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)/

Question

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)/

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 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

4 views Answered 2023-02-05 15:29