CHAP. 19 lecture 3:

19.8 ELECTROLYSIS

ELECTROLYTIC CELLS:  electrical energy from an outside
source causes nonspontaneous chemical reactions to
occur. 

WE ARE SUPPLYING THE ENERGY TO MAKE THIS REACTION
GO!!  (Ex.:  electroplating metals).


19.8  ELECTROLYSIS

ELECTROLYSIS OF MOLTEN SODIUM CHLORIDE

Take molten NaCl and electrolyze it:
At the anode (oxidation):  2 Cl-(l)  ---->  Cl2(g)  +  2e-
At the cathode (reduction):  2Na+(l)  +  2e-  --->  2Na(l)

Overall:  2Na+(l)  +  2 Cl-(l)  --->  2Na(l)  +  Cl2(g)
Theoretical estimates show that Eo for this process is
about -4V, therefore, we have to SUPPLY more than 4V
to make this reaction occur.

ELECTROLYSIS OF WATER

Take water and electrolyze it:
     2H2O(l)  --->  2H2(g)  +  O2(g)     DHo = 474.4 kJ

If we try to electrolyze with electricity, nothing happens
because there just are not ions (remember that
H+ = 1 x 10-7 M in water!).  Therefore, if we make the
solution 0.1M H2SO4, then there are enough ions to
conduct electricity, and the reaction proceeds.
At the anode (oxidation): 
                     2H2O(l)  ---->  O2(g)  + 4H+(aq) + 4e-
At the cathode (reduction): 
                     H+(aq)  +  e-  --->  1/2H2(g)

Overall:  2H2O(l)  ---->  2H2(g) + O2(g)
   The 0.1M H2SO4 is not consumed in this reaction.



ELECTROLYSIS OF AQUEOUS NaCl SOLUTION

Possible reactions at the anode could be:

            2 Cl-(aq)   --->  Cl2(g) +  2e- 
            2H2O(l)    --->   O2(g) + 4H+(aq) + 4e-

By experiment, we find that the gas liberated at
       the anode is chlorine and not oxygen.

Possible reactions at the cathode could be:

     2H+(aq)  +  2e-  --->  H2(g)     Eo = 0.00V       
     2H2O(l)  + 2e-  --->   H2(g) + 2OH-(aq)   Eo = -0.83V
      Na+(aq)  + e- --->  Na(s)                          Eo = -2.71V

The last reaction can be ruled out because it has such
a negative standard reduction potential.  Since, in water,
H+ = 1 x 10-7 M then the 1st reaction is very improbable.

Therefore, the 1/2-cell reactions for the electrolysis
of aqueous NaCl are:
 Anode (oxidation)  2 Cl-(aq)   --->  Cl2(g) +  2e- 
Cathode (reduction) 2H2O(l) + 2e- ---> H2(g) + 2OH-(aq)

Giving:  2H2O(l) + 2 Cl-(aq) ---> H2(g) + 2OH-(aq) + Cl2(g)


Quantitative Aspects of Electrolysis

The FARADAY is the amount of electricity that corresponds
to the gain or loss of 6.022 x 1023 electrons, or 1 mole of
electrons.

         1 Ampere x sec = 1 Coulomb;    1(A)(s)  = 1 C
         1 Faraday = 1 F = 6.022 x 1023 e-= 96500 C
 

If 1 Faraday of electricity is transferred, that is
    = 96500 Coulombs and that is = 1 mol of electrons,

Ag+(aq) + e- -----> Ag(s)   1 mol e- transferred in this
                                                  1/2 reaction
               Therefore, 1 mol Ag produced = 107.87g

Cu2+(aq) + 2e- -----> Cu(s)  2 mol e- transferred in this
                                                    1/2 reaction
Therefore, we would require 2 F of electricity per mole 
 of Cu produced; since we only have 1 F in our example,  
          1/2 mol Cu produced = 63.54/2 g = 31.77g

Cr3+(aq) + 3e- -----> Cr(s)  3 mol e- transferred in this
                                                    1/2 reaction

 Therefore, we would require 3 F of electricity per mole
  of Cr produced; since we only have 1 F in our example,
         1/3 mol Cr produced = 52.0/3 g = 17.33g

2Cl-(aq)  -----> Cl2(g) + 2e-  2 e- transferred in this
                                                        1/2 reaction
            1/2 mol Cl2 produced = 70.90/2 g = 35.45g
 

                      q = (i)(t) = amps x time(sec)

    number of Coulombs transferred = current x time

 Since 1 coulomb = 1 amp-s,
                   1 mole of e- transferred = 96500 C
 

Calculate the mass of Cu(s) produced when a current of
2.05 amperes is passed for 10.0 minutes through
molten CuSO4
                        Cu2+(aq) + 2e- -----> Cu(s)

q = i t = amps x time(sec)=(2.05A)(10.0 min)(60s/min)
    q = 1230 C

# moles of e- = (1230 C)x(1 mol e-/96500 C)
                       = 0.01275 mol e-

# moles of Cu = (0.01275 mol e-)(1 mol Cu/2 mol e-)
                        =  0.006373 mol Cu
mass of Cu = (0.006373)(63.55 g/mol) = 0.405 g Cu
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Current of 1.20 Amps is passed through molten NaCl and
  then molten SrCl2 in series for 1.00 hour and 20 minutes.
  Calculate the mass of Nao, Sro and Cl2(g) in each cell and
  the total mass of Cl2(g) produced.

     Na+(l) + e- -----> Nao

     Sr2+(l) + 2e- -----> Sro

     2Cl-  -----> Cl2 + 2e-

q = i t = amps x time(sec)=(1.20 A)(80.0 min x 60s/min)
    q = 5760 C

# moles of e- = (5760 C)(1 mol e-/96500 C)
                       = 0.05979 mol e-
# moles of Na = (0.05979 mol e-)(1 mol Na/1 mol e-)
     mass of Na = (0.05979)(22.99 g/mol) = 1.37 g Na

# moles of Sr = (0.05979 mol e-)(1 mol Sr/2 mol e-)
     mass of Sr = (0.02989)(87.62 g/mol) = 2.62 g Sr

# moles of Cl2 = (0.05979 mol e-)(1 mol Cl2/2 mol e-)
     mass of Cl2 = (0.02989)(70.906 g/mol)
                         = 2.12 g Cl2 in cell 1

<>Also, same current and time in cell 2, therefore another
    2.12 g Cl2 made in cell 2; therefore, total mass of <>
    chlorine <>in both cells = 4.24 g Cl2
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --

In electrolytic cells, the "+" electrode is the ANODE (OX)
and the "-" electrode is the CATHODE (RED).

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