निम्न सेल अभिक्रिया के लिए
`Pb(s)+Hg_(2)SO_(4) hArr PbSO_(4)(s)+2Hg(l), E_("सेल")^(@)= 0.92V` है। तब `K_(sp) (PbSO_(4)) = 2xx10^(-8), K_(sp) (Hg_(2)SO_(4)) = 1xx10^(-5), E_("सेल")^(@)` होगा

Options

A. 0.92 V
B. 0.89 V
C. 1.04 V
D. 0.95 V

Answer

D

Explanation

`Q= ([Pb^(2+)])/([Hg_(2)^(2+)])= (sqrt(K_(sp)(PbSO_(4)))/(sqrt(K_(sp)(Hg_(2)SO_(4)))))`
`= (sqrt(2xx10^(-8)))/(sqrt(1 xx 10^(-6)))= sqrt(2xx 10^(-2))= 0.14`
`E_("सेल")= E_("सेल")^(@)- (0.059)/(2)log0.14= 0.92 - 0.0295 log 0.14= 0.95V`