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Calcualte the cell EMP in mV for Pt|H(...

Calcualte the cell EMP in mV for
`Pt|H_(2)(1 atm) | HCl(0.01 M)|| AgCl(s)| Ag(s)` at 298 K
If `DeltaG_(f)^(@)` values are at `25^(@)C`,
`-109.56 (kJ)/(mol) for AgCl(s)` and
`-130.79 (kJ)/(mol)` for `(H^(+) + Cl^(-)(aq)`, Take 1F = 96500 C

A

`456mV`

B

`654mV`

C

`546mV`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`triangleG_("cell reaction")^(0)=2(-130.79)-2(-109.56)=-42.46kJ//"mole"`
(for `H_(2)+2AgClto2Ag+2H^(+)+2Cl^(-))`
`thereforeE_(cell)^(0)=(-42460)/(-2xx96500)=+0.220V`
Now `E_(cell)=+0.220+(0.059)/(2)log((1)/((0.01)^(4)))=0.456V=456mV`.
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