1.

Acetic acid has K_a=1.8xx10^(-5) while formic acid has K_(a)=2.1xx10^(-4).What would be the magnitude of the emf of the cell Pt(H_(2))|0.1 M acetic acid+ 0.1 M sodium acetate ||0.1 M formic acid + 0.1 M sodium formate|Pt(H_(2)) at 25^(@)C ?

Answer»

0.032 volt
0.063 volt
0.0456 volt
0.055 volt

Solution :Reaction is `H_C^+ overset(1e^(-))to H_A^+`
`therefore E=(0.059)/1 "LOG" (2.1xx10^(-4))/(1.8xx10^(-5))=0.0629 V=0.063 V`


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