In the cell `Pt (s) | H_(2) (g, 1 " bar")| HCl (aq)| AgCl(s) | Ag(s) | Pt(s)` the cell potential is 0.92 V when a `10^(-6)` molal HCl solution is used. The standard electrode potential of `(AgCl//Ag, Cl^(-))` electrode is
{Given, `(2.303RT)/(F) = 0.06V` at 298 K}

Options

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

Answer

A

Explanation

Pt (s) `|H_(2)` (g) , 1 bar |HCl (aq) `|AgCl (s) |Pt (s)`

`E = E^(@) - (0.06)/(2) "log" [H^(+)]^(2) [Cl^(-)]^(2)`
`0.92 = E^(@) - (0.06)/(2) log (10^(-6))^(2) (10^(-6))^(2)`
`0.92 = [E^(@) (AgCl|AgCl^(-))] - [E^(@) (H_(2)|H^(+))]`
`-0.03 log 10^(-24)`
`0.92 = E^(@) (AgCl |Ag, Cl^(-)) - 0.03 xx (-24)`
`E^(@) (AgCl|Ag , Cl^(-)) = 0.92 - 0.72 = 0.2 V`