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Ag|Ag^(+), KI||Ag I| Ag emf is E, then K...

`Ag|Ag^(+), KI||Ag I| Ag` emf is E, then `K_(sp)` of AgI is given as:

A

`K_(sp) = (nF)/(2.303 RT) "log " E^(@)`

B

In `K_(sp) = nF [(delta E^(@))/(delta T) - E^(@)]`

C

In `K_(sp) = (nF)/(E^(@))`

D

log `K_(sp) = (nFE^(@))/(2.303RT)`

Text Solution

Verified by Experts

The correct Answer is:
D
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