calculate `E^(@)` of the following half -cell reaction at 298 K: `Ag(NH_(3))_(2)^(+)+e^(-) rarr Ag+2NH_(3)` `{:(Ag^(+)+e^(-) rarr Ag,,E_(Ag^(+)//Ag)^(

Question

calculate `E^(@)` of the following half -cell reaction at 298 K: `Ag(NH_(3))_(2)^(+)+e^(-) rarr Ag+2NH_(3)` `{:(Ag^(+)+e^(-) rarr Ag,,E_(Ag^(+)//Ag)^(

calculate `E^(@)` of the following half -cell reaction at 298 K:
`Ag(NH_(3))_(2)^(+)+e^(-) rarr Ag+2NH_(3)`
`{:(Ag^(+)+e^(-) rarr Ag,,E_(Ag^(+)//Ag)^(@)=0.80 V),(Ag(NH_(3))_(2)^(+) hArr Ag^(+)+2NH_(3),,K=6xx10^(-8)):}`

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

Correct Answer - 0.373 volt
`{:("Anode "Ag(s) rarr Ag^(+)+e^(-)," "E^(@)=0.80" volt"),("Cathode "[Ag(NH_(3))_(2)]^(+)+e^(-) rarr Ag(s)+2NH_(3)," "E^(@)=V" volt"),(ulbar(Ag(s)+[Ag(NH_(3))_(2)]^(+) hArr Ag(s)+Ag^(+)+2NH_(3))):}`
`Q=([Ag^(+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)]^(+))=6xx10^(-8) (n=1) E_(cell)^(@)=(V-0.80)`
At equilibrium, `E=0`
`:. E=E^(@)-0.0591/n log Q`
`0=(V-0.80)-0.0591/1 log (6xx10^(-8))`
`V=0.373` volt