Electro - Chemistry-8

Q.1. Standard potential of following cell is 0.23V at 15⁰C and 0.21V at 35⁰C
            Pt H₂(g)|HCl(ag.)|AgCl(s)|Ag(s)
(i) Calculate n⁰ & S⁰ for cell rxn . by assuming that these quantities remains unchanged in range 15⁰ & 35⁰C.
(ii) Calculate solubility of AgCl in H₂O at 25⁰C. Given, standard reduction potential of Ag⁺(ag.)/Ag(s) couple is 0.8v at 25⁰C.

Ans:   Pt H₂(g)|HCl(ag.)|AgCl(s)|Ag(s)
         H₂ H⁺ + e^-           (Anode)
         AgCl + e^- Ag + Cl^-   (Cathode)
        
        H₂ + AgCl n⁺ + Ag + Cl^-
      - G⁰ = nE⁰F = 1 x 0.23 x 96500 = 2219 SJ (at 25⁰ c)
      - G⁰ = nE⁰F = 1 x 0.21 x 96500 = 2026 SJ (at 35⁰ c)

        G⁰ = H⁰ - TS⁰
      - 22195 = H⁰ - 298 x S⁰ ................................. (i)
      - 20265 = n⁰ - 308 x S⁰ ................................. (ii)
    On solving
        S⁰ = -96.5 J    &    H⁰ = 49.987 KJ

(ii)   Ag Ag⁺ + e^-                E⁰_OP = - 0.8 V
       Agcl + e^- Ag + Cl^-          =
      
        AgCl Ag⁺ + Cl^-
    E_cell = log[AG⁺] + +
At eqm,
      E_cell = 0
    + = log[Ag⁺][Cl^-] = 0.059 logK_{SP AgCl}
    - 0.8 + 0.22 = 0.059 logK_SP
      K_SP = 1.47 x 10^-10
      Solubility of AgCl =
                              = 1.21 x 10^-5 mol/L


Q.2. Overall formation constant for reaction of 6 mole of CN^- with cobalt (II) is 1 x 10¹⁹. Calculate formation constant for rxn. of 6 mole of CN^- with cobalt (III). Given that

      CO + e^- CO    = - 0.83 V
      CO³⁺ + e^- CO²⁺            = 1.82 V

Ans: CO CO + e^-    = 0.83V
       CO³⁺ + e^- CO²⁺              = 1.82 V
     
       CO + CO³⁺ CO²⁺ + CO
      E_cell = E⁰_cell - log₁₀
Also, 6CN^- + CO²⁺ CO
        
&      6CN^- + CO³⁺ CO
        
    E_cell = E⁰_cell + log₁₀
At eqm.    E_cell = 0
              0 = 1.82 - (- 0.82) + log₁₀
               = 8.23 x 10⁴⁴
               = 8.23 x 10⁶³


Q.3. Show that potential areadditive for process in which half reaction are added to yeild on overall rxn. but they arenot additive when added to yeild a third half rxn.

Ans: Case - I: When two half rxnx. are added to give an over all rxn., no. of moles of electrons involved in each half rxn. & over all rxn. are necessarily same.
    M₁ + n₁e           - = n₁F
     + n₂e^- M₂          - = n₂F

    M₁ + + M₂    - = n₃F

  = +
    n₃ = n₁ + n₂E⁰
     =
Since   n₁ = n₂ = n₃
  = +

Case - II: When n₁ n₂ n₃
    M₁ + n₁e^-                - = n₁F
     + (n₂ - h₁)e^-      - = (n₂ - n₁)F
  
    M₁ + n₂e^-                - = n₂F
  = +
    n₂F = n₁ F + (n₂ - n₁) F

  =


Key Words

Electrolysis.
Faraday's law of electrolysis.
Transport No.
Conductance.
Specific Conductance.
Equivalent Conductance.
Molar Conductance.
Kohirausch Law.
EMF.
NErst Eq.
Cellpotential.