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Nernst Equation

Given the fol...

Question

Given the following cell at $25^{\circ} C$ . What will be the potential of the cell ? Given $p K_{a}$ of $C H_{3} C O O H = 4.74$ :
$P t ∣ ∣ ∣ \begin{matrix} H_{2} (1 a t m) \end{matrix} ∣ ∣ ∣ \begin{matrix} C H_{3} C O O H (10^{- 3} M) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} N a O H (10^{- 3} M) \end{matrix} ∣ ∣ ∣ \begin{matrix} H_{2} (1 a t m) \end{matrix} ∣ ∣ ∣ P t$

- 0.42 V

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0.42 V

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- 0.19 V

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0.19 V

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Solution

The correct option is A $- 0.42 V$
It is concentration cell, therefore, $E_{c e l l}^{⊖} = 0$ .
pH of $W_{A} = \frac{1}{2} (p K_{a} - l o g C)$
$= \frac{1}{2} (4.74 - l o g 10^{- 3}) = 3.87$
$p H$ of $N a O H$
$= 14 - 3$
$= 11$
$∴ E = - 0.059 (p H_{c} - p H_{a})$
$= - 0.059 (11 - 3.87)$
$= - 0.42 V$

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