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The emf of the cell, Pt|H(2)(1atm)|H^(+)...

The emf of the cell, `Pt|H_(2)(1atm)|H^(+) (0.1M, 30mL)||Ag^(+) (0.8M)Ag` is `0.9V`. Calculate the emf when `40mL` of `0.05M NaOH` is added.

Text Solution

Verified by Experts

The correct Answer is:
`E_(sp) = 1.16 xx 10^(-16)`
`K_(sp) = [Ag^(+)] [0.05]`
`E_(cell) = E_(OP_(Ag)) +E_(RP_(Ag))`
`E_(cell) = (E_(OP_(Ag))^(@) -(0.0591)/(1)log[Ag^(+)]_(L.H.S))`
`+(E_(RP_(Ag))^(@) +(0.0591)/(1)log[Ag^(+)]_(R.H.S))`
`E_(cell) =(0.0591)/(1)log.([Ag^(+)]_(R.H.S))/([Ag^(+)]_(L.H.S))`
`0.788 = 0.0591 log.(0.05)/([Ag^(+)]_(L.H.S))`
`[Ag^(+)]L.H.S = 2.203 xx 10^(-15)`
`K_(sp) = 2.203 xx 10^(-5) xx 0.05 = 1.1 xx 10^(-16)`
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