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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)`A. `-0.92 V`B. `-0.46 V`C. `-0.35 V`D. `-0.20 V` |
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Answer» Correct Answer - B 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` |
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