NEET UG 2013 — Chemistry Physical Chemistry

For hydrogen electrode, oxidation half reaction is $\underset{(1 \mathrm{~atm})}{\mathrm{H}_2} \longrightarrow \underset{(\text{At } \mathrm{pH~} 10)}{2 \mathrm{H}^{+}}+2 \mathrm{e}^{-}$(1 atm)H2​​⟶(At pH 10)2H+​+2e− If $\mathrm{pH}=10$pH=10 $\mathrm{H}^{+}=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10}$H+=1×10−pH=1×10−10 From Nernst equation, $E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.0591}{2} \log \frac{\left[\mathrm{H}^{+}\right]^2}{P_{\mathrm{H}_2}}$Ecell ​=Ecell 0​−20.0591​logPH2​​[H+]2​ For hydrogen electrode, $E_{\text {cell }}^{\circ}=0$Ecell ∘​=0 $$\begin{aligned} E_{\text {cell }} & =-\frac{0.0591}{2} \log \frac{\left(10^{-10}\right)^2}{1} \\ & =+\frac{0.0591 \times 2}{2} \log \frac{1}{10^{-10}} \\ & =0.0591 \log 10^{10} \\ & =0.0591 \times 10=0.591 \mathrm{~V} \end{aligned}$$Ecell ​​=−20.0591​log1(10−10)2​=+20.0591×2​log10−101​=0.0591log1010=0.0591×10=0.591 V​