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) Calculaten⁰&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+ AgCln⁺+ Ag + Cl^-      -G⁰= nE⁰
F = 1 x 0.23 x 96500 = 2219 SJ (at 25⁰c)      -G⁰= nE⁰
F = 1 x 0.21 x 96500 = 2026 SJ (at 35⁰c)


G⁰=H⁰- TS⁰       - 22195 =H⁰- 298 xS⁰................................. (i)
- 20265 =n⁰- 308 xS⁰................................. (ii)
On solving        S⁰= -96.5 J    &   H⁰= 49.987 KJ

(ii)   AgAg⁺+ e^-               E⁰_OP= - 0.8 V       Agcl + e^- Ag + Cl^-        =




AgClAg⁺+ 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^-5mol/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 10
⁶³

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



M
₁ + 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.