Question

1. What is the cell emf (Eo cell) for electrolysis of molten nickel (II) iodide? (Use Eo values from Appendix E of your lecture text).

2. What reactions would take place at the anode and cathode during electrolysis of an aqueous solution of nickel (II) iodide?

3. Use the Nernst Equation (Equation 7) to calculate Ecell at 25o C for electrolysis of aqueous sodium chloride if the concentration of chlorine in the brine solution is 4.5 M and the partial pressure of Cl2 is 0.001 atmospheres.

4. Chlorine will spontaneously react with iodide to produce chloride and iodine. What is Eo for this reaction.

Answer 1

1) NiI₂     Ni⁺²  +   2 I^-

In this reaction Two half reactions Are involved

At anode( Oxidation) Ni ---> Ni ⁺²+2e^- E⁰ = 0.25 V

   At Cathode ( Reduction ) I₂+2e^- ----> 2I^-  E⁰   = 0.53 V

    E⁰_cell =E⁰  _Anode+E⁰_Cathode

              = 0.25 V+0.53 V

                =0.78 V

2) At Anode Oxidation takes place Here Ni is convertedin to Ni²⁺

       At cathode ReductionTakesplace .Here I₂ is converted in to 2I^-.

3) According to Nernest Equation E_cell=E⁰_cell - 0.0592/n*logQ

        For ConcentrationCells E⁰_cell = Zero

   ∴   E_cell =0- 0.0592/2*log[Na][Cl^-]/[NaCl]

                    -0.0592/2 *log 4.5 M

                  = -0.0296*0.6532

                   = - 0.01933 V

4)At Anode (Oxidation ) 2I^- ---->I₂ + 2e^- E⁰_cell =-0.53 V

     At Cathode ( Reducuion)Cl₂ +2e^- --> 2Cl^- E⁰_cell = 1.36V

         E_cell = 1.36V+(-0.53V)

              E_cell   = 0.83 V