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Question

The dissociation constant for $$CH_{3}COOH$$ is $$1.8\times 10^{-5}$$ at $$298\ K$$. The electrode potential for the half-cell: $$Pt|H_{2}\ (1\ bar)\|0.5\ M\ CH_{3}COOH$$, at $$298\ K$$ is:

$$(log\ 2 = 0.3\ ;\ log\ 3 = 0.48\ ;\ 2.303\ RT/F = 0.06)$$

A
$$+0.3024\ V$$
B
$$-0.3024\ V$$
C
$$-0.1512\ V$$
D
$$+0.1512\ V$$
Solution
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Correct option is D. $$+0.1512\ V$$

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