The potential for the given half cell at 298 K is (-) ____ $\times 10^{-2}$ V. $2\mathrm{H}^{+}_{(\mathrm{aq})}+2\mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{g})$ $[\mathrm{H}^{+}]=1$ M, $\mathrm{P}_{\mathrm{H}_{2}}=2$ atm (Given: 2.303 RT/F=0.06 V, $\log 2=0.3$)

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Redox Reactions and Electrochemistry

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

\times 1 0^{- 2}

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

[H^{+}] = 1

P_{H_{2}} = 2

atm
(Given: 2.303 RT/F=0.06 V,

lo g 2 = 0.3

)

Answer: 1

We need to calculate the electrode potential for the hydrogen half-cell under non-standard conditions using the Nernst equation.

Given:

Nernst equation:

E = E^{\circ} - \frac{2.303 R T}{n F} lo g \frac{[ products ]}{[ reactants ]}

For the given half-cell:

E = E_{H^{+} / H_{2}}^{\circ} - \frac{0.06}{2} lo g \frac{P _{H_{2}}}{[ H ^{+} ] ^{2}}

Substituting values:

E = 0.00 - \frac{0.06}{2} lo g \frac{2}{( 1 ) ^{2}}

E = - 0.03 \times lo g 2

E = - 0.03 \times 0.3

E = - 0.009 V

E = - 0.9 \times 1 0^{- 2} V

The magnitude is

0.9 \times 1 0^{- 2}

V, which rounds to

1 \times 1 0^{- 2}

V.

Therefore, the answer is 1.

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer