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Nernst Equation

The potential...

Question

The potential of the cell at $25^{\circ} C$ is
$P t | H_{2} (1 atm) | C H_{3} C O O N a (0.1 M) + C H_{3} C O O H (0.01 M) | | N H_{4} C l (0.2 M) + N H_{4} O H (0.1 M) | H_{2} (1 atm) | P t$
Given: $p K_{a}$ of $C H_{3} C O O H$ and $p K_{b}$ of $N H_{4} O H = 4.74$

A

- 0.04 V

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B

0.04 V

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C

- 0.189 V

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D

0.189 V

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Solution

The correct option is C $- 0.189 V$
This is an example of a concentration cell where anode is an acidic buffer and cathode is a basic buffer.
At anode,
$p H$ = $p K a + l o g \frac{s a l t}{a c i d}$
= $4.74 + l o g \frac{0.1}{0.01}$
= $4.74 + l o g 10$
= 5.74
At cathode,
$p O H$ = $p K b + l o g \frac{s a l t}{a c i d}$
= $4.74 + l o g \frac{0.2}{0.1}$
= $4.74 + l o g 2$
= 5.04
$P t | H_{2} (1 atm) | C H_{3} C O O N a (0.1 M) + C H_{3} C O O H (0.01 M) | | N H_{4} C l (0.2 M) + N H_{4} O H (0.1 M) | H_{2} (1 atm) | P t$
Which is same as,
$P t | H_{2} (1 atm) | H_{o x}^{+} | | H_{r e d}^{+} | H_{2} (1 atm) | P t$
$H_{2} \to 2 H_{o x}^{+} + 2 e^{-}$
$2 H_{r e d}^{+} + 2 e^{-} \to H_{2}$
$E = - \frac{0.059}{2} l o g \frac{[H_{o x}^{+}]^{2}}{[H_{r e d}^{+}]^{2}}$
$= - \frac{0.059}{2} l o g \frac{[H_{o x}^{+}]}{[H_{r e d}^{+}]^{2}}$
$= - 0.059 l o g \frac{[H_{o x}^{+}]}{[H_{r e d}^{+}]}$
$= - 0.059 l o g \frac{[H_{o x}^{+}] [O H_{r e d}^{-}]}{K w}$

$= 0.059 [- l o g [H_{o x}^{+}] - l o g [O H_{r e d}^{-}] - l o g [\frac{1}{K w}]]$
= $0.059 \times (5.74 + 5.04 - 14)$
= $0.059 \times (- 3.22)$
= -0.189

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Similar questions

Q. The potential of the cell at

25^{\circ} C

P t | H_{2} (1 atm) | C H_{3} C O O N a (0.1 M) + C H_{3} C O O H (0.01 M) | | N H_{4} C l (0.2 M) + N H_{4} O H (0.1 M) | H_{2} (1 atm) | P t

p K_{a}

C H_{3} C O O H

p K_{b}

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Q. The potential of the cell at

25^{\circ} C

p K_{a}

C H_{3} C O O H

p K_{b}

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P t ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} H_{2} (1 a t m) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} C H_{3} C O O N a (0.1 M) + C H_{3} C O O H (0.01 M) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} N H_{4} C l (0.1 M) + N H_{4} O H (0.1 M) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} H_{2} (1 a t m) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ P t

Q. The potential of the cell at

25^{\circ} C

p K_{b}

N H_{4} O H = 4.74

P t ∣ ∣ ∣ \begin{matrix} H_{2} (1 a t m) \end{matrix} ∣ ∣ ∣ \begin{matrix} H H_{4} 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

Q. The observed emf of the cell,

p t | H_{2} (1 a t m) | H^{+} (3 \times 10^{- 4} M) ∥ H^{+} (M_{1}) | H_{2} (1 a t m) | P t

0.154 V

. Calculate the value of

M_{1}

p H

of cathodic solution.

Given the following cell at $25^{\circ} C$ . What will be the potential of the cell ? Given $p K_{a}$ of $C H_{3} C O O H = 4.74$ :

$P t ∣ ∣ ∣ \begin{matrix} H_{2} (1 a t m) \end{matrix} ∣ ∣ ∣ \begin{matrix} C H_{3} C O O H (10^{- 3} M) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} N a O H (10^{- 3} M) \end{matrix} ∣ ∣ ∣ \begin{matrix} H_{2} (1 a t m) \end{matrix} ∣ ∣ ∣ P t$

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