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In the cell Pt (s) | H(2) (g, 1 " bar")|...

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}

A

(a)0.20V

B

(b)0.076V

C

(c)0.040V

D

(d)0.94V

Text Solution

Verified by Experts

The correct Answer is:
1
`underset("Anode")(ubrace(Pt(s)|H_(2)(1"bar")|HCl(aq)))"||" underset("Cathode")(ubrace(MCl(s)|M(s)|Pt(s)))`
At `A:H_(2) to 2H^(+) +2e^(-)`
At `C: (MCl+e^(-) to Ag+Cl^(-))xx2`
`" "H_(2)+2MCl to 2HCl+2M`
`1.72 = E_("cell")^(0)-0.06 log(10^(-6))^(2)`
`1.72=E_("cell")^(0)-0.06(-12)`
`1.72 = E_("cell")^(0)+0.72`|
`E_("cell")^(0)=1.0`
`(E_(c)^(0)-E_(A)^(0))_(r)=1.0`
`(because E_(A)^(0)=0)`
`[E_(C)^(0)]_(r)=1.0V`
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