The potential for the given half cell at $298 K$ is $(-)$ ............ $\times 10^{- 2} V .$

$2 H_{(aq)}^{+} + 2 e^{-} \to H_{2} (g)$
$[H^{+}] = 1 M, P_{H_{2}} = 2 atm$
(Given $2.303 RT / F = 0.06 V, \log 2 = 0.3$ )

Question

The potential for the given half cell at 298 K is - ............ × 10 - 2 V . 2 H ( aq ) + + 2 e - → H 2 ( g ) H + = 1 M , P H 2 = 2 atm (Given 2 . 303 RT / F = 0 . 06 V , log 2 = 0 . 3 )

TestHub · Accepted Answer

The Nernst equation for the given cell reaction $2 H^{+} + 2 e^{-} \to H_{2}$ is

$E = E^{°} - \frac{0.06}{2} \times \log \frac{p_{H_{2}}}{{[H^{+}]}^{2}}$

$E = 0 - \frac{0.06}{2} \log \frac{2}{{[1]}^{2}}$

$E = - 0.03 \times 0.3 = - 0.009 = - 9 \times 10^{- 3}$

$E = - 0.9 \times 10^{- 2} V$

Chemistry - Electrochemistry Question with Solution | TestHub

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