Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
How is the potential of hydrogen electrode obtained?
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Answer
Hydrogen gas electrode is represented as, $H ^{+}( aq )\left| H _2\left( g , P _{ H _2}\right)\right| Pt$ Electrode reduction reaction is, $2 H ^{+}{ }_{( aq )}+2 e ^{-} \rightarrow H _{2( g )}$ By Nernst equation, the reduction potential is, $\begin{aligned} & E_{ H ^{+} / H _2}=E_{ H ^{+} / H _2}^0-\frac{0.0592}{2} \log _{10} \frac{P_{ H _2}}{\left[ H ^{+}\right]^2} \\ & \because E_{ H ^{+} / H _2}^0=E_{ SHE }=0.0 V , \\ & E_{ H ^{+} / H _2}=-\frac{0.0592}{2} \log _{10} \frac{P_{ H _2}}{\left[ H ^{+}\right]^2} \end{aligned}$ If $H _2$ gas is passed at $1 atm$, then $P _{ H _2}=1 atm$ $\begin{aligned} E_{ H ^{+} / H _2} & =-\frac{0.0592}{2} \log _{10} \frac{1}{\left[ H ^{+}\right]^2} \\ & =-\frac{0.0592}{2} \log _{10}\left[ H ^{+}\right]^2 \\ & =-0.0592 \log _{10}\left[ H ^{+}\right] \\ & =0.0592\left[-\log _{10}\left( H ^{+}\right)\right] \\ & =0.0592 pH \\ \therefore pH & =\frac{E_{ H ^{+} / H _2}}{0.0592}\end{aligned}$
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