Calculate the standard free energy change for the reaction occuring in the cell:
Zn(s) | Zn2+ (1 M)|| Cu2+ (1M) | Cu(s)
[Given E°Zn2+/Zn = – 0.076 V, E°Cu2+/Cu = + 0.34 V, F = 96500 C mol–1]

Given:
EZn2+/Zn 0= – 0.076 V

ECu2+/Cu0 = + 0.34 V

F = 96500 C mol–1
from the reaction 
n=2

Ecell0 = ECu2+/Cu0 -  EZn2+/Zn 0


Ecell0 =0.34 -(-0.076)
        = 0.416

We know that
G = -nFEcell0

    =  -2 x 96500 x 0.416
     =–802.88 kJ mol
–1

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Calculate the pH of the following half reactions: Pt. H2 (1 atom) / HCl, E = 0.25 V.


H2 ----> 2H++2e-
E° = Ecell -0.0592log[H+][PH2]or0.25 =0-0.0592log [H+]21-logH+ =4.237 orPH =4.237
Ans.  pH = 4.237

274 Views

Calculate the molar conductivity at infinite dilution of acetic acid from the following data:
Λm (HCl) = 426 ohm–1 cm2 mol–1, Λm CH3COONa = 91 ohm–1 cm2 mol–1 and Λm(NaCl) = 126 ohm–1 cm2 mol–1.


Acetic acid is weak electrolyte than HCl and NaCl  which is strong electrolyte.
Acoording to kohlransch's law
λm(HCl) =λH+ +λCl- λm(NaCl) = λNa+ +λCl-λm(CH3COONa) =λCH3COO- +λNa+ λm(CH3COOH) =λCH3COO- +λH+ λm(CH3COOH) =λCH3COO- +λNa+-λNa+ +λCl-+λH+ +λCl- 

=91-126+426 =391

Ans.  391 ohm–1 cm2 mol-1

1909 Views

Electrolytic conductivity of 0.20 mol L–1 solution of KCl at 298 k is 2.48 x 10–2 ohm–1cm–1. Calculate its molar conductivity.

we have given that
electrolyitc conductivity =0.20mol/L
conductivity = 2.48 x 10-2 ohm-1cm-1
thus apply the formula

Λm= k×1000M

M=k×1000ΛmM= 2.48×10-2×10000.20

here M is molar conductivity.

Ans.  124 ohm–1 cm2 mol–1

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Calculate the equilibrium constant of reaction at 25°C:
Ni(s) + Cu2+ (aq) → Cu(s) + Ni2+ (aq)
Given : E°Ni2+/Ni = -0.25 V, E°Cu2+/Cu = + 0.34 V, R = 8.314 J K–1 mol–1, F = 96500 C mol–1.


We have given 
 E°Ni2+/Ni = -0.25 V
Cu2+/Cu = + 0.34 V,

Ecell
0 =Ecell0 =ECu2+Cu0 - ENi2+Ni0  


Ecell0 =
0.34-(-0.25) 

Ecell0 
= 0.34+0.25 =0.59

R = 8.314 J K–1 mol–1
F = 96500 C mol–1.
T=25celcius =273+25 =298 Kelvin
n= 2 
K =Antilog nFEcell02.303×R×T



 

 

K =Antilog 2×96500×0.592.303×8.314×298



= Antilog[19.956]

Ans.  9.07 x 1019

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