Cu64 (Half life=22.8 hours) decay by beta emission (38%), beta+ emission (19%) and electron capture (43%). write the decay product and calculate Half lives for each of the decay process. [IIT 2002]

SOLUTION:  

(1):

 (2)  Above are parallel reactions occurring from Cu⁶⁴   

T₁ T₂ and T₃ are the corresponding partial half lives and also K= K₁ + K₂ + K₃ (for parallel reaction) 

Write correct Nernst equation for the given galvanic cell containing both the electrodes are M/M (insoluble) salt at 25°c.: Ag(s)|AgCl(s)| KCl(M) || KBr (M)|AgBr(s)|Ag(s): [Given Ksp(AgCl) and Ksp(AgCl)]

Important thing :

For concentrated cell both electrodes are same so E not cell is always zero  and Qc is always less than one.

You should first assume any one is the cathode and the other one  is the anode. Then you can calculate the theoretical emf:

(1)  If the emf > 0, then that reaction will be spontaneous and it means you got the right cathode and right anode.

 (2) If the calculated emf < 0, it means that that reaction is not spontaneous, which means that the reverse reaction is spontaneous; that happens if you exchange the cathode and anode.

For the galvanic cell: Ag|AgCl(s)| KCl (0.2M) || KBr (0.001M) |AgBr(s) |Ag : calculate the e.m.f. generated and assign correct polarity to each electrode for a spontaneous process after taking an account of cell reaction at 25°c. [Given Ksp(AgCl) = 2.8×10^-10 and Ksp(AgCl) = 3.3×10^-13]

Thus to get positive cellphone potential (Ecell) , the polarity of the cell should be reversed i.e cell is represented as 

Ag(s)|AgBr(s)| KBr(0.002M) || KCl (0.2M)|AgCl(s)|Ag(s):

Where 

Ag(s)|AgBr(s)| KBr(0.002M) act as Anode while KCl (0.2M)|AgCl(s)|Ag(s): act as cathode

How to determine anode and cathode of concentration cell instead of both electrodes are same ?

Important thing :

For concentrated cell both electrodes are same so E not cell is always zero  and Qc is always less than one.

You should first assume any one is the cathode and the other one  is the anode. Then you can calculate the theoretical emf:

(1)  If the emf > 0, then that reaction will be spontaneous and it means you got the right cathode and right anode.

 (2) If the calculated emf < 0, it means that that reaction is not spontaneous, which means that the reverse reaction is spontaneous; that happens if you exchange the cathode and anode

The conductivity of 0.0011028 mol per liter acetic acid is 4.95 × 10–5 S per cm. Calculate its dissociation constant if ^°m for acetic acid is 390.5 S cm2 per mol–1 .