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If the dissociation constant for \mathrm{CH}_3 \mathrm{COOH} is 1.5 \times 10^{-4} at 298 \mathrm{k}. The electrode potential for the half cell \mathrm{Pt} \mid \mathrm{H}_2 (1 bar) \mid 10.5 \mathrm{M}-\mathrm{CH}_3 \mathrm{COOH}, at 298 \mathrm{~K} is.

Option: 1

0.123


Option: 2

1.286


Option: 3

2.679


Option: 4

1.786


Answers (1)

best_answer

\mathrm{\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}(\mathrm{eq})+2 e^{-}}

\mathrm{E_{\mathrm{H}_2 / \mathrm{OH}_3 \mathrm{COOH}}=E_{\mathrm{H}_2 / \mathrm{H}^{+}}^0=E_{\mathrm{H}_2 / \mathrm{H}^{+}}^0-\frac{0.06}{2} \log \frac{{\left(\mathrm{H}^{+}\right)}^{2}}{\mathrm{pH}_2}}

\mathrm{=0-\frac{0.06}{2} \log \frac{\left(\sqrt{\mathrm{Ka} \times c^2}\right)}{pH_2}}

\mathrm{=\frac{-0.06}{2} \log \frac{1.5 \times 10^{-4} \times 0.5}{1}}

\mathrm{\text { Electrode potential }=0.123}

Posted by

Ajit Kumar Dubey

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