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 fH₂at 27⁰
c & 1atm pressure i sobtained during process ?

Ans:    W = ZQ = ZTt =It =
Q          2Cl^- Cl₂+ 2e^-


=

Q =
= 20 FQ = 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 CL
₂
at NTP produced during electrolysis of MgCl
₂
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 gAt NTP    PV =RT    1 x v =x 0.0821 x 273

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^- 2A
l  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 = 4Q.5. Electrolysis of a sol. of MnSO₄in ag. H₂SO₄is method of preparation of MnO₂.

(ag.) + 2H₂OMnO₂(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 10⁴cm²


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 spontanousif Zn²⁺is 1 M          
= 0.35 v     = - 0.76


Ans: At Anode   ZnZn²⁺+ 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 vAns: 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 vAns: 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