Verified Solution • 2025–26 Session moderate

QJEEM24Chdb08: JEE Main 2024 Chemistry - Class 12 JEEM24Chdb08 Solution: The hydrogen electrode is dipped in a solution of at . The…

The hydrogen electrode is dipped in a solution of pH=3\mathrm{pH}=3 at 25∘C25^{\circ} \mathrm{C} . The potential of the electrode will be _________ ×10−2 V\times 10^{-2} \mathrm{~V} .

(2.303RTF=0.059 V)\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)

Step-by-Step Verified Solution

amp;2H(aq.)++2e−→H2( g)amp;Ecell =Ecell 0−0.0592log⁡PH2[H+]2amp;=0−0.059×3=−0.177 volts. =−17.7×10−2 V\begin{aligned}& 2 \mathrm{H}_{(\mathrm{aq} .)}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g}) \\& \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2} \\& =0-0.059 \times 3=-0.177 \text { volts. }=-17.7 \times 10^{-2} \mathrm{~V}\end{aligned}

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