The potential for the given half cell at $298K$ is \left(-\right) $............$ \times {10}^{- 2}V . $</p><p>$ 2{H}_{( aq )}^{+} + 2{e}^{-} \to {H}_{2} ( g ) $</p><p>\left[{H}^{+}\right] = 1M, {P}_{{H}_{2}} = 2atm$
(Given $2 . 303RT / F = 0 . 06V, \log 2 = 0 . 3$ )

Question

The potential for the given half cell at

298K

is \left(-\right)

............

\times {10}^{- 2}V .

</p><p>

2{H}_{( aq )}^{+} + 2{e}^{-} \to {H}_{2} ( g )

</p><p>\left[{H}^{+}\right] = 1M, {P}_{{H}_{2}} = 2atm

(Given $2 . 303RT / F = 0 . 06V, \log 2 = 0 . 3$ )

Quantrex Academy · Accepted Answer

Correct answer: 1

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

$E = {E}^{°} - \dfrac{0 . 06}{2} \times log\dfrac{{p}_{{H}_{2}}}{{\left[{H}^{+}\right]}^{2}}$

$E = 0 - \dfrac{0 . 06}{2}log\dfrac{2}{{\left[1\right]}^{2}}$

$E = - 0 . 03 \times 0 . 3 = - 0 . 009 = - 9 \times {10}^{- 3}$

$E = - 0 . 9 \times {10}^{- 2}V$

The potential for the given half cell at 298K is (-) ............ ×...