity (C) = Current (A) x Time (s)
                                9650C = I x 2 x 3600s               (1hr = 3600s)
                                          I = 9650/7200
                                            = 1.34A
Note that we chose to work with the quantity of electricity directly in Coulombs because it will be faster to calculate the value of the current from it than using Faraday.
Question 4
What quantity of aluminium is deposited when a current of 10A is passed through a solution of an aluminium salt for 1930s? [Al = 27, F = 96500 C mol^-1]
Answer
Current passed = 10A; Time taken = 1930s
Quantity of electricity passed = current x time
                                                   = 10 x 1930
                                                   = 19300C
Aluminium ions are reduced by gaining 3 electrons each at the cathode,
                       Al3+(aq) + 3e- ----> Al(s)
                       1mol           3F            1mol
                       27g             3F             27g
If, 3 x 96500C of electricity deposit 27g of Al, then
          19300C of electricity will deposit xg of Al
                                x = 19300 x 27/289500
                                   = 1.8g of Al
Question 5
0.444g of a divalent metal is deposited when a current of 0.45A is passed through a solution of its salt for 50mins. Calculate the relative atomic mass of the metal.
Answer
Quantity of electricity passed = Current x Time
                                                    = 0.45A x 50 x 60
                                                    = 1350C
Being a divalent metal, 2F of electricity will be required to deposit 1 mole of the metal according to the equation:
                       M2+(aq) + 2e- ----> M(s)
                       1mol          2F             1mol
If, 1350C of electricity deposit 0.444g of the metal, then
2 x 96500C of electricity will deposit x g of the metal
                                 x = 193000 x 0.444/1350
                                    = 63.48g
Faraday's 2nd Law of Electrolysis
This law states that when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the element discharged are inversely proportional to the charges on the ions of the element.
That is to say, if the same quantity of electricity is passed through two electrolytic cells containing silver trioxonitrate (V), AgNO3, solution and copper (II) tetraoxosulphate (VI), CuSO4, solution respectively, 2 moles of Ag will be deposited in the cell containing AgNO3, while 1 mole of Cu will be deposited in the one containing CuSO4 solution, as illustrated below:
                    Ag+(aq)  +   e- ----> Ag(s)
                    1mol            1F           1mol
                  Cu2+(aq) + 2e- ----> Cu(s)
                   1mol           2F             1mol
From the above, 1F of electricity is required to deposit 1 mole of Ag because Ag+ is univalent. Similarly, 2F of electricity are required to deposit 1 mole of Cu because Cu2+ is divalent. Since the same quantity of electricity is passed, the same number of moles of electrons must be transferred. So, balancing the number of electrons involved, the above equation becomes:
                2Ag+(aq)  +   2e- ----> 2Ag(s)
                2mol               2F             2mol
                 Cu2+(aq) +   2e- ----> Cu(s)
                 1mol              2F             1mol
At the end of the electrolysis:
Ag - Charge of ion : No of Moles deposited = +1: 2
Cu - Charge of ion : No of Moles deposited  = +2:1
(Please check your textbooks for the verification of Faraday's 2nd Law of Electrolysis)
Examples
Question 6
A copper and silver voltameter are connected in series. If a current of 4A flow for 1hr 30mins, calculate the masses of copper and silver deposited [Cu = 63.5, Ag = 108, F = 96500C mol^-1]
Answer
Quantity of electricity passed = Current x Time
                                                   = 4 x [(1 x 3600) + (30 x 60)]
                                                   = 4 x (3600  + 1800)
                                                   = 4 x 5400
                                                   = 21600C
                     Cu2+(aq) +  2e- ----> Cu(s)
                     1mol             2F             1mol
                     63.5g            2F             63.5g
From the above,
2F of electricity are required to deposit 1 mole of copper, i.e.,
2 x 96500C of electricity are required to deposit 63.5g of copper
      21600C of electricity will deposit x g of copper
                                 x = 21600 x 63.5/193000
                                    = 7.11g of copper
                      2Ag+(aq) + 2e- ----> 2Ag(s)
                      2mol            2F            2mol
                      2x108g        2F            2x108g
Conversely,
2F of electricity are required to deposit 2 moles of silver
2 x 96500C of electricity are required to deposit 2 x 108g of silver
      21600C of electricity will deposit x g of silver
                                  x = 21600 x 216/193000
                                     = 24.17g of silver
Note that if we convert the masses of copper and silver deposited into moles, we will obtain 0.11 and 0.22 moles respectively. This gives a ratio of 1:2, compared to the ratio of their charges, which is 2:1.
Question 7
A voltameter containing silver trioxonitrate (V) solution was connected in series to another containing copper (II) tetraoxosulphate (VI) solution. When a current of 0.2amp was passed through the solutions, 0.780g of silver was deposited. Calculate (i) the mass of copper deposited in the copper voltameter. (ii) the quantity of electricity used. (iii) the time of current flow [Cu = 63.5, Ag = 108, 1F = 96500C]
Answer
                      2Ag+(aq) + 2e- ----> 2Ag(s)
                      Cu2+(aq) + 2e- ----> Cu(s)
From the above equations, 2F of electricity are required to deposit 1 mole of Cu and 2 moles of Ag.
(i) For every 63.5g of Cu deposited, 2 x 108g of Ag are deposited, so
                           xg of Cu will be deposited when 0.780g of Ag are deposited
                                x = 0.780 x 63.5/216
                                   = 0.229g of Cu
(ii) If 2 x 108g of Ag are deposited by 2 x 96500C of electricity, then
            0.780g of Ag will be deposited by xC of electricity
                                x = 0.780 x 193000/216
                                   = 696.94C
(iii) Quantity of electricity = Current x Time
                              696.94C = 0.2A x t
                                           t = 696.94/0.2
                                             = 3484.7s
but, 60s = 1min
                                             = 58.07mins
Question 8
A current is passed through three electrolytic cells connected in series containing solutions of silver trioxonitrate (V), copper (II) tetraoxosulphate (VI) and brine respectively. If 38.1g of copper are deposited in the second electrolytic cell, calculate
(a) the mass of silver deposited in the first cell,
(b) the volume of chlorine liberated in the third cell at 27°C and 900mmHg pressure. [Ag = 108, Cu = 63.5, 1F = 96500C, Molar Gas Volume at s.t.p. = 22.4dm^-3]
Answer
                      2Ag+(aq) + 2e- ----> 2Ag(s)
                      Cu2+(aq) + 2e- ----> Cu(s)
                       2Cl-(aq) ----> Cl2(g)  +  2e-
                       2mol              22.4dm^3   2F
From the above equations, 2F of electricity are required to deposit 1 mole of copper, 2 moles of silver and liberate 1 mole (22.4dm^3) of chlorine gas.
(a) For every 63.5g of Cu deposited, 2 x 108g of Ag are deposited, so
                for 38.1g of Cu deposited, x of Ag will be deposited
                                 x = 38.1 x 216/63.5
                                    = 129.6g of Ag
(b) For every 63.5g of Cu deposited, 22.4dm^3 of Cl2 are liberated at s.t.p., so
                for 38.1g of Cu deposited, x dm^3 of Cl2 will be liberated
                               x = 38.1 x 22.4/63.5
                                  = 13.44dm^3 of chlorine gas at s.t.p.
We will now convert this volume to that at 27°C and 900mmHg, using the general gas equations:
                             P1V1/T1 = P2V2/T2
where P1 = 760mmHg, T1 = 273K, V1 = 13.44dm^3, P2 = 900mmHg, T2 = 27 + 273 = 300K, V2 = ?
                               V2 = P1V1T2/P2T1
                                     = 760 x 13.44 x 300/900 x 273
                                     = 3064320/245700
                                     = 12.47dm^3
Therefore, the volume of chlorine gas liberated in the third cell at 27°C and 900mmHg is 12.47dm^3.
Do These
Question 1
Calculate the volume of oxygen liberated at 570K and 1.82 x 10^5 Nm^-2 when a current of 2.5A is passed through dilute sodium chloride solution for 3mins. [Molar Volume of a Gas = 22.4dm^3; Standard Pressure = 1.01 x 10^5 Nm^-2; Standard Temperature = 273K; 1F = 96500C]
Question 2
A solution of copper (II) tetraoxosulphate (VI) was electrolysed between pure copper electrodes and the following results were obtained:
Mass of anode before experiment      =  14.40g
Mass of anode after experiment         =    8.00g
Mass of cathode before experiment   =  11.50g
From the information provided,
(i) calculate the mass of the cathode after experiment
(ii) write an equation for the reaction at the  I. anode II. cathode
(iii) state whether the colour of the solution would change during the electrolysis. Give a reason for your answer.
(iv) if the electrolysis was carried out for 2 hours 40 minutes with a current of 2.0 Amperes, determine the value of the Faraday.
Question 3
State Faraday's Laws of electrolysis and show how you would verify the Laws
Question 4
If three electrolytic cells with each containing a solution of gold trioxonitrate (V), zinc tetraoxosulphate (VI) and brine respectively are connected together and a given quantity of electricity passed through them. If 26g of zinc are deposited in the second electrolytic cell, calculate
(a) the mass of gold deposited in the first cell,
(b) the volume of chlorine liberated in the third cell at 17°C and 800mmHg pressure. [Au = 197, Zn = 65, 1F = 96500C, Molar Gas Volume at s.t.p. = 22.4dm^-3]
(c) the quantity of electricity passed
(d) the current flow if the electrolysis took 3 hours 50 minutes.
Follow us on Twitter: @gmtacademy
WhatsApp: 07034776117
Like our Page on Facebook: www.facebook.com/greatermindstutors