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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 CA. 456 mVB. 654 mVC. 646 mVD. None of these |
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Answer» 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 |
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