Electro - Chemistry-5

(3) To predict spontancity of redox rxn.:

For redox rxn. to be spotaneous, EMF of cell must be +ve.


Illustration: Cen Nickel vessel used to store CuSO₄ sol ?
     = - 0.25 v
     = 0.34 v

Ans: NI²⁺/Ni || Cu²⁺/Cu
    E⁰_cell = - = 0.34 - (- 0.25) = 0.59 v
EMF comes out to be +ve. So, CuSO₄ will reacts with Ni. So, cannot store in nickel vessel.


Nerst's Equation:

    Mⁿ⁺ + ne^- M
Nerst eq. as follows
    E = E⁰ -
    E electrode potential at given conc. of Mⁿ⁺ ions & Temp. T.
    R gas constant.
    T TEmp. in K.
    F 1 Faraday.
    n no. of electrons involved in electrode rxn.
For pure solids or liquids or gases, at 1 atm, molar conc. taken as unity.
i.e. [M) = 1
So,   E = E⁰ -
Putting value of R, T & F        R = 8.314,   T = 298 K,   F = 965000 coulom
   ......................................... (i) At 298 K Temp.


Nerts eq. for EMF of cell:

    Zn(s) + Cu²⁺(ag) Zn²⁺(ag) + Cu(s)
For Zn(s) | Zn²⁺ (ag.) electrode,
    Zn²⁺(ag) + 2e^- Zn(s)
     = + [Zn²⁺(ag.)]
For Cu/Cu²⁺ electrode
    Cu²⁺ + 2e^- Cu(s)
     = + ln[Cu²⁺(ag.)]
Since oxidation takes place at Zn & reduction takes place at Cu electrode.
    E_cell = E_cathode - E_anode

          = -
          = { + ln[Cu²⁺(ag.)]} - { + ln[Zn²⁺(ag.)]}
          = ( - ) + n
          = E⁰_cell - ln
    
For cell rxn,
    aA + bB xX + yY
    E_cell = E⁰_cell - ln
    At 296 K
  


Illustration: Calculate emf of cell
                  Cr|Cr³⁺(0.1 M) || Fe²⁺(0.01 M) | Fe
                   = - 0.75 v
                   = - 0.45 v

Ans: 2Cr + 3Fe²⁺ 2Cr²⁺ + 3Fe      n = 6
    E_cell = E⁰_cell -
          = ( - ) -
          = (- 0.45 + 0.75 v) -
    E_cell = 0.2606 v


Eqm. constant form Nerst Eq.:

    Zn + Cu²⁺ Zn²⁺ + Cu
    eqn. const = = KC
Eqn. occurs when E_cell = 0
    0 = E⁰_cell - log
    E⁰_cell = logK_c
At 296 k,

E0cell = logKc

Illustration: Calculate eqm. constant for rxn.
    Cu(s) + 2Ag⁺ Cu²⁺ + 2Ag

Ans: Here    n = 2
    E⁰_cell = logK_c
     - = logK_c
    0.8 - 0.34 = logK_c
    K_c = 3.688 x 10¹⁵


Dibb's Free Energy & Cell Potential:

    - G⁰ = nFE⁰_cell
But for rxn. at eqm
    E⁰_cell = ln K_c
    G⁰ = - nF x ln K_c

G0 = - 2.303 RT ln Kc

Illustration: Calculate standard free energy change
    Zn(s)|Zn²⁺ || Cu²⁺(1M)|CU(s)
     = - 0.76 v      = + 0.34 v      F = 96500 C mol^-1
Also calculate eqm constant for rxn.

Ans: mrxn.    n = 2
    G⁰ = - nFE⁰_cell

    E⁰_cell = 0.34 - (- 0.76) = 1.1 v

    G⁰ = - 2 x 96500 x 1.1 = - 212.3 KJ/mol

    G⁰ = - 2.303 RT logK_c

    - 212300 = - 2.303 xn 8.314 x 298 x logK_c
    K_c = 1.6 x 10³⁷