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H2(1atm)| 2.26M HCOOH || 0.222M CH3COOH |(1atm)H2Ka(HCOOH)=1.77*10^-4Ka(CH3COOH)=1.8*10^-5Emf of the cell is(neglect the liquid liquid junction potential)

Answer» H2(1atm)| 2.26M HCOOH || 0.222M CH3COOH |(1atm)H2
Ka(HCOOH)=1.77*10^-4
Ka(CH3COOH)=1.8*10^-5
Emf of the cell is(neglect the liquid liquid junction potential)


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