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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

456 mV

B

654 mV

C

646 mV

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`(H_2+2 AgCl to 2Ag+2H^+ + 2Cl^(-))`
`DeltaG_("cell reaction")^(@)=2(-130.79)-2(-109.56)=-42.46 kJ//"mole" =-2fxxE_("cell")^(@)`
`therefore E_("cell)^(@)=-42460/(-2xx96500)=+ 0.220 V`
Now `E_("cell")=+0.220+0.059/2 "log" 1/(0.01)^4 =0.456 V` =456 mV
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