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How much electricity is required in coulomb for the oxidation of
1 mol of FeO to Fe2O3

2FeO+H2OFe2O3+2H++2e-2 mol                                                2 mol or 2F                                                 (oxidation reaction)
Charge on Fe in compound FeO        = + 2
Charge on Fe in compound Fe2O= + 3
Change in charge  n = 1
Electricity required          = nF
 = 96487 Coulombs
 = 9.65 × 10Coulombs
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Conductivity of 0.00241 M acetic acid solution is 7.896 x 10–5 S cm-1. Calculate its molar conductivity in this solution. If Λ°m for acetic acid be 390.5 S cm2 mol–1, what would be its dissociation constant?

Given that ,        
 κ = 7.896 × 10−5 S m−1
C=M= 0.00241 mol L−1
The formula of molar conductivity,
Λ= (k  × 1000)/M
Plug the value we get
Λm = (7.896 × 10−5  ×  1000)/ 0.00241
      = 32.76S cm2 mol−1
The formula of degree of dissociation
α    = Λm/ Λom
Plug the value we get
α    = 32.76S/390.5
                         = 0.084
The formula of dissociation constant
K = Cα/(1 – α)
Plug the values we get
K = 0.00241 × 0.084/(1– 0.084)
                      = 1.86 × 10−5 mol L−1
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Three electrolytic cells A, B and C containing solutions of ZnSO4(zinc sulphate), AgNO(silver nitrate) and CuSO4  (copper sulphate),  respectively are connected in series. A steady current of 1.5 ampere was passed through them until 1.45 g of silver is deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? (Atomic masses: Ag = 108, Zn = 65.4, Cu = 63.5, all in amu). 

Ag++e-  Ag  1 mol            1 mol ( = 108 g)              (Cathode reaction in cell B)
108 g of silver is deposited at cathode when 1 mol of electrons are passed or 108 g of silver deposit needs = 1 Faraday = 96,500 coulombs. Therefore, 1.45 g of silver needs
                     = 96,500×1.45108  = 1295.6 columbs
       But quantity of electricity passed
                    = current x time
                    = 1295.6 C  = 1.5 A x time (in sec.)
       or time for which current is passed
               = 1295.61.5  seconds = 863.7 s = 14.40 min.
          The cathode reaction in copper sulphate cell is
                         Cu2++2e-    Cu     2 mol                   1 mol (63.5 g)    (2×96500 c)
              2 x 96500 coulombs gives a deposit of 63.5 g of Cu.
          Therefore, 1295.6 coulombs will deposit of 
                       = 63.5 × 1295.62×96, 500 = 0.4263 g
Similarly,
                        Zn2++2e-  Zn     2 mol              1 mol (65.4 g)   (2×96500)
               (cathode reaction in zinc sulphate cell)
Mass of zinc deposited
                         =65.4 x 1295.62×96, 500 g = 0.44 g.
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Using the standard electrode potentials given in the table 3.1(in NCERT), predict if the reaction between the following is feasible:
(i) Fe3+ (aq) and I (aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br(aq)
(iv) Ag(s) and Fe3+(aq)
(v) Br2(aq) and Fe2+(aq)


From the table, standard electrode potents at 298 k are:
(E°Fe3+/VF = 0.77 V, E°I2/I– = 0.54 V)
(E°Ag+/Ag = E°Cu2+/Cu = 0.34)
(E°Fe3+/Fe= 0.77 V, E°Br2/Br- = 1.08 V)
(E°Ag+/Ag= 0.8 V E°Fe3+/Fe2+ = 0.77 V)
(E°Fe3+/Fe2+= 0.77 V,E°Br2/Br- = 1.08 V)
(a) 
From the table, standard electrode potents at 298 k are:(E°Fe3+/VF Fe to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space straight I to the power of minus left parenthesis aq right parenthesis space rightwards arrow space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis plus 1 half straight I subscript 2 left parenthesis straight g right parenthesis
In this reaction, Fe3+ is reduced to Fe2+ and I is oxidised to I2. The cell giving above reaction will be
straight I subscript 2 left parenthesis straight s right parenthesis space vertical line space straight I to the power of minus space left parenthesis aq right parenthesis space vertical line vertical line space Fe to the power of 3 plus end exponent space left parenthesis aq right parenthesis space vertical line space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis
therefore space straight E to the power of 0 subscript cell space equals space straight E to the power of 0 subscript cathode space minus space straight E to the power of 0 subscript anode space equals space straight E to the power of 0 subscript right space minus space straight E to the power of 0 subscript left
space space space space space space space space space space space space space space space equals space 0.77 space straight V space minus space 0.54 space straight V space equals space plus space 0.23 space straight V
As E0 is positive, the reaction between Fe3+ (aq) and I (aq) occurs as indicated by possible reaction given above.

(b) 2 space Ag to the power of plus left parenthesis aq right parenthesis space plus space Cu left parenthesis straight s right parenthesis space rightwards arrow space 2 space Ag left parenthesis straight s right parenthesis space plus space Cu to the power of 2 plus end exponent space left parenthesis aq right parenthesis
Here, in this reaction, Ag+ is reduced to Ag (i.e., it should be cathode) and Cu(s) is oxidised to Cu2+(aq) (i.e., it should be anode).
The cell can be represented as
           Cu left parenthesis straight s right parenthesis space vertical line space Cu to the power of 2 plus end exponent left parenthesis aq right parenthesis vertical line vertical line space Ag to the power of plus space left parenthesis aq right parenthesis space vertical line space Ag left parenthesis straight s right parenthesis
space space space space space space space space space space space space space space space space space space space space space space anode space space space space space space space space space space space space space space cathode
straight E to the power of 0 subscript cell space space equals space straight E to the power of 0 subscript cathode space minus space straight E to the power of 0 subscript anode
space space space space space space space space space space space space equals space 0.80 space straight V space minus space 0.34 space straight V space equals space plus space 0.46 space straight V
As E°cell is positive, the reaction between (Ag(aq) and Cu(s) occurs as indicated by possible reaction given above.

(c) Fe to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space Br to the power of minus left parenthesis aq right parenthesis space rightwards arrow space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space 1 half Br subscript 2 left parenthesis aq right parenthesis
 In this reaction Fe3+ is reduced to Fe2+ (i.e., Fe3/Fe2+ electrode should be cathode) and Br is oxidised to Br2 (i.e., Br2/Br electrode should be anode.
The cell can be represented as:
Br subscript 2 left parenthesis aq right parenthesis vertical line Br to the power of minus left parenthesis aq right parenthesis vertical line vertical line Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis space vertical line space Fe to the power of 3 plus end exponent left parenthesis aq right parenthesis
straight E to the power of 0 subscript cell space equals space straight E to the power of 0 subscript cathode space minus space straight E to the power of 0 subscript anode space equals space straight E to the power of 0 subscript right space end subscript minus straight E to the power of 0 subscript left
space space space space space space space space space space space equals space 0.77 space straight V space minus space 1.08 space straight V space equals space minus 0.31 space straight V
As E°cell is negative, no reaction will occur between Fe3+ (aq) and Br(aq).
(d) Ag left parenthesis straight s right parenthesis space plus space Fe to the power of 3 plus end exponent space left parenthesis aq right parenthesis space rightwards arrow space Ag to the power of plus left parenthesis aq right parenthesis space plus space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis     
Two half-cell reactions can be expressed as:
Ag left parenthesis straight s right parenthesis space rightwards arrow space Ag to the power of plus left parenthesis aq right parenthesis space plus space straight e to the power of minus
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis oxidation space anode right parenthesis
Fe to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space straight e to the power of minus space rightwards arrow space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis reduction comma space cathode right parenthesis
straight E to the power of 0 subscript cell space equals space straight E to the power of 0 subscript cathode space minus space straight E to the power of 0 subscript anode
space space space space space space space space space space space equals space 0.77 space straight V space minus space 0.80 space straight V space equals space minus 0.3 space straight V
As E°cell is negative, no reaction occurs between Fe3+(aq) and Ag(s).

(e) 1 half Br subscript 2 left parenthesis aq right parenthesis space plus space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis space rightwards arrow space Fe to the power of 3 plus end exponent space left parenthesis aq right parenthesis space plus space Br to the power of minus space left parenthesis aq right parenthesis
The two half-cell reactions are
   1 half Br subscript 2 left parenthesis aq right parenthesis space plus space straight e to the power of minus space rightwards arrow space Br to the power of minus space left parenthesis aq right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis reduction comma space cathode right parenthesis
space space space space space space Fe to the power of 2 plus end exponent left parenthesis aq right parenthesis space rightwards arrow space Fe to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space straight e to the power of minus
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis oxidation space anode right parenthesis
straight E to the power of 0 subscript cell space equals space straight E to the power of 0 subscript cathode space minus space straight E to the power of 0 subscript anode
space space space space space space space space space space space equals space 1.80 space straight V space minus space 0.77 space straight V space equals space plus 0.31 space straight V
As E°cell is positive, the reaction is feasible, i.e., reaction between Br2(aq) and Fe2+ (aq) occurs as indicated by possible reaction given above.

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A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Given that ,
Current, I = 5A
Time, t = 20 × 60 = 1200 s (1 min = 60 second)
Charge = I × t
            = 5 × 1200
  = 6000 Coulombs
Charge on Ni
No has –1 charge always so that
Ni +2(NO3) =0
Ni + 2(–1)  = 0
Ni             = +2
Charge required to deposited 1 mol of Ni = nF
                                              = 2 × 96487 Coulombs
                                              = 192974 Coulombs
1 mol of Ni = 58.7 g (use periodic table to get this value)
192974 Coulombs will generate = 58.7 g of Ni
6000 Coulombs will generate     = 58.7 g × 6000 Coulombs /192974 Coulombs
                                               = 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.
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