JEE Main 20241 Feb, Morning shiftChemistryHardRedox Reactions and Electrochemistry138× since 2002The potential for the given half cell at 298 K298 \mathrm{~K}298 K is (-) __________ ×10−2 V\times 10^{-2} \mathrm{~V}×10−2 V 2H(aq)++2e−⟶H2( g)[H+]=1M,PH2=2 atm\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}2H(aq)++2e−⟶H2( g)[H+]=1M,PH2=2 atm (Given : 2.303RT/F=0.06 V,log2=0.32.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.32.303RT/F=0.06 V,log2=0.3 )Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution
