QJEEM06Ch8978: JEE Main 2006 Chemistry - Class 12 JEEM06Ch8978 Solution: Given the data at 25 o C, Ag + I - AgI + e - , E o = 0.152 V Ag …

Question

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Given the data at 25^oC,
Ag + I^-

\to

AgI + e^- , E^o = 0.152 V
Ag

\to

Ag⁺ + e^-, E^o = -0.800 V
What is the value of log K_sp for AgI? (2.303 RT/F = 0.059 V)

Accepted Answer

\left( i \right)\,\,\,Ag \to A{g^ + } + {e^ - }\,\,

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = - 0.800\,V

\left( {ii} \right)\,\,\,Ag + {{\rm I}^ - } \to Ag{\rm I} + {e^ - }

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = 0.152\,V

From

(i)

and

(ii)

we have,

Ag{\rm I} \to A{g^ + } + {{\rm I}^ - }

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = - 0.952\,\,V

E_{cell}^ \circ = {{0.059} \over n}\log \,K

\therefore

- 0.952 = {{0.059} \over 1}\,\log \left[ {A{g^ + }} \right]\left[ {{{\rm I}^ - }} \right]

[ as

\,\,\,k = \left[ {A{g^ + }} \right]\left[ {{{\rm I}^ - }} \right]

or

\,\,\, - {{0.952} \over {0.059}} = \log \,{K_{sp}}

or

\,\,\, - 16.13 = \log \,{K_{sp}}

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