Electro - Chemistry-7

Q.10. Standard reduction potential for Cu2+/Cu is +0.34 v. Calculate reduction potential at PH = 14
    Ksp of Cu(OH)2 is 1.0 x 10-19

Ans: For Cu(OH)2,      Ksp = [Cu2+][on-]2
 PH = 14 So, [H+) = 10-14
    Kw = [H+)[OH-)   [OH-) = = 1
So, [Cu2+] = = 1.0 x 10-19 = 1 x 10-19
     Ecell = E0cell + log10
           = 0.34 + log10[Cu2+]
           = 0.34 + log10[1 x 10-19]
     Ecell = - 0.2205 v at PH = 14


Medium Type


Q.1. In a fuel cell H2 & O2 react to produceelectricity. In process H2 gas is oxidized at anode& O2 at cathode. If 67.26 of H2 atSTP reacts in 15 min. What is avg. current produced ? If entire current is used for electrodeposition of Cu from Cu2+, how many gm of Cu are deposited ?
    Cathode:   O2 + 2H2O + 4e- 4OH-
    Anode:      H2 + 2OH- + 4e- 2H2O + 2e-

Ans: Moles of H2 reacting = = 3
  Eq. of H2 used = Moles x n-factor
                         = 3 x 2 = 6
    
i = 643.33 A
Also, Eq. of H2 = Eq. of Cu formed
  wCu = 6 x = 190.5 g
wt. of Cu deposited = 190.5 g


Q.2. 19 g fused SnCl2 was electrolysed using inert electrodes. 0.119 g Sn was deposited at cathode. If nothing was given out during electrolysis, calculate ratio of wt. of SnCl2 & SnCl2 in fused state after electrolysis (at. et. Sn = 119).

Ans: electrolysis SnCl2 yieldes:
    Anode:      2Cl- Cl2 + 2e-
    Cathode:   Sn2+ + 2e- Sn
Further Cl2 formed at anode reacts with SnCl2 to give SnCl4
    SnCl2 + Cl2 SnCl4
Eq. of SnCl2 lost during electrolysis = Eq. of Cl2 formed during electrolysis = Eq. of Sn formed.
Eq. of Cl2 formed = = 2 x 10-3
  Eq. of SnCl4 formed = 2 x 10-3 ( whole Cl2 react to from SnCl4)
Now, total loss in Eq. of SnCl4 during complete course
    = Eq. of SnCl2 lost during electrolysis + Eq. of SnCl2 lost during reaction with Cl2
    = 2 x 10-3 + 2 x 10-3 = 4 x 10-3
Initial eq. of SnCl2 left in sol. = 2 x 10-1 - 4 x 10-3 = 0.196
Eq. of SnCl4 formed = 2 x 10-3 = 0.002



Q.3. Calculate emf of given cell rxn. and Pb(s) + Hg2So4 PbSO4(s) + 2Hg(l)
Given   = 0.126          = - 0.789 v
KSP of PbSO4 = 2.43 x 10-8    KSP of Hg2SO4 = 1.46 x 10-6.

Ans: Oxidation potential = - REduction Potential
    Anode:        Pb Pb2+ + 2e2e-
    Cathode:     Hg22+ + 2e- HG2
    Ecell = E0cell - log
          = ( - ) - log
          = 0.789 - (- 0.126) - log
  KSP = [Pb2+][SO42-] = [Pb2+]2     [ [Pb2+] = [SO42-]]
    [Pb2+] =
Similarly,
    
    Ecell = 0.789 + 0.126 - log = 0.941


Q.4. Two weak acid sol. HA1 & HA2 each with same conc. & having PKa values 3 & 5 are placed in contact with Hydrogen electrode (1atm, 250c) and are interconnected through salt bridge. Find emf of cell.

Ans: Cell is
       Pt H2(1 atm)|HA2||HA1|(H2)(1 atm) Pt
At L.H.S.,
      
       = 0 - 0.059 (PH)2
At R.H.S.,
      
            = - 0.059 (PH)1
For acid HA1,
       HA1 H+ + A1-
       [H+) = c = if is small)
       (PH)1 = PKa - log10c
Similarly,
       (PH)2 = PKa - logc
       Ecell = - 0.059(PH)1 + 0.059(PH)2
             = 0.059[P - P] = [5 - 3]
             = 0.059 v


Q.5. REduction potential diagram for Cu in acid solution is:


Calculate value of X.

Ans: Given,    Cu2+ + e- Cu+;     = 0.15 v ;   ........................... (i)
                    Cu+ + e- Cu+ ,     = 0.5 v ;     ........................... (ii)
                    Cu2+ + 2e- Cu ;   = ? ;           ........................... (iii)
                     = - nF = - 1 x 0.15 F = - 0.15 F
                     = -n F = - 1 x 0.5 x F = - 0.5 F
  Adding
     + =
    - 0.15 F + (- 0.5 F) =
     = - 0.65 F
    - nF = - 0.65 F
     = = 0.325 volt.
    X = 0.325 Volt









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