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Neglecting the liquid-liquid junction po...

Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at `25^(@)C`
`H_(2) (1 atm)|0.5 M HCOOH|| 1M CH_(3)COOH | (1 atm) H_(2)`
`K_(a) " for " HCOOH and CH_(3) COOH` are `1.77 xx 10^(-4) and 1.8 xx 10^(-5)` respectively.

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To calculate the EMF of the given electrochemical cell, we can follow these steps: ### Step 1: Write down the given information We have the following data: - Cell: \( H_2 (1 \text{ atm}) | 0.5 \text{ M } HCOOH || 1 \text{ M } CH_3COOH | H_2 (1 \text{ atm}) \) - \( K_a \) for \( HCOOH = 1.77 \times 10^{-4} \) - \( K_a \) for \( CH_3COOH = 1.8 \times 10^{-5} \) ...
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