दी गयी सेल `Pt|H_(2)(p_(1))|H^(+)(aq)||H_(2)(p_(2))|Pt` के लिए विद्युत् वाहक बल क्या होगा ?

Options

A. `(RT)/(2F) log_(e)""(p_(1))/(p_(2))`
B. `(RT)/(F) log_(e)""(p_(1))/(p_(2))`
C. `(RT)/(F)log_(e)""(p_(2))/(p_(1))`
D. इनमे से कोई नहीं

Answer

A

Explanation

ऐनोड अभिक्रिया `H_(2)(p_(1)) rarr 2H^(+)`
कैथोड अभिक्रिया `2H^(+) rarr H_(2)(p_(2))`
`E_("कैथोड") = -(RT)/(2F)ln""(p^(2))/([H^(+)]^(2))`
`E_("ऐनोड") = -(RT)/(2F)ln""([H^(+)]^(2))/(p_(1))`
`R_("सेल") = E_("ऐनोड") + E_("कैथोड")`
`= -(RT)/(2F) ln ""([H^(+)]^(2))/(p_(1)) - (RT)/(2F) ln ""(p_(2))/([H^(+)]^(2))`
`= -(RT)/(2F) ln ""(p_(2))/(p_(1))`
`= (RT)/(2F) ln ""(p_(1))/(p_(2))`