Electro - Chemistry-6

Q.1. Calculate quantity of electricity that will be required to liberate 710 g of Cl₂ gas by electrolysing a conc. sol. of NaCl what weight of NaOH & what volume o fH₂ at 27⁰c & 1atm pressure i sobtained during process ?

Ans:    W = ZQ = ZTt = It = Q
          2Cl^- Cl₂ + 2e^-
        =
          Q = = 20 F
          Q = 1930000 coulomb.
  1 F gives 1 eg. or 40g NaOH
  20 F gives 20 eg. or 40 x 20 g = 800 NaOH
Also   1 F gives 1g eg. or 1 g H₂
  20 F gives 20 g eg. or 20 g H₂


    PV = RT
    1 x v = x 0.0821 x 300
     = 246.3 L


Q.2. Calculate volume of CL₂ at NTP produced during electrolysis of MgCl₂ which produces 6.5 g Mg. At wt. o fMG = 24.3.

Ans: At Cathode:      Mg²⁺ + 2e^- Mg
       At Anode:         2Cl^- Cl₂ + 2e^-
Eq. of Mg formed at cathode = Eq. of Cl₂ formed at anode
    
  = 18.99 g
At NTP
    PV = RT
    1 x v = x 0.0821 x 273

Vol. of Cl2 = 5.99 L

Q.3. How long would it taice to deposit 100g ofAl for electrolytic cell containing Al₂O₃ using a current of 125 A ?

Ans: 2Al³⁺ + 6e^- 2Al
  Eq. wt of Al = = 9
    w = ZIt = It
    100 = x t
    t = 8577.77 second.


Q.4. 3 ampere current was passed through an ag. solution of unknown salt of Pd for 1hr 2.977g of Pdⁿ⁺ was deposited. atcathode. Find n(At. wt. of Pd = 106.4)

Ans: Pdⁿ⁺ + ne^- Pd
       w = It
      
    n = 4


Q.5. Electrolysis of a sol. of MnSO₄ in ag. H₂SO₄ is method of preparation of MnO₂.
    (ag.) + 2H₂O MnO₂(s) + 2H⁺(ag.) + H₁(g)
Passing a current of 27A for 24 hr. gives 1 Kg of MnO₂. What is value of current efficiency.

Ans:   w =                              Mn²⁺ Mn⁴⁺ + 2e^-
         1000 =             eq. wt. =
         i = 25.6 A
       Current efficiency = x 100 = 94.8 %


Q.6. How many gms. ofsilver could be plated out on serving tray by elctrolysis of sol. containing Ag in +1 oxidation state for period of 8 hr at current o f8.46 A ? What is area of tray if thichness of silver plating is 0.00254 cm ? Density of silver is 10.5 g/cm³.

Ans:   w_Ag =
    Eq wt. of Ag = = 107.8
    w_Ag = = 272.18 g
  Volume of Ag deposited = = 25.92 ml
  Surface area = = 1.02 x 10⁴ cm²


Q.7. Calculate
    Zn|Zn²⁺(ag) || Cu²⁺(ag.)|Cu
    min. conc. of Cu²⁺ at which cell rxn.
    Zn + Cu²⁺(ag.) Zn²⁺(ag.) + Cu   will be spontaneous
if Zn²⁺ is 1 M            = 0.35 v      = - 0.76

Ans: At Anode   Zn Zn²⁺ + 2e^-
       At cathode Cu²⁺ + 2e^- Cu
        
              Zn + Cu²⁺ Zn²⁺ + Cu
       E⁰_cell = - = E⁰_cathode - E⁰_anode
       E⁰_cell = 0.35 - (- 0.76) = 1.1 v
       E_cell = E⁰_cell + log₁₀
       For rxn. to be spontaneous,   E_cell = + ve
       > - 1.11
       log₁₀[Cu²⁺] >
       [Cu²⁺] > 2.36 x 10^{- 38} M


Q.8. Calculate PH of following half cells solution
     Hcl        E = 0.25 v

Ans: H₂ 2H⁺ + 2e^-

    
    0.25 = 0 - log[H⁺]²
  - log[⁺] = 4.237
  PH = 4.237


Q.9. Calculate eqm. constant for rxn:
    Fe²⁺ + Ce⁴⁺ Fe³⁺ + Ce³⁺        = 1.44 v
                                                               = 0.68 v

Ans: At eqm.   E_cell = 0
    E_cell = E⁰_cell - log₁₀K_c
    E⁰_cell = log₁₀K_c
    E⁰_cell = E⁰_cathode - E⁰_anode = 1.44 - 0.68 = 0.76 v
  log₁₀K_c = = 12.8814
  K_c = 7.6 x 10¹²