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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

`(0.40V)`

B

`(0.76V)`

C

`0.20V`

D

`0.94V`

Text Solution

Verified by Experts

`pt (s) H_2(1"bar")HCL(aq)||AcCL(s)Ag(s)|pt(s)`
At `A:H_2to 2H^(+)+2e^(-)`
At `C: (AgCl+e^(-)toAg+cl^(-))xx2`
`H_2+2AgC1to2HCl+2Ag`
`0.92=E_(cell)^(0)-0.06 log(10^(-6))^(2)implies 0.92=E_(cell)^(0)-0.06(-12) implies0.92=E_(cell)^(0)+0.72`
`E_(cell)^(0)=0.20implies (E_c^(0)-E_A^(0))=0.2implies(Q E_A^(0)=0)implies[E_c^(0)]_r=0.2V`
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For the following electrochemical cell at 298 K, pt (s) |H_(2) (g, 1 " bar" )|H^(+) (aq.), M^(2+) (aq.)| Pt (s) E_("cell") = 0.92 V when ([M^(2+) (aq.)])/([M^(4+) (aq.)]) = 10^(x) Given : E_(M^(4+)//M^(2+))^(@) = 0.151 V, 2.303 (RT)/(F) = 0.059V The value of x is :