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Given the data at 25 °C Ag+IAgI+e E=0.152V AgAg++e E=0.800V

What is the value of log Kspfor Agl? (2.303 RT/IF = 0.059 V)

a
-16.13
b
-8.12
c
+8.612
d
-37.83

detailed solution

Correct option is A

E° for the reactionAgIAg++I  is  Ecell 0=(0.8000.152)V=0.952V 

logKsp=EcellRT/F=0.952V0.052V=16.13

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