Chemistry · 2024
JEE Main · 1 February 2024, Shift 1 · Q85
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…
The potential for the given half cell at $\displaystyle 298$ K is ( - ) $\displaystyle \_\_\_\_$ $\displaystyle \times 10^{-2} \mathrm{~V}$.
$$\begin{aligned}
& 2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g}) \\
& {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}}
\end{aligned}
$$
(Given : $\displaystyle 2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3$ )
Official answer
From NTA’s final answer key for this paper.
1
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JEE Main 2024 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.