Cell potential of
`Pt_((S))|H_(2)(g," 1 bar")|HCl_((aq))|AgCl_((S))` is 0.92 V.
Concentration of HCl solution is `10^(-6)` molal, then find out standard potential of `(AgCl//AgCl^(-))` electrode. (at 298K temperature, `((2.303RT)/(F)=0.06V)`.

Options

A. 0.94 V
B. 0.20 V
C. 0.76 V
D. 0.40 V

Answer

B

Explanation

Anode : `H_(2) to 2H^(+) +2overline(e)`
Cathode : `([overline(e)+AgCl_((S))toAg_((S))+Cl^(-)]xx2)/(2AgCl+H_(2) to 2H^(+)+2Ag+2Cl^(-))`
`E_(cell)=E_(cell)^(@)-(0.06)/(2)log[H^(+)]^(2)[Cl^(-)]`
`0.92=[E_(AgCl|AgCl^(-))^(@)-E_(H^(+)|H^(2))^(@)]-0.03log(10^(-6))^(2)(10^(-6))^(2)`
`0.92=[E_(AgCl|AgCl^(-))^(@)-0.03log10^(-24)]`
`E_(AgCl|AgCl^(-))^(@)=0.92+0.03(-24)`
`=0.20V`