Calculate the e.m.f. of the cell
`Pt|H_(2)(1.0atm)|CH_(3)COOH (0.1M)||NH_(3)(aq,0.01M)|H_(2)(1.0atm)|Pt`
`K_(a) (CH_(3)COOH) =1.8 xx 10^(-5), K_(b),(NH_(3)) = 1.8 xx 10^(-5)`

Options

A. `-0.92 V`
B. `-0.46 V`
C. `-0.35 V`
D. `-0.20 V`

Answer

B

Explanation

Pt, `H_(2)(1atm)//CH_(3)COOH(0.1M)"//" NH_(3)(aq,0.01M)//H_(2)(1atm)Pt`
`K_(a)(CH_(3)COOH)=1.8xx10^(-5)`
`K_(b)(CH_(3)=1.8xx10^(-5)`
`{:(A:(1)/(2)H_(2)-e-rarrH^(+)" "E^(@)=0),(C:H^(+)+e-rarr(1)/(2)H_(2)(g)" "E_("cell")^(@)=0),(-------------),(E_("cell")=E_("cell")^(@)-(0.06)/(1)"log"([H^(+)]_(A))/([H^(+)]_(C ))):}`
`[OH^(-)]^(2)=0.01xx1.8xx10^(-5)`
`[OH^(-)]=4.2xx10^(-4),[H^(+)]_(C )=(10^(-14))/(4.2xx10^(-4))`
Similarly
`[H^(+)]^(2)=1.8xx10^(-5)xx0.1`
`[H^(+)]_(A)=sqrt(1.8xx10^(-6))`
`E_("cell")=-0.0591xx7.78=-0.46 v`