The potential for the given half cell at 298 ~K is (-) __________ 10^-2 ~V $ aligned & 2 H_(aq)^++2 e^- H_2(~g) \\ & [H^+]=1 M, P_H_2=2 ~atm aligned $ (Given : 2.303 RT / F=0.06 ~V, 2=0.3 )

The potential is given by the Nernst equation: E=E^-0.062 (P_H_2[H^+]^2) Substituting the values: E=0-0.062 (21^2) \\ E=-0.03 0.3=-0.9 10^-2 ~V

The potential for the given half cell at 298 K is x 102 V 2 : (Chemistry) | DivineJEE

Exam

Chemistry

The potential for the given half cell at

298 \mathrm{~K}

is (-) __________

\times 10^{-2} \mathrm{~V}

\begin{aligned} & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g}) \\ & {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}} \end{aligned}

(Given :

2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3

)

The potential is given by the Nernst equation:

$E=E^{\circ}-\frac{0.06}{2} \log \left(\frac{P_{\mathrm{H}_2}}{\left[H^{+}\right]^2}\right)$

Substituting the values:

$E=0-\frac{0.06}{2} \log \left(\frac{2}{1^2}\right) \\$

$E=-0.03 \times 0.3=-0.9 \times 10^{-2} \mathrm{~V}$

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