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Question

For the galvanic cell : Ag|AgCl(s)|KCl(0.2M)||KBr(0.001M)|AgBr(s)|Ag,Calculate the EMF in V, generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at 25C.

[Ksp=2.8×1010;Ksp(AgBr)=3.3×1013] The answer is -m.n ×102 Then n-m =

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Solution

At anode :
Ag+ClAgCl+e
At cathode:
AgBr+eAg+Br
E=EAg/AgCl/Cl+EBr/AgBr/Ag0.0591log[Br][Cl]
....(i)
For reaction :
Ag+ClAgCle
Ag++eAg
E=O=EAg/AgCr/Cl+EAg+/Ag0.0591
log1[Ag+][Cl]
EAg/AgCr/Cl=EAg/Ag+0.0591logKsp(AgCl) ....(ii)
for reaction
AgBr+eAg+Br
AgAg++e
E=0=EBr/AgBr/Ag+EAg/Ag+0.0591log[Ag+][Br]
EBr/AgBr/Ag=EAg+/Ag+0.0591logKsp(AgBr) ...(iii)
from equation (1), (2) & (3)
E=0.0591logKsp(AgBr)Ksp(AgCl)0.0591log[Br][Cl]
E=0.0591log3.3×1013×0.22.8×1010×0.001=0.037V

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