The `pK_(sp)` of `AgI` is `16.07` if the `E^(@)` value for `Ag^(+)//Ag` is `0.7991V`, find the `E^(@)` for the half cell reaction `AgI(s) +e^(-) rarr Ag+I^(-)`

Dissociation constant for Ag(NH_(3))_(2)^(+) into Ag^(+) and NH_(3) is 6 xx 10^(-14) . Calculate E^(@) for the half reaction. Ag(NH_(3))_(2)^(+) + e rarr Ag + 2NH_(3) Given, Ag^(+) + e rarr Ag has E^(@) = 0.799 V

Dissociation constant for Ag(NH_(3))_(2)^(+) into Ag^(+) and NH_(3) is 6 xx 10^(-14) . Calculate E^(@) for the half reaction. Ag(NH_(3))_(2)^(+) + e rarr Ag + 2NH_(3) Given, Ag^(+) + e rarr Ag has E^(@) = 0.799 V

The cell Pt(H_(2))(1atm) |H^(+) (pH =?) T^(-) (a=1)AgI(s), Ag has emf, E_(298KK) =0 . The electrode potential for the reaction AgI +e^(-) rarr Ag + I^(Theta) is -0.151 volt. Calculate the pH value:-

The cell Pt(H_(2))(1atm) |H^(+) (pH =?) T^(-) (a=1)AgI(s), Ag has emf, E_(298KK) =0 . The electrode potaneial for the reaction AgI +e^(-) rarr Ag + I^(Theta) is -0.151 volt. Calculate the pH value:-

The cell Pt(H_(2))(1atm) |H^(+) (pH =?) T^(-) (a=1)AgI(s), Ag has emf, E_(298KK) =0 . The electrode potaneial for the reaction AgI +e^(-) rarr Ag + I^(Theta) is -0.151 volt. Calculate the pH value:-

K_(d) for dissociation of [Ag(NH_(3))_(2)]^(+) into Ag^(+) and NH_(3) is 6 xx 10^(-8) . Calculae E^(@) for the following half reaction. Ag(NH_(3))_(2)^(+) +e^(-) rarr Ag +2NH_(3) Given Ag^(+) +e^(-) rarr Ag, E^(@) = 0.799V

K_(d) for dissociation of [Ag(NH_(3))_(2)]^(+) into Ag^(+) and NH_(3) is 6 xx 10^(-8) . Calculae E^(@) for the following half reaction. AG(NH_(3))_(2)^(+) +e^(-) rarr Ag +2NH_(3) Given Ag^(+) +e^(-) rarr Ag, E^(@) = 7.99V

K_(d) for dissociation of [Ag(NH_(3))_(2)]^(+) into Ag^(+) and NH_(3) is 6 xx 10^(-8) . Calculae E^(@) for the following half reaction. AG(NH_(3))_(2)^(+) +e^(-) rarr Ag +2NH_(3) Given Ag^(+) +e^(-) rarr Ag, E^(@) = 7.99V

Ag^(+) + e^(-) rarr Ag, E^(0) = + 0.8 V and Zn^(2+) + 2e^(-) rarr Zn, E^(0) = -076 V . Calculate the cell potential for the reaction, 2Ag + Zn^(2+) rarr Zn + 2Ag^(+)