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HARD TYPE
HARD TYPE



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-               E0OP= - 0.8 V       Agcl + e- Ag + Cl-        =




AgClAg++ Cl-


  Ecell=log[AG+] ++
At eqm,      Ecell= 0 
+=log[Ag+][Cl-] = 0.059 logKSP AgCl

- 0.8 + 0.22 = 0.059 logKSP       KSP= 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      
Ecell= E0cell-log10

Also, 6CN-
+ CO2+ CO



&      6CN
-
+ CO3+ CO        


  Ecell= E0cell+log10
At eqm.    Ecell= 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.

 

 
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