1.

Given the data at 25^(@)C, {:(AgI^(-) rarr AgI+e^(-),,E^(@) = 0.152 V),(Ag rarr Ag^(+) + e^(-),,E^(@) = -0.800 V):} What is the value of log K_(sp) for AgI (2.303 RT/F = 0.059 V)

Answer»

`-8.12`
`+8.612`
`-37.83`
`-16.13`

SOLUTION :APPLYING
`E_(I^(-)|AgI|Ag)^(@) = E_(Ag^(+)|Ag)^(@) + 0.059 log K_(SP) (AgI)`
`log K_(sp) (AgI) = (E_(I^(-)|AgI|Ag)-E_(Ag^(+)|Ag))/(0.09) = (-0.152-0.8)/(0.059) = -16.13`


Discussion

No Comment Found

Related InterviewSolutions