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