The standard reaction potential for Cu2+/Cu is + 0.34 V. Calculate the reduction potential at pH = 14 for the above couple. Ksp of Cu(OH)2 is 1.0 x 10–19.

reduction potential is given by:

E =E° - RT2Fln 1Cu2+/M
From the given data PH =14 and KsP (Cu(OH2) =1.0 x 10-19 


we get 
H+ =10-14 M thus,

[OH]- =OH- =Kw[H+] =10-14M10-14M =1M[Cu2+]  =KspOH- =1.0×10-191 = 1.0 x 10-19ME =0.34 -0.0592log11.0 x10-190.34-0.059×192 = 0.34 -0.56 =-0.22
ans -0.22V

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The conductivity of 0.1 M KCl solution at 298 K is 0.0129 s cm–1. The resistance of this solution in a conductivity cell is found to be 58 ohms. What is the cell constant of the cell? The 0.1 M AgNO3 solution at 298 K in the same conductivity cell offered a resistance of 60.5 ohms. What is the conductivity of 0.1 M AgNO3 solution?

KKCl = 0.0129 s cm–1
RKCl = 58 Ω
cell constant= k x R
cell constant= 0.0129 x 58
Cell constant = 0.7482 cm–1
As AgNO3 is also in the same conductivity cell, the cell constant remains same. Therefore,
Conductivity of AgNO3 solution = cell constantRAgNO3

                                   KAgNO3 = 0.748260.5KAgNO3 = 0.124 s cm-1

 

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The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K. is 1.46 x 10–6 s cm–1.

We ave given that 

R = 1500 Ω
K = 1.46 x 10–4 Ω–1 cm–1

Cell constant = K x R

= 1.46 x 10–4 x 1500 = 0.219

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Consider the cell Zn/Zn2+ (aq) (1.0 M) || Cu2+ (aq) (0.1 M) | Cu The standard reaction potentials are + 0.35 V for 2e + Cu2+ (aq) → Cu and – 0.763 V for 2e + Zn2+ (aq) → Zn
(i) Write down the cell reaction.
(ii) Calculate the emf of the cell.
(iii) Is the cell reaction spontaneous or not?


(I)Cell Reaction:
Oxidation: Zn----> Zn2+ +2e-
Reduction: Cu2+ +2e------> Cu
Therefore overall reaction 
Zn +Cu2+ ------> Zn2+ +Cu
Ecell0 =ECu2+Cu0 -EZn2+Zn0
(ii) Ecell0=+0.35-(-0.763)
=1.113 volts

(iii) Since EMF of cell is positive, it is a spontaneous reaction.

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The standard reduction potential of the reaction at 25°C
2H2O + 2e H2(g) + 2OH is – 0.8277 V
Calculate equilibrium constant for the reaction
2H2O H3O++OH- at 25°C.


Consider the given reaction as the net cell reaction 
two half reaction :

Oxidation:
H2O +1/2H2 ----> H3O+ +e-  Ecell0 =0


reduction: 
H2O +e- -----> 1/2H2(g) + OH- Ecell0 = -0.8277V

It is evident from the cell reaction that it involves the transfer of one electron so that n= 1
logKc =nEcell00.059Kc = antilog 10.059(-0.8277) =--14.028 = 15.972Kc =Antilog (15.972)  =9.376×10-15

 9.88 × 10-15

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