Electro - Chemistry-7

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

Ans: For Cu(OH)₂,      K_sp = [Cu²⁺][on^-]²
 P^H = 14 So, [H⁺) = 10^-14
    K_w = [H⁺)[OH^-)   [OH^-) = = 1
So, [Cu²⁺] = = 1.0 x 10^-19 = 1 x 10^-19
     E_cell = E⁰_cell + log₁₀
           = 0.34 + log₁₀[Cu²⁺]
           = 0.34 + log₁₀[1 x 10^-19]
     E_cell = - 0.2205 v at P^H = 14




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
  w_Cu = 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) + 2H_g(l)
Given   = 0.126          = - 0.789 v
K_SP of PbSO₄ = 2.43 x 10^-8    K_SP of Hg₂SO₄ = 1.46 x 10^-6.

Ans: Oxidation potential = - REduction Potential
    Anode:        Pb Pb²⁺ + 2e^2e-
    Cathode:     Hg₂²⁺ + 2e^- HG₂
    E_cell = E⁰_cell - log
          = ( - ) - log
          = 0.789 - (- 0.126) - log
  K_SP = [Pb²⁺][SO₄^2-] = [Pb²⁺]²     [ [Pb²⁺] = [SO₄^2-]]
    [Pb²⁺] =
Similarly,
    
    E_cell = 0.789 + 0.126 - log = 0.941


Q.4. Two weak acid sol. HA₁ & HA₂ each with same conc. & having P^Ka 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) P_t
At L.H.S.,
      
       = 0 - 0.059 (P^H)₂
At R.H.S.,
      
            = - 0.059 (P^H)₁
For acid HA₁,
       HA₁ H⁺ + A₁^-
       [H⁺) = c = if is small)
       (P^H)₁ = PK_a - log₁₀c
Similarly,
       (P^H)₂ = PK_a - logc
       E_cell = - 0.059(P^H)₁ + 0.059(P^H)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