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calculate the emf of the following cell:...

calculate the emf of the following cell:
`Pt(H_(2)" 1 atm")|CH_(3)CH_(2)COOH(0.15 M) ||0.01 M NH_(4)OH|H_(2)" (1 atm)"Pt`
`K_(a)` for `CH_(3)CH_(2)COOH=1.4xx10^(-5)`
`K_(b)` for `NH_(4)OH =1.8xx10^(-5)`

Text Solution

Verified by Experts

The correct Answer is:
`-0.4603` volt
`[H^(+)]` in `CH_(3)CH_(2)COOH=sqrt(CxxK_(a))=sqrt(0.15xx1.4xx10^(-5))`
`=1.449xx10^(-3)`
`[OH^(-)]` in `NH_(4)OH=sqrt(C xxK_(b))=sqrt(0.01xx1.8xx10^(-5))=0.4242xx10^(-3)`
`[H^(+)]` in `NH_(4)OH=10^(-14)/(0.4242 xx 10^(-3))=2.3573xx10^(-11)`
`E_(cell)=0.0591 "log"([H^(+)]_(RHS))/([H^(+)]_(LHS))`
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